StraightLinesandPairofStraightLines
Question1
Theportionoftheline4x+5y=20inthefirstquadrantistrisectedby
thelinesL1andL2passingthroughtheorigin.Thetangentofanangle
betweenthelinesL1andL2is:
[27-Jan-2024Shift1]
Options:
A.
8/5
B.
25/41
C.
2/5
D.
30/41
Answer:D
Solution:
Question2
LetRbetheinteriorregionbetweenthelines3x−y+1=0andx+2y
−5=0containingtheorigin.Thesetofallvaluesofa,forwhichthe
points(a2,a+1)lieinR,is:
[27-Jan-2024Shift2]
Options:
A.
B.
C.
D.
Answer:B
Solution:
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Question3
LetAandBbetwofinitesetswithmandnelementsrespectively.The
totalnumberofsubsetsofthesetAis56morethanthetotalnumberof
subsetsofB.ThenthedistanceofthepointP(m,n)fromthepointQ(−2,
−3)is
[27-Jan-2024Shift2]
Options:
A.
10
B.
6
C.
4
D.
8
Answer:A
Solution:
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Question4
Ifthesumofsquaresofallrealvaluesofα,forwhichthelines2x−y+
3=0,6x+3y+1=0andαx+2y−2=0donotformatriangleisp,
thenthegreatestintegerlessthanorequaltopis........
[27-Jan-2024Shift2]
Answer:32
Solution:
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Question5
Ina△ABC,supposey=xistheequationofthebisectoroftheangleB
andtheequationofthesideACis2x−y=2.If2AB=BCandthepoint
AandBarerespectively(4,6)and(α,β),thenα+2βisequalto
[29-Jan-2024Shift1]
Options:
A.
42
B.
39
C.
48
D.
45
Answer:A
Solution:
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Question6
Let(5,a/4),bethecircumcenterofatrianglewithverticesA(a,−2),
B(a,6)andC(a/4,-2).Letαdenotethecircumradius,βdenotethearea
andγdenotetheperimeterofthetriangle.Thenα+β+γis
[29-Jan-2024Shift1]
Options:
A.
60
B.
53
C.
62
D.
30
Answer:B
Solution:
Question7
Thedistanceofthepoint(2,3)fromtheline2x−3y+28=0,
measuredparalleltotheline√3x−y+1=0,isequalto
[29-Jan-2024Shift2]
Options:
A.
4√2
B.
6√3
C.
3+4√2
D.
4+6√3
Answer:D
Solution:
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Question8
LetAbethepointofintersectionofthelines3x+2y=14,5x−y=6
andBbethepointofintersectionofthelines4x+3y=8,6x+y=5.
ThedistanceofthepointP(5,−2)fromthelineABis
[29-Jan-2024Shift2]
Options:
A.
13/2
B.
8
C.
5/2
D.
6
Answer:D
Solution:
SolvinglinesL1(3x+2y=14)andL2(5x−y=6)togetA(2,4)andsolvinglinesL3(4x+3y=8)andL4(6x+y=5)to
getB(1/2,2).
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Question9
AlinepassingthroughthepointA(9,0)makesanangleof30∘withthe
positivedirectionofx-axis.IfthislineisrotatedaboutAthroughan
angleof15∘intheclockwisedirection,thenitsequationinthenew
positionis
[30-Jan-2024Shift1]
Options:
A.
B.
C.
D.
Answer:A
Solution:
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Question10
Ifx2−y2+2hxy+2gx+2fy+c=0isthelocusofapoint,whichmoves
suchthatitisalwaysequidistantfromthelinesx+2y+7=0and2x−
y+8=0,thenthevalueofg+c+h−fequals
[30-Jan-2024Shift2]
Options:
A.
14
B.
6
C.
8
D.
29
Answer:A
Solution:
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Question11
Letα,β,γ,δ∈ZandletA(α,β),B(1,0),C(γ,δ)andD(1,2)bethe
verticesofaparallelogramABCD.IfAB=√10andthepointsAandClie
ontheline3y=2x+1,then2(α+β+γ+δ)isequalto
[31-Jan-2024Shift1]
Options:
A.
10
B.
5
C.
12
D.
8
Answer:D
Solution:
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Question12
LetA(a,b),B(3,4)and(−6,−8)respectivelydenotethecentroid,
circumcentreandorthocentreofatriangle.Then,thedistanceofthe
pointP(2a+3,7b+5)fromtheline2x+3y−4=0measuredparallel
tothelinex−2y−1=0is
[31-Jan-2024Shift2]
Options:
A.
B.
C.
D.
Answer:C
Solution:
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Question13
LetA(−2,−1),B(1,0),C(α,β)andD(γ,δ)betheverticesofa
parallelogramABCD.IfthepointClieson2x−y=5andthepointD
lieson3x−2y=6,thenthevalueof|α+β+γ+δ|isequalto____
[31-Jan-2024Shift2]
Answer:32
Solution:
Question14
LetABCbeanisoscelestriangleinwhichAisat(−1,0),∠A=2π/3,AB
=ACandBisonthepositivex-axis.IfBC=4√3andthelineBC
intersectstheliney=x+3at(α,β),thenβ4/α2is:
[1-Feb-2024Shift2]
Answer:36
Solution:
Question15
LetPQRbeatriangle.ThepointsA,BandCareonthesidesQR,RP
andPQrespectivelysuchthat QA
AR =RB
BP =PC
CQ =1
2.Then Area (△PQR)
Area (△ABC)isequal
to
[24-Jan-2023Shift1]
Options:
A.4
B.3
C.2
D.−
Answer:B
Solution:
LetPis→
0,Qis→
qandRis→
r
Ais 2→
q+→
r
3,Bis 2→
r
3andCis →
q
3
Areaof△PQRis = 1
2
→
q×→
r
Areaof△ABCis 1
2
→
AB ×→
AC
→
AB =→
r−2
→
q
3,→
AC = −→
r−→
q
3
Areaof△ABC =1
6
→
q×→
r
Area (△PQR)
Area(△ABC)=3
||
||
||
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Question16
TheequationsofthesidesABandACofatriangleABCare
(λ+1)x+λy =4andλx + (1−λ)y+λ=0respectively.ItsvertexAison
they-axisanditsorthocentreis(1,2).Thelengthofthetangentfrom
thepointCtothepartoftheparabolay2=6xinthefirstquadrantis
[24-Jan-2023Shift2]
Options:
A.√6
B.2√2
C.2
D.4
Answer:B
Solution:
AB : (λ+1)x+λy =4
AC :λx + (1−λ)y+λ=0
VertexAisony-axis
⇒x=0
Soy=4
λ,y=λ
λ−1
⇒4
λ=λ
λ−1
⇒λ=2
AB :3x +2y =4
AC :2x −y+2=0
⇒A(0,2)Let C(α,2α +2)
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Question17
Theequationsoftwosidesofavariabletrianglearex =0andy =3,and
itsthirdsideisatangenttotheparabolay2=6x.Thelocusofits
circumcentreis:
[25-Jan-2023Shift2]
Options:
A.4y2−18y −3x −18 =0
B.4y2+18y +3x +18 =0
C.4y2−18y +3x +18 =0
D.4y2−18y −3x +18 =0
Answer:C
Solution:
Solution:
Now(SlopeofAltitudethroughC) − 3
2= −1
2α
α−1−3
2= −1⇒α= − 1
2
SoC−1
2,1
LetEquationoftangentbey=mx +3
2m
m2+2m −3=0
⇒m=1, −3
SotangentwhichtouchesinfirstquadrantatTis
T≡a
m2,2a
m
≡3
2,3
⇒CT = √4+4=2√2
( )
( ) ( )
( )
( )
( )
y2=6x&y2=4ax
⇒4a =6⇒a=3
2
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Question18
AtriangleisformedbyX-axis,Y-axisandtheline3x +4y =60.Then
thenumberofpointsP(a,b)whichliestrictlyinsidethetriangle,where
aisanintegerandbisamultipleofa,is________.
[25-Jan-2023Shift2]
Answer:31
Solution:
y=mx +3
2m; (m≠0)
h=6m −3
4m2,k=6m +3
4m ,Noweliminatingmandweget
⇒3h =2(−2k2+9k −9)
⇒4y2−18y +3x +18 =0
Ifx=1,y=57
4=14.25
(1,1)(1,2) − (1,14) ⇒ 14 pts.
Ifx=2,y=27
2=13.5
(2,2)(2,4).. . (2,12) ⇒ 6pts.
Ifx=3,y=51
4=12.75
(3,3)(3,6) − (3,12) ⇒ 4pts.
Ifx=4,y=12
(4,4)(4,8) ⇒ 2pts.
Ifx=5.y=45
4=11.25
(5,5), (5,10) ⇒ 2pts.
Ifx=6,y=21
2=10.5
(6,6) ⇒ 1 pt
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Question19
Alightrayemitsfromtheoriginmakinganangle30∘withthepositive
x-axis.Aftergettingreflectedbythelinex +y=1,ifthisrayintersects
x-axisatQ,thentheabscissaofQis
[29-Jan-2023Shift1]
Options:
A. 2
(√3−1)
B. 2
3+ √3
C. 2
3− √3
D. √3
2(√3+1)
Answer:B
Solution:
Solution:
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Question20
LetBandCbethetwopointsontheliney +x=0suchthatBandCare
symmetricwithrespecttotheorigin.SupposeAisapointony −2x =2
suchthat△ABCisanequilateraltriangle.Then,theareaofthe△ABCis
[29-Jan-2023Shift1]
Options:
A.3√3
B.2√3
Ifx=7,y=39
4=9.75
(7,7) ⇒ 1 pt.
Ifx=8,y=9
(8,8) ⇒ 1 pt.
Ifx=9y =33
4=8.25⇒nopt.
Total = 31 pts.
Slopeofreflectedray = tan 60∘= √3
Liney=x
√3intersecty+x=1at √3
√3+1,1
√3+1
Equationofreflectedrayis
y−1
√3+1= √3 x −√3
√3+1
Puty=0⇒x=2
3+ √3
( )
( )
C. 8
√3
D. 10
√3
Answer:C
Solution:
Solution:
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Question21
Atriangleisformedbythetangentsatthepoint(2,2)onthecurves
y2=2xandx2+y2=4x,andthelinex +y+2=0.Ifristheradiusofits
circumcircle,thenr2isequalto_______.
[29-Jan-2023Shift2]
Answer:10
Solution:
AtA x =y
Y−2x =2
(−2, −2)
Heightfromlinex+y=0
h=4
√2
Areaof∆ = √3
4
h2
sin 260 =8
√3
S1:y2=2x S2:x2+y2=4x
P(2,2)iscommonpointonS1&S2
T1istangenttoS1atP⇒T1:y⋅2=x+2
⇒T1:x−2y +2=0
T2istangenttoS2atP⇒T2:x⋅2+y⋅2=2(x+2)
⇒T2:y=2
L3:x+y+2=0isthirdline
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Question22
AstraightlinecutsofftheinterceptsOA =aandOB =bonthepositive
directionsofx-axisandy−axisrespectively.Iftheperpendicularfrom
originOtothislinemakesanangleof π
6withpositivedirectionofy-axis
andtheareaof△OABis 98
3√3,thena2−b2isequalto:
[30-Jan-2023Shift1]
Options:
A. 392
3
B.196
C. 196
3
D.98
Answer:A
Solution:
Solution:
PQ =a= √20
QR =b= √8
RP =c=6
Area (∆PQR) = ∆ = 1
2×6×2=6
∴r=abc
4∆=√160
4= √10 ⇒r2=10
Equationofstraightline: x
a+y
b=1
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Question23
Iftheorthocentreofthetriangle,whoseverticesare(1,2), (2,3)and
(3,1)is(α,β),thenthequadraticequationwhoserootsareα +4βand
4α +β,is
[1-Feb-2023Shift1]
Options:
A.x2−19x +90 =0
B.x2−18x +80 =0
C.x2−22x +120 =0
D.x2−20x +99 =0
Answer:D
Solution:
Orx cos π
3+ysin π
3=p
x
2+y√3
2=p
x
3p +y
2p =1
Comparingboth:a=2p,b=2p
√3
Nowareaof△OAB =1
2⋅ab =98
3⋅ √3
1
2⋅2p⋅2p
√3=98
3⋅ √3
p2=49
a2−b2=4p2−4p2
3=2
34p2
=8
3⋅49 =392
3
HeremBH ×mAC = −1
β−3
α−2
1
−2= −1
β−3=2α −4
β=2α −1
mAH ×mBC = −1
⇒β−2
α−1(−2) = −1
( ) ( )
( )
-------------------------------------------------------------------------------------------------
Question24
Thecombinedequationofthetwolinesax +by +c=0and
a′x+b′y+c′=0canbewrittenas(ax +by +c)(a′x+b′y+c′) = 0
Theequationoftheanglebisectorsofthelinesrepresentedbythe
equation2x2+xy −3y2=0is
[1-Feb-2023Shift1]
Options:
A.3x2+5xy +2y2=0
B.x2−y2+10xy =0
C.3x2+xy −2y2=0
D.x2−y2−10xy =0
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question25
Thestraightlines11and12passthroughtheoriginandtrisecttheline
segmentofthelineL :9x +5y =45betweentheaxes.Ifm1andm2are
theslopesofthelines11and12,thenthepointofintersectionofthe
liney = (m1+m2)xwithLlieson.
[6-Apr-2023shift1]
Options:
⇒2β −4=α−1
⇒2(2α −1) = α+3
⇒3α =5
α=5
3,β=7
3⇒H5
3,7
3
α+4β =5
3+28
3=33
3=11
β+4α =7
3+20
3=27
3=9
x2−20x +99 =0
( )
Equationofthepairofanglebisectorforthehomogenousequationax2+2hxy +b2=0isgivenas
x2−y2
a−b=xy
h
Herea=2,h=1∕2&b= −3
Equationwillbecome
x2−y2
2− (−3)=xy
1∕2
x2−y2=10xy
x2−y2−10xy =0
A.6x +y=10
B.6x −y=15
C.y −2x =5
D.y −x=5
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question26
LetA(0,1), B(1,1)andC(1,0)bethemid-pointsofthesidesofa
trianglewithincentreatthepointD.Ifthefocusoftheparabola
→Px=2×5+1×0
1+2=10
3
Py=0×2+9×1
1+2=3
P:10
3,3
Similarly→Qx=1×5+2×0
1+2=5
3
Qy=1×0+2×9
1+2=6
Q:5
3,6
Nowm1=3−0
10
3−0
=9
10
m2=6−0
5
3−0
=18
5
from(1)&(2)
9x +5y =45
9x −2y =0
−+−
7y =45 ⇒y=45
7
⇒x=10
7
whichsatisfyy−x=5Ans.4
( )
( )
y2=4 axpassingthroughDis(α+β√3,0),whereαandβarerational
numbers,then α
β2isequalto
[8-Apr-2023shift2]
Options:
A.6
B.8
C. 9
2
D.12
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question27
TheareaofthequadrilateralABCDwithvertices
A(2,1,1), B(1,2,5), C(−2, −3,5)andD(1, −6, −7)isequalto
[8-Apr-2023shift2]
Options:
A.54
B.9√38
a=OP =2 b =OQ =2 c =PQ =2√2
D4
2+2+2√2,4
2+2+2√2≡D2
2+ √2,2
2+ √2
y2=4 ax ⇒2
2+ √2
2=4a ⋅2
2+ √2
∴4a =2
2+ √2∴a=1
2⋅2− √2
4−2=1
4(2− √2)
∴α=2
4=1
2β=−1
4
∴α
β2=8 Ans.
( ) ( )
( ) ( )
C.48
D.8√38
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question28
AlinesegmentABoflengthλmovessuchthatthepointsAandB
remainontheperipheryofacircleofradiusλ.Thenthelocusofthe
point,thatdividesthelinesegmentABintheratio2 :3,isacircleof
radius:
[10-Apr-2023shift1]
Options:
A. 2
3λ
B. √19
7λ
C. 3
5λ
D. √19
5λ
Answer:D
VectorArea =→
v
=1
2
→
AB ×→
AC +1
2
→
AC ×→
AD
=1
2
→
AB −→
AD ×→
AC
→
AB = −^
i+^
j+4^
k
→
AD = −^
i−7^
j−8^
k
→
AC = −4^
i−4^
j+4^
k
=1
28^
j+12^
k× (−4)^
i+^
j−^
k
=1
2
^
i^
j^
k
0 8 12
1 1 −1
= (−2) −20^
i+12^
j−8^
k
=8 5^
i−3^
j+2^
k
∴Area =→
v=8√25 +9+4=8√38 Ans.
( ) ( )
( ) ( )
| |
( )
( )
| |
Solution:
-------------------------------------------------------------------------------------------------
Question29
LetAbethepoint(1,2)andBbeanypointonthecurvex2+y2=16.If
thecentreofthelocusofthepointP,whichdividesthelinesegmentAB
intheratio3 :2isthepointC(α,β)thenthelengthofthelinesegment
ACis
[10-Apr-2023shift2]
Options:
A. 6√5
5
B. 2√5
5
C. 3√5
5
D. 4√5
5
Answer:C
SinceOABformequilateral∆
∴ ∠OAP =60∘
AP =2λ
5
cos 60∘=OA2+AP2−OP2
2 OA ⋅AP
⇒1
2=
λ2+4λ2
25 −OP2
2λ 2λ
5
⇒2λ2
5=λ2+4λ2
25 −OP2
⇒OP 2 =19λ2
25
⇒OP =√19
5λ
Therefore,locusofpointPis √19
5λ
( )
Solution:
-------------------------------------------------------------------------------------------------
Question30
LettheequationsoftwoadjacentsidesofaparallelogramABCDbe
2x −3y = −23and5x +4y =23.IftheequationofitsonediagonalACis
3x +7y =23andthedistanceofAfromtheotherdiagonalisd ,then
50d2isequalto_______.
[10-Apr-2023shift2]
Answer:529
Solution:
12 cos θ +2
5=h⇒12 cos θ =5h −2
sq&add
144 = (5h −2)2+ (5k −4)2
x−2
5
2+y−4
5
2=144
25
Centre ≡2
5,4
5≡ (α,β)
AC =1−2
5
2+2−4
5
2
=9
25 +36
25 =√45
5=3√5
5
( ) ( )
( )
√( ) ( )
√
A&Cpointwillbe(−4,5)&(3,2)
midpointofACwillbe − 1
2,7
2
equationofdiagonalBDis
y−7
2=
7
2
−1
2
x+1
2
⇒7x +y=0
DistanceofAfromdiagonalBD
=d=23
√50
( )
( )
-------------------------------------------------------------------------------------------------
Question31
LetRbearectanglegivenbythelinex =0,x=2,y=0andy =5.Let
A(α,0)andB(0,β), α∈ [0,2]andβ ∈ [0,5],besuchthattheline
segmentABdividestheareaoftherectangleRintheratio4 :1.Then,
themidpointofABliesona:
[11-Apr-2023shift1]
Options:
A.straightline
B.parabola
C.circle
D.hyperbola
Answer:D
Solution:
-------------------------------------------------------------------------------------------------
ar(OPQR)
or(OAB)=4
1
LetMbethemid-pointofAB.
M(h,k) ≡ α
2,β
2
⇒
10 −1
2αβ
1
2αβ
=4
⇒5
2αβ =10 ⇒αβ =4
⇒ (2h)(2K) = 4
∴LocusofMisxy = 1
Whichisahyperbola.
( )
Ifthelinel 1:3y −2x =3istheangularbisectoroftheline
l2:x−y+1=0andl 3:ax +βy +17,thenα2+β2−α −βisequalto
_______.
[11-Apr-2023shift2]
Answer:348
Solution:
-------------------------------------------------------------------------------------------------
Question33
Ifthepoint α,7√3
3liesonthecurvetracedbythemid-pointsofthe
linesegmentsofthelinesx cos θ +ysin θ =7,θ∈0,π
2betweenthe
co-ordinatesaxes,thenαisequalto
[12-Apr-2023shift1]
Options:
A.7√3
B.-7
C.7
D.−7√3
Answer:C
Solution:
( )
( )
Pointofintersectionofℓ1:3y −2x =3
ℓ2:x−y+1=0isP≡ (0,1)
Whichliesonℓ3:αx −βy +17 =0,
⇒β= −17
ConsiderarandompointQ≡ (−1,0)
onℓ2:x−y+1=0,imageofQabout
ℓ2:x−y+1=0,isQ′≡−17
13 ,6
13 whichiscalculatedbyformulae
x− (−1)
2=y−0
−3=2−2+3
13
Now,Q′liesinℓ3:αx +βy +17 =0
⇒α=7
Now,α2+β2−α−β=348
( )
( )
pt α,7√3
3
( )
Question32
-------------------------------------------------------------------------------------------------
Question34
Let(α,β)bethecentroidofthetriangleformedbythelines
15x −y=82,6x −5y = −4and9x +4y =17.Thenα +2βand2α −βare
therootsoftheequation
[13-Apr-2023shift2]
Options:
A.x2−13x +42 =0
B.x2−10x +25 =0
C.x2−7x +12 =0
D.x2−14x +48 =0
Answer:A
Solution:
x cos θ +ysin θ =7
x -intercept =7
cos θ
y−intercept =7
sin θ
A:7
cos θ,0 B :0,7
sin θ
LocusofmidptM:(h,k)
h=7
2 cos θ,k=7
2sin θ
7
2sin θ =7√3
3⇒sin θ =√3
2⇒θ=π
3
α=7
2 cos θ =7
( ) ( )
-------------------------------------------------------------------------------------------------
Question35
If(α,β)istheorthocenterofthetriangleABCwithvertices
A(3, −7), B(−1,2)andC(4,5),then9α −6β +60isequalto
[15-Apr-2023shift1]
Options:
A.30
B.35
C.40
D.25
Answer:D
Solution:
Solution:
Centroid (α,β) = 6+1+5
3,8−7+2
3= (4,1)
α+2β =4+2=6
2α −β=8−1=7
Quadraticequation
x2− (6+7)x+ (6×7) = 0
⇒x2−13x +42 =0
( )
AltitudeofBC :y+7=−5
3(x−3)
3y +21 = −5x +15
5x +3y +6=0
AltitudeofAC:y−2=−1
12 (x+1)
12y −24 = −x−1
x+12y =23
-------------------------------------------------------------------------------------------------
Question36
FromthetopAofaverticalwallABofheight30m,theanglesof
depressionofthetopPandbottomQofaverticaltowerPQare15∘and
60∘respectively,BandQareonthesamehorizontallevel.IfCisapoint
onABsuchthatCB =PQ,thenthearea(inm2)ofthequadrilateral
BCPQisequalto:
[6-Apr-2023shift1]
Options:
A.200(3− √3)
B.300(√3+1)
C.300(√3−1)
D.600(√3−1)
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question37
α=−47
19 ,β=121
57
9α −6β +60 =25
AB
BQ =tan 60∘
BQ =30
√3=10√3=y
& △ ACP
AC
CP =tan 15∘⇒(30 −x)
y= (2− √3)
30 −x=10√3(2− √3)
30 −x=20√3−30
x=60 −20√3
Area =x⋅y=20(3− √3) ⋅ 10√3
=600(√3−1)
TheangleofelevationofthetopPofatowerfromthefeetofone
personstandingdueSouthofthetoweris45∘andfromthefeetof
anotherpersonstandingduewestofthetoweris30∘.Iftheheightof
thetoweris5meters,thenthedistance(inmeters)betweenthetwo
personsisequalto
[11-Apr-2023shift2]
Options:
A.10
B.5√5
C. 5
2√5
D.5
Answer:A
Solution:
Solution:
TowerAB =5m
∠APB =45∘
∠PAB =90∘
tan 45∘=AB
AP
1=AB
AP
AP =5m
tan 30∘=AB
AQ
-------------------------------------------------------------------------------------------------
Question38
LetA 3
√a, √a,a>0,beafixedpointinthexy-plane.TheimageofAiny-axisbeBandtheimageofBinx-axisbeC.
IfD(3 cos θ,a sin θ)isapointinthefourthquadrantsuchthatthemaximumareaof△ACDis12squareunits,thenais
equalto____
[24-Jun-2022-Shift-1]
LetA 3
√a, √a,a>0,beafixedpointinthexy-plane.TheimageofAin
y-axisbeBandtheimageofBinx-axisbeC.IfD(3 cos θ,a sin θ)isa
pointinthefourthquadrantsuchthatthemaximumareaof△ACDis12
squareunits,thenaisequalto____
[24-Jun-2022-Shift-1]
Answer:8
Solution:
( )
( )
1
1√3=5
AQ
AQ =5√3
AP2+AQ2=PQ2
PQ2=52+ (5√3)2
PQ2=25 +75 =100
PQ =10 cm
Option(A)10 cmcorrect.
-------------------------------------------------------------------------------------------------
Question39
LettheareaofthetrianglewithverticesA(1,α), B(α,0)andC(0,α)be
4sq.units.Ifthepoints(α, −α ), (−α,α)and(α2,β)arecollinear,then
βisequalto
[24-Jun-2022-Shift-2]
Options:
A.64
B.−8
C.−64
D.512
Answer:C
Solution:
-------------------------------------------------------------------------------------------------
Question40
LetRbethepoint(3,7)andletPandQbetwopointsontheline
x+y=5suchthatPQRisanequilateraltriangle.Thentheareaof
△PQRis:
[26-Jun-2022-Shift-1]
Options:
A. 25
4√3
B. 25√3
2
C. 25
√3
D. 25
2√3
Answer:D
Solution:
Question41
InanisoscelestriangleABC,thevertexAis(6,1)andtheequationof
thebaseBCis2x +y=4.LetthepointBlieonthelinex+3y =7.If
(α,β)isthecentroidof△ABC,then15(α+β)isequalto:
[27-Jun-2022-Shift-1]
-------------------------------------------------------------------------------------------------
Question42
ArectangleRwithendpointsofoneofitssidesas(1,2)and(3,6)is
inscribedinacircle.Iftheequationofadiameterofthecircleis
2x −y+4=0,thentheareaofRis
[27-Jun-2022-Shift-1]
Answer:16
Solution:
2x +y=4
2x +6y =14 y=2,x=3
B(1,2)
LetC(k,4−2k)
NowAB2=AC2
52+ (−1)2= (6−k)2+ (−3+2k)2
⇒5k2−24k +19 =0
(5k −19)(k−1) = 0⇒k=19
5
C19
5, − 18
5
Centroid(α,β)
α=
6+1+19
5
3=18
5
β=
1+2−18
5
3= − 1
5
Now15(α+β)
15 17
5=51
}
( )
( )
Options:
A.39
B.41
C.51
D.63
Answer:C
Solution:
-------------------------------------------------------------------------------------------------
Question43
ArayoflightpassingthroughthepointP(2,3)reflectsonthex-axisat
pointAandthereflectedraypassesthroughthepointQ(5,4).LetRbe
thepointthatdividesthelinesegmentAQinternallyintotheratio2 :1.
Lettheco-ordinatesofthefootoftheperpendicularMfromRonthe
bisectoroftheanglePAQbe(α,β).Then,thevalueof7α +3βisequal
to
[28-Jun-2022-Shift-1]
Answer:31
Solution:
Asslopeoflinejoining(1,2)and(3,6)is2givendiameterisparalleltoside
∴a= (3−1)2+ (6−2)2= √20
andb∕2=4
√5⇒b=8
√5
Area =ab =2√5⋅8
√5=16
√
4
5−α=3
α−2⇒4α −8=15 −3α
α=23
7
A=23
7,0 Q = (5,4)
R=
10 +23
7
3,8
3
=31
7,8
3
BisectorofanglePAQisX=23
7
⇒M=23
7,8
3
So,7α +3β =31
( )
( )
( )
( )
Question44
Letatrianglebeboundedbythelines
L1:2x +5y =10;L2: −4x +3y =12andthelineL3,whichpasses
throughthepointP(2,3),intersectsL2atAandL1atB.IfthepointP
dividestheline-segmentAB,internallyintheratio1 :3,thenthearea
ofthetriangleisequalto:
[28-Jun-2022-Shift-2]
Options:
A. 110
13
B. 132
13
C. 142
13
D. 151
13
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
L1:2x +5y =10
L2: −4x +3y =12
SolvingL1andL2weget
C≡−15
13 ,32
13
Now,LetA x1,1
3(12 +4x1) and
B x2,1
5(10 −2x2)
∴3x1+x2
4=2
and
(12 +4x1) + 10 −2x2
5
4=3
So,3x1+x2=8and10x1−x2= −5
So,(x1,x2) = 3
13,95
13
A=3
13,56
13 andB=95
13,−12
13
=1
2
3
13
−44
13
−56
13
110
13 +12860
169
=132
13 sq.units
( )
( )
( )
( )
( ) ( )
| ( ( ) ( ) ( ) ) |
Question45
ThedistancebetweenthetwopointsAandA′whichlieony =2such
thatboththelinesegmentsABandA′B(whereBisthepoint(2,3))
subtendangle π
4attheorigin,isequalto:
[29-Jun-2022-Shift-1]
Options:
A.10
B. 48
5
C. 52
5
D.3
Answer:C
Solution:
-------------------------------------------------------------------------------------------------
Question46
Thedistanceoftheoriginfromthecentroidofthetrianglewhosetwo
sideshavetheequationsx −2y +1=0and2x −y−1=0andwhose
orthocenteris 7
3,7
3is:
[29-Jun-2022-Shift-2]
Options:
A.√2
B.2
C.2√2
( )
M1=3∕2 M2=2∕x
tan π ∕4=3∕2−2∕x
1+6∕2x =1
⇒x1=10,x2= −2∕5
⇒AA1=52 ∕5
| |
D.4
Answer:C
Solution:
-------------------------------------------------------------------------------------------------
Question47
LetABandPQbetwoverticalpoles,160mapartfromeachother.LetC
bethemiddlepointofBandQ,whicharefeetofthesetwopoles.Let π
8
andθbetheanglesofelevationfromCtoPandA,respectively.Ifthe
heightofpolePQistwicetheheightofpoleAB,thentan2θisequalto
[28-Jun-2022-Shift-1]
ForpointA,
2x −y−1=0
x−2y +1=0
Solving(1)and(2),weget
x=1,y=1
∴PointA= (1,1)
AltitudefromBtolineACisperpendiculartolineAC.
∴EquatorofaltitudeBHis
2x +y+λ=0
ItpassesthroughpointH7
3,7
3soitsatisfytheequation(3).
14
3+7
3+λ=0
⇒α= −7
∴AltitudeBH =2x +y−7=0
Solvingequation(1)and(4),weget
x=2,y=3
∴PointB= (2,3)
AltitudefromCtolineABisperpendiculartolineAB.
∴EquationofaltitudeCHis
x+2y +λ=0
ItpassesthroughpointH7
3,7
3soitsatisfyequation(5).
7
3+14
3+λ=0
⇒λ= −7
∴AltitudeCH =x+2y −7=0
Solvingequation(2)and(6),weget
x=3,y=2
∴PointC= (3,2)
CentroidG(x,y)oftriangleA(1,1), B(2,3)andC(3,2)is
x=1+2+3
3=2,y=1+2+3
3=2
NowdDistanceofpointG(2,2)fromcenterO(0,0)is
OG =22+22=2√2
( )
( )
√
Options:
A. 3−2√2
2
B. 3+ √2
2
C. 3−2√2
4
D. 3− √2
4
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question48
Fromthebaseofapoleofheight20meter,theangleofelevationofthe
topofatoweris60∘.Thepolesubtendsanangle30∘atthetopofthe
tower.Thentheheightofthetoweris
[29-Jun-2022-Shift-2]
Options:
A.15√3
B.20√3
C.20 +10√3
D.30
Answer:D
Solution:
l
80 =tan θ.... (i)
2l
80 =tan π
8.... (ii)
From(i)and(ii)
1
2=tan θ
tan π
8
⇒tan2θ=1
4tan2π
8
⇒tan2θ=√2−1
4(√2+1)=3−2√2
4
Solution:
-------------------------------------------------------------------------------------------------
Question49
Aline,withtheslopegreaterthanone,passesthroughthepointA(4,3)
andintersectsthelinex −y−2=0atthepointB.Ifthelengthofthe
linesegmentABis √29
3,thenBalsoliesontheline:
[25-Jul-2022-Shift-1]
Options:
A.2x +y=9
B.3x −2y =7
C.x +2y =6
D.2x −3y =3
Answer:C
Solution:
HereABisatowerandCDisapole.
IntriangleABC,tan 60∘=AB
AC =20 +h
x...... . (1)
IntriangleBED,tan 30∘=h
x..... . (2)
Divideequation(1)byequation(2),weget
tan 60∘
tan 30∘=20 +h
x×x
h
⇒√3
1
√3
=20 +h
h
⇒3=20 +h
h
⇒3h =20 +h
⇒h=10m
∴Heightoftower =20 +10 =30m
-------------------------------------------------------------------------------------------------
Question50
LetthepointP(α,β)beataunitdistancefromeachofthetwolines
L1:3x −4y +12 =0,andL2:8x +6y +11 =0.IfPliesbelowL1and
aboveL2,then100(α+β)isequalto
[25-Jul-2022-Shift-2]
Options:
A.−14
B.42
C.−22
D.14
Answer:D
Solution:
Solution:
Letinclinationofrequiredlineisθ,
SothecoordinatesofpointBcanbeassumedas
4−√29
3cos θ,3−√29
3sin θ
Whichsatisficesx−y−2=0
4−√29
3cos θ −3+√29
3sin θ −2=0
sin θ −cos θ =3
√29
Bysquaring
sin 2 θ =20
29 =2 tan θ
1+tan2θ
tan θ =5
2only(becauseslopeisgreaterthan1)
sin θ =5
√29,cos θ =2
√29
PointB:10
3,4
3
Whichalsosatisfiesx+2y =6
( )
( )
-------------------------------------------------------------------------------------------------
Question51
ApointPmovessothatthesumofsquaresofitsdistancesfromthe
points(1,2)and(−2,1)is14.Letf (x,y) = 0bethelocusofP,which
intersectsthex-axisatthepointsA,Bandthey-axisatthepointsC,D.
ThentheareaofthequadrilateralACBDisequalto:
[26-Jul-2022-Shift-1]
Options:
A. 9
2
B. 3√17
2
C. 3√17
4
D.9
Answer:B
Solution:
Solution:
L1:3x −4y +12 =0
L2:8x +6y +11 =0
EquationofanglebisectorofL1andL2ofanglecontainingorigin
2(3x −4y +12) = 8x +6y +11
2x +14y −13 =0....... (i)
3α −4β +12
5=1
⇒3α −4β +7=0 .....(ii)
Solutionof2x +14y −13 =0and3x −4y +7=0givestherequiredpointP(α,β), α=−23
25 ,β=53
50100(α+β) = 14
LetpointP: (h,k)
(h−1)2+ (k−2)2+ (h+2)2+ (k−1)2=14
2h2+2k2+2h −6k −4=0
LocusofP:x2+y2+x−3y −2=0
Intersectionwithx-axis,
x2+x−2=0
-------------------------------------------------------------------------------------------------
Question52
TheequationsofthesidesAB,BCandCAofatriangleABCare
2x +y=0,x+py =15aandx −y=3respectively.Ifitsorthocentreis
(2,a), − 1
2<a<2,thenpisequalto_________.
[26-Jul-2022-Shift-1]
Answer:3
Solution:
Solution:
-------------------------------------------------------------------------------------------------
⇒x= −2,1
Intersectionwithy-axis,
y2−3y −2=0
⇒y=3± √17
2
AreaofthequadrilateralACBDis
=1
2(|x1|+ | x2|)(|y1|+ | y2|)
=1
2×3× √17 =3√17
2
SlopeofAH =a+2
1
SlopeofBC = − 1
p
∴p=a+2...... (i)
CoordinateofC=18p −30
p+1,15p −33
p+1
SlopeofH C =
15p −33
p+3−a
18p −33
p+1−2
=15p −33 − (p−2)(p+1)
18p −30 −2p −2
=16p −p2−31
16p −32
∵16p −p2−31
16p −32 ×−2= −1
∴p2−8p +15 =0
∴p=3 or 5
Butifp=5thena=3notacceptable
∴p=3
( )
Question53
LetA(1,1), B(−4,3), C(−2, −5)beverticesofatriangleABC,Pbea
pointonsideBC,and∆1and∆2betheareasoftrianglesAPBandABC,
respectively.If∆1: ∆2=4:7,thentheareaenclosedbythelinesAP,AC
andthex-axisis
[27-Jul-2022-Shift-1]
Options:
A. 1
4
B. 3
4
C. 1
2
D.1
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question54
TheequationsofthesidesAB,BCandCAofatriangleABCare
2x +y=0,x+py =39andx −y=3respectivelyandP(2,3)isits
circumcentre.ThenwhichofthefollowingisNOTtrue?
[27-Jul-2022-Shift-2]
Options:
∆1
∆2
=
1
2×BP ×AH
1
2×BC ×AH
=4
7
P−20
7,−11
7
LineAC :y−1=2(x−1)
Intersectionwithx-axis = 1
2,0
LineAP :y−1=2
3(x−1)
Intersectionwithx-axis −1
2,0
Verticesare(1,1), 1
2,0and −1
2,0
Area = 1
2sq.unit
( )
( )
( )
( ) ( )
A.(AC)2=9p
B.(AC)2+p2=136
C.32 <area(∆ABC) < 36
D.34 <area(△ABC) < 38
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question55
Fort ∈ (0,2π),ifABCisanequilateraltrianglewithvertices
A(sin t, −cos t), B(cost,sin t)andC(a,b)suchthatitsorthocentrelieson
acirclewithcentre 1,1
3,then(a2−b2)isequalto:
[28-Jul-2022-Shift-1]
Options:
A. 8
3
B.8
C. 77
9
D. 80
9
Answer:B
Solution:
( )
Intersectionof2x +y=0andx−y=3:A(1, −2)
EquationofperpendicularbisectorofABis
x−2y = −4
EquationofperpendicularbisectorofACis
x+y=5
PointBistheimageofAinlinex−2y +4=0whichisobtainedasB−13
5,26
5
SimilarlyvertexC: (7,4)
EquationoflineBC :x+8y =39
So,p=8
AC =(7−1)2+ (4+2)2=6√2
AreaoftriangleABC =32.4
( )
√
Solution:
-------------------------------------------------------------------------------------------------
Question56
LetthecircumcentreofatrianglewithverticesA(a,3), B(b,5)and
C(a,b), ab >0beP(1,1).IfthelineAPintersectsthelineBCatthe
pointQ(k1,k2),thenk1+k2isequalto:
[29-Jul-2022-Shift-1]
Options:
A.2
B. 4
7
C. 2
7
D.4
Answer:B
Solution:
Solution:
LetP(h,k)betheorthocentreof△ABC
Then
h=sint +cos t +a
3,k=−cos t +sint +b
3
(orthocentrecoincidewithcentroid)
∴(3h −a)2+ (3k −b)2=2
∴h−a
3
2+k−b
3
2=2
9
∵orthocentreliesoncirclewithcentre 1,1
3
∴a=3,b=1
∴a2−b2=8
( ) ( )
( )
-------------------------------------------------------------------------------------------------
Question57
Letm1,m2betheslopesoftwoadjacentsidesofasquareofsideasuch
thata2+11a +3(m1
2+m2
2) = 220.Ifonevertexofthesquareis
(10(cos α −sin α), 10(sin α +cos α)),whereα ∈0,π
2andtheequation
ofonediagonalis(cos α −sin α)x+ (sin α +cos α)y−10,then
72(sin4α+cos4α) + a2−3a +13isequalto:
[29-Jul-2022-Shift-2]
Options:
A.119
B.128
C.145
D.155
Answer:B
Solution:
Solution:
( )
LetDbemid-pointofAC,then
b+3
2=1⇒b= −1
LetEbemid-pointofBC,
5−b
b−a⋅
(3+b)
2
a+b
2−1
= −1
Onputtingb= −1,wegeta=5or−3
Buta=5isrejectedasab >0
A(−3,3), B(−1,5), C(−3, −1), P(1,1)
LineBC ⇒y=3x +8
LineAP ⇒y=3−x
2
Pointofintersection −13
7,17
7
( )
m1m2= −1
a2+11a +3 m1
2+1
m1
2=220
Eq.ofAC
AC = (cos α −sin α) + (sin α +cos α)y=10
BD = (sin α −cos α)x+ (sin α −cos α)y=0
(10(cos α −sin α), 10(sin α −cos α))
SlopeofAC =sin α −cos α
sin α +cos α =tan θ =M
( )
( )
-------------------------------------------------------------------------------------------------
Question58
LetA(α, −2), B(α,6)andC α
4, −2 beverticesofa△ABC.If 5,α
4is
thecircumcentreof△ABC,thenwhichofthefollowingisNOTcorrect
about△ABC?
[29-Jul-2022-Shift-2]
Options:
A.areais24
B.perimeteris25
C.circumradiusis5
D.inradiusis2
Answer:B
Solution:
( ) ( )
Eq.oflinemakinganangleπ4withAC
m1,m2=
m±tan π
4
1±m tan π
4
=m+1
1−mor m−1
1+m
sin α −cos α
sin α +cos α +1
1−sin α −cos α
sin α +cos α
,
sin α −cos α
sin α +cos α −1
1+sin α −cos α
sin α +cos α
m1,m2=tan α,cot α
midpointofAC&BD
=M(5(cos α −sin α), 5(cos α +sin α))
B(10(cos α −sin α), 10(cos α +sin α))
a=AB = √2BM = √2(5√2) = 10
a=10
∵a2+11a +3 m1
2+1
m12=220
100 +110 +3(tan2α+cot2α) = 220
Hencetan2α=3,tan2α=1
3⇒α=π
3or π
6
Now72(sin 4α+cos4α) + a2−3a +13
=72 9
16 +1
16 +100 −30 +13
=72 5
8+83 =45 +83 =128
( )
( )
( )
( )
-------------------------------------------------------------------------------------------------
Question59
AtowerPQstandsonahorizontalgroundwithbaseQontheground.
ThepointRdividesthetowerintwopartssuchthatQR =15m.Iffroma
pointAonthegroundtheangleofelevationofRis60∘andthepartPR
ofthetowersubtendsanangleof15∘atA,thentheheightofthetower
is:
[25-Jul-2022-Shift-1]
Options:
A.5(2√3+3)m
B.5(√3+3)m
C.10(√3+1)m
D.10(2√3+1)m
Answer:A
Solution:
Solution:
Circumcentreof△ABC
=
α+α
4
2,6−2
2
=5α
8,2
=5,α
4
⇒α=8
area(△ABC) = 1
2⋅3α
4×8=24sq.units
Perimeter = 8+3α
4+82+3α
4
2 = 8+6+10 =24
Circumradius = 10
2=5
r=∆
s=24
12 =2
( )
( )
( )
√( )
-------------------------------------------------------------------------------------------------
Question60
LetaverticaltowerABofheight2hstandsonahorizontalground.Let
fromapointPonthegroundamancanseeuptoheighthofthetower
withanangleofelevation2α.WhenfromP,hemovesadistanced in
thedirectionof →
AP,hecanseethetopBofthetowerwithanangleof
elevationα.Ifd = √7h,thentan αisequalto
[27-Jul-2022-Shift-1]
Options:
A.√5−2
B.√3−1
C.√7−2
D.√7− √3
Answer:C
Solution:
For△AQR,
tan 60∘=15
x..... . (1
From△AQP,
tan 75∘=h
x
⇒(2+ √3) = h
x[∵tan 75∘=2+ √3]
⇒h= (2+ √3)x
= (2+ √3)15
√3[From (1)]
= (2+ √3)× 15√3
3
= (2+ √3) × 5√3
=5(2√3+3)m
-------------------------------------------------------------------------------------------------
Question61
TheangleofelevationofthetopPofaverticaltowerPQofheight10
fromapointAonthehorizontalgroundis45∘.LetRbeapointonAQ
andfromapointB,verticallyaboveR,theangleofelevationofPis60∘.
If∠BAQ =30∘,AB =dandtheareaofthetrapeziumPQRBisα,then
theorderedpair(d,α)is:
[27-Jul-2022-Shift-2]
Options:
A.(10(√3−1), 25)
B. 10(√3−1), 25
2
C.(10(√3+1), 25)
D. 10(√3+1), 25
2
Answer:A
Solution:
Solution:
( )
( )
△APM gives
tan 2 α =h
x.....(i)
△AQBgives
tan 2 α =2h
x+d=2h
x+h√7......(ii)
From(i)and(ii)
tan 2 α =2⋅tan 2 α
1+ √7⋅tan 2 α
Lett=tan α
⇒t=
22t
1−t2
1+ √7⋅2t
1−t2
⇒t2−2√7t+3=0
t= √7−2
LetBR =x
-------------------------------------------------------------------------------------------------
Question62
AhorizontalparkisintheshapeofatriangleOABwithAB =16.A
verticallamppostOPiserectedatthepointOsuchthat
∠PAO = ∠PBO =15∘and∠PCO =45∘,whereCisthemidpointofAB.
Then(OP)2isequalto
[28-Jul-2022-Shift-2]
Options:
A. 32
√3(√3−1)
B. 32
√3(2− √3)
C. 16
√3(√3−1)
D. 16
√3(2− √3)
Answer:B
Solution:
Solution:
x
d=1
2⇒x=d
2
10 −x
10 −x√3= √3⇒10 −x=10√3−3x
2x =10(√3−1)
x=5(√3−1)
d=2x =10(√3−1)
α=1
2(x+10)(10 −x√3) = Area(PQRB)
=1
2(5√3−5+10)(10 −5√3(√3−1))
=1
2(5√3+5)(10 −15 +5√3) − 1
2(75 −25) = 25
-------------------------------------------------------------------------------------------------
Question63
TheangleofelevationofthetopofatowerfromapointAduenorthof
itisαandfromapointBatadistanceof9unitsduewestofAis
cos−13
√13 .IfthedistanceofthepointBfromthetoweris15units,
thencot αisequalto:
[29-Jul-2022-Shift-1]
Options:
A. 6
5
B. 9
5
C. 4
3
D. 7
3
Answer:A
Solution:
Solution:
( )
OP =OA tan 15 =OB tan 15.......(i)
OP =OC tan 45 ⇒OP =OC.......(ii)
OA =OB......(iii)
OC2+82=OA2
OP2+64 =OP2√3+1
√3−1
2
64 =OP2(√3+1)2− (√3−1)2
(√3−1)2
=OP24√3
(√3−1)2
OP2=64(√3−1)2
4√3=32
√3(2− √3)
( )
[ ]
( )
-------------------------------------------------------------------------------------------------
Question64
Theintersectionofthreelinesx −y=0,x+2y =3and2x +y=6isa
[2021,26Feb.Shift-1]
Options:
A.rightangledtriangle
B.equilateraltriangle
C.isoscelestriangle
D.Noneofthese
Answer:C
Solution:
Solution:
N A =152−92=12
h
15 =tan θ =2
3
h=10units
cot α =12
10 =6
5
√
Givenlines,x−y=0,x+2y =3,2x +y=6
-------------------------------------------------------------------------------------------------
Question65
Ifthecurvex2+2y2=2intersectsthelinex +y=1attwopointsPand
Q,thentheanglesubtendedbythelinesegmentPQattheoriginis
[2021,25Feb.Shift-II]
Options:
A. π
2+tan−11
4
B. π
2−tan−11
4
C. π
2+tan−11
3
D. π
2−tan−11
3
Answer:A
Solution:
Solution:
( )
( )
( )
( )
Theonlytrianglewhichincludeallthreelinesis△ABC.
Now,AB = (2−1)2+ (2−1)2= √2
AC = (2−3)2+ (2−0)2= √5
BC = (3−1)2+ (0−1)2= √5
⇒AC =BC(twosidesareequal)
⇒ ∆ ABCisisoscelestriangle.
√
√
√
Curvex2+2y2=2intersectthelinex+y=1atpointsPandQ.Firstwehavetofindanycommonrelationbetween
thesetwocurves.
Usesubstitutionforthesameasfollows,
-------------------------------------------------------------------------------------------------
Question66
Theimageofthepoint(3,5)inthelinex −y+1=0,lieson
[2021,25Feb.Shift-1]
Options:
A.(x−2)2+ (y−2)2=12
B.(x−4)2+ (y+2)2=16
C.(x−4)2+ (y−4)2=8
D.(x−2)2+ (y−4)2=4
Answer:D
Solution:
Solution:
x2+2y2=2⋅⋅⋅⋅⋅⋅ ⋅ (i)
x+y=1,then (x+y)2=12
⇒x2+y2+2xy =1⋅⋅⋅⋅⋅⋅ ⋅ (ii)
WecanwriteEq.(i)as,
x2+2y2−2(1)2=0
⇒x2+2y2−2(x+y)2=0 [usingEq.(ii)inEq.(i)]
⇒x2+2y2−2x2−2y2−4xy =0
⇒ − x2−4xy =0⇒ −x(x+4y) = 0
Given,x=0andx+4y =0ory=−1
4x
Drawtheliney=−1
4xongraphandtake
arbitrarypoint(anyone)asfollows,
Fromgivengraph,
tan θ =1
4⇒θ=tan−11
4
Wehavetwolines,y= − 1
4xandx=0(i.e.Y-axis).
Thus,anylinejoiningthesetwocurvesmakesanangle π
2+θatorigin.
∴Answeris π
2+tan−11
4.
( )
( )
ImageofP(3,5)onthelinex−y+1=0is
x−3
1=y−5
−1=−2(3−5+1)
2
-------------------------------------------------------------------------------------------------
Question67
Amaniswalkingonastraightline.Thearithmeticmeanofthe
reciprocalsoftheinterceptsofthislineonthecoordinateaxesis 1
4.
ThreestonesA,BandCareplacedatthepoints(1,1), (2,2)and(4,4)
respectively.Thenwhichofthesestonesis\/areonthepathoftheman?
24Feb2021Shift1
Options:
A.Aonly
B.Conly
C.Allthethree
D.Bonly
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question68
LetA(−1,1), B(3,4)andC(2,0)begiventhreepoints.Aline
y=mx,m>0intersectslinesACandBCatpointPandQ1respectively.
LetA1andA2betheareasof△ABCand△PQC1respectively,suchthat
A1=3A2,thenthevalueofmisequalto
[2021,16MarchShift-II]
Options:
⇒x−3
1=y−5
−1=1
⇒x−3
1=1 and y−5
−1=1
x=4,y=4
∴Requiredimageisat(4,4).
Clearly,thispointlieson
(x−2)2+ (y−4)2=4as
(4,4)satisfiesthisequation.
Letthelinebey=mx +c
∴x-intercept:− c
m
y-intercept:c
A.M.ofreciprocalsoftheintercepts:
−m
c+1
c
2=1
4⇒2(1−m) = c
Line:y=mx +2(1−m) = c
⇒(y−2) − m(x−2) = 0
⇒linealwayspassesthrough(2,2)
A. 4
15
B.1
C.2
D.3
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question69
Given,pointsA(−1,1), B(3,4), C(2,0)
EquationofAC =y−1
x+1=0−1
2+1=−1
3
⇒3y −3= −x−1⇒x+3y =2⋅⋅⋅⋅⋅⋅ ⋅ (i)
OnsolvingEq.(i)andy=mx,weget
P2
3m +1,2m
3m +1
EquationofBC =y−0
x−2=4−0
3−2
⇒y=4x −8⋅⋅⋅⋅⋅⋅ ⋅ (ii)
Similarly,onsolvingEq.(ii)andy=mx ,
weget08
4−m,8m
4−m
Areaof△ABC =3Areaof△PQC(given)
1
2
−11 1
2 0 1
3 4 1
=3×1
2
2 0 1
8
4−m
8m
21
2
3m +1
2m
3m +11
⇒13 =31
4−m
1
3m +1
2 0 1
8 8m 4−m
2 2m 3m +1
⇒13 =3
4+11m −3m2× (52m2)
⇒15m2−11m −4=0
⇒m=1,−4
15 [butm>0]
⇒m=1
( )
( )
| | | |
( ) ( ) | |
Ina△PQR,thecoordinatesofthepointsPandQare(−2,4)and
(4, −2),respectively.IftheequationoftheperpendicularbisectorofPR
is2x −y+2=0,thenthecentreofthecircumcircleofthe△PQRis
[2021,17MarchShift-1]
Options:
A.(−1,0)
B.(−2, −2)
C.(0,2)
D.(1,4)
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question70
Thenumberofintegralvaluesofm,sothattheabscissaofpointof
LetObethecentreofthecircumcircle.
AndTbethemid-pointofPR.
So,equationofOTisgivenas
2x −y+2=0
LetSbethemid-pointofPQ.
Now,Swillbe
−2+4
2,4−2
2= (1,1)
EquationofOS =y−1
x−1=−1
mPQ
mPO =−2−4
4+2= −1
∴OS =y−1=1(x−1)
y=x
Now,coordinatesofOwillbetheintersectionoflinesOSandOT.
y=x
2x −y+2=0.
⇒2x −x+2=0⇒x= −2
∴y= −2⇒0= (−2, −2)
( )
{
intersectionoflines3x +4y =9andy =mx +1isalsoaninteger,is
[2021,18MarchShift-1]
Options:
A.1
B.2
C.3
D.0
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question71
Theequationofoneofthestraightlineswhichpassesthroughthepoint
(1,3)andmakesanangletan−1(√2)withthestraightline,y +1=3√2x
is
[2021,18MarchShift-1]
Options:
A.4√2x+5y − (15 +4√2) = 0
B.5√2x+4y − (15 +4√2) = 0
C.4√2x+5y −4√2=0
D.4√2x−5y − (5+4√2) = 0
Given, y=mx +1
and3x +4y =9
FromEqs.(i)and(ii),
3x +4(mx +1) = 9
⇒3x +4mx +4=9
⇒x(3+4m) = 5
⇒x=5
3+4m
Given,thattheabscissaofpointofintersectionofEqs.(i)and(ii)i.e.x=5
3+4misaninteger.
∴Possiblevaluesofxare
x=1, −1,5, −5
i.e. 5
4m +3=1or 5
4m +3= −1
or 5
4m +3=5or 5
4m +3= −5
⇒4m =2or−4m =8
or4m = −2or−4m =4
⇒m=1
2, −2, − 1
2, −1
∵ − 1
2,1
2∉1
∴m= {−1, −2} ∈ 1
∴Numberofintegralvaluesofmare2.
{ }
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question72
ConsideratrianglehavingverticesA(−2,3), B(1,9)andC(3,8).Ifaline
Lpassingthroughthecircumcentreof△ABC,bisectslineBC,and
intersectsY -axisatpoint 0,α
2,thenthevalueofrealnumberαis
[2021,20JulyShift-II]
( )
MethodlLetm= Slopeofrequiredline
∴Equationofrequiredline
y−3=m(x−1)
Given,equationoflineis
3√2x−y−1=0
Since,angleθbetweenEqs.(i)and(ii)istan−1(√2)
i.e.tan θ = √2
⇒m−3√2
1+3√2m= √2
(∵SlopeofEq.(i)=mandslopeofEq.(ii) = 3√2)
Squaringonbothsides,
m2−6√2m+18 =2(1+18m2+6√2m)
⇒35m2+18√2m−16 =0
∴m=−18√2± √648 +2240
70
=−18√2±38√2
70
⇒m=2√2
7, − 4
5√2
Form=2√2
7,equationofrequiredline
willbe
y−3=2√2
7(x−1)
⇒2√2x−7y +21 −2√2=0
(optionsarenotmatchingso,neglectthis)
Form=−4√2
5,equationofrequiredline
willbe
y−3=−4√2
5(x−1)
⇒5y −15 = −4√2x+4√2
⇒4√2x+5y −15 −4√2=0
⇒4√2x+5y − (15 +4√2) = 0
| |
Answer:9
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question73
Lettheequationofthepairoflines,y =pxandy =qxcanbewrittenas
(y−px)(y−qx) = 0Then,theequationofthepairoftheanglebisectors
ofthelinex2−4xy −5y2=0is
[2021,25JulyShift-II]
Options:
A.x2−3xy +y2=0
B.x2+4xy −y2=0
C.x2+3xy −y2=0
AB = (1+2)2+ (9−3)2= √45
BC = (3−1)2+ (8−9)2= √5
AC = (3+2)2+ (8−3)2= √50
∴ (√50)2= (√45)2+ (√5)2
(AC)2= (AB)2+ (BC)2
∠B=90∘
⇒ABCisrightangledtriangle.
Circumcentre=Mid-pointofhypotenuse
=Mid-pointofAC
=1
2,11
2
Mid-pointoflineBC =2,17
2
LinepassingthroughcircumcentreandbisectlineBCwillbe
y−11
2=
17
2−11
2
2−1
2
x−1
2
⇒y−11
2=3×2
3x−1
2
⇒y−11
2=2 x −1
2
Itpassesthrough 0,α
2.
∴α
2−11
2=2 0 −1
2⇒α−11 =4−1
2
⇒α=11 −2=9
⇒α=9
√
√
√
( )
( )
( )
( )
( )
( )
( ) ( )
D.x2−3xy −y2=0
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question74
Twosidesofaparallelogramarealongthelines4x +5y =0and
7x +2y =0.Iftheequationofoneofthediagonalsoftheparallelogram
is11x +7y =9,thenotherdiagonalpassesthroughthepoint
[2021,27JulyShift-II]
Options:
A.(1,2)
B.(2,2)
C.(2,1)
D.(1,3)
Answer:B
Solution:
Solution:
Equationofanglebisectorofhomogeneousequationofpairofstraightlineax2+2hxy +by2is
x2−y2
a−b=xy
h
Forx2−4xy −5y2=0
a=1,h= −2,b= −5
So,equationofanglebisectoris
x2−y2
1− (−5)=xy
−2⇒ x2−y2= −3xy⇒x2+3xy −y2=0
So,combinedequationofanglebisectoris
x2+3xy −y2=0.isx2+3xy −y2=0.
Given,twosidesofparallelogramare
r4x +5y =0
7x +2y =0
and 7x +2y =0
Bothlinesarepassingthroughorigin.
Thus,pointA= (0,0)
-------------------------------------------------------------------------------------------------
Question75
LetAbeafixedpoint(0,6)andBbeamovingpoint(2t,0).LetM be
themid-pointofABandtheperpendicularbisectorofABmeetstheY -
axisatC.
Thelocusofthemid-pointPofM Cis
[2021,27Aug.Shift-1]
Options:
A.3x2−2y −6=0
B.3x2+2y −6=0
C.2x2+3y −9=0
D.2x2−3y +9=0
Answer:C
Solution:
Solution:
Theequationofdiagonalis11x +7y =9
PointDisthepointofintersectionof
4x +5y =0
and11x +7y =9
So,coordinateofD=5
3, − 4
3
Also,pointBisthepointofintersectionof7x +2y =0and11x +7y =9
So,coordinateofpointB= − 2
3,7
3
Weknowthat,diagonalsofparallelogrambisectseachother.LetPisthemiddlepointofBD.
So,coordinateof
P=
5
3+ −2
3
2,
−4
3+7
3
2=1
2,1
2
Now,equationofdiagonalAC
y−0=
1
2−0
1
2−0
(x−0)
⇒y=
1
2
1
2
x⇒y=x
∴DiagonalACpassesthrough(2,2).
( )
( )
(( ) )( )
Given,A(0,6)andB(2t,0)
Mid-pointofAB =M(t,3)
Equationofperpendicularbisectorof
ABpassesthroughM.
∴y−3=t
3(x−t) ⋅⋅⋅⋅⋅⋅ ⋅ (i)
So,C 0,3−t2
3
IntersectionofEq.(i)onY-axis
C 0,3−t2
3
( )
( )
-------------------------------------------------------------------------------------------------
Question76
LetABCbeatrianglewithA(−3,1)and∠ACB =θ,0<θ<π
2.Ifthe
equationofthemedianthroughBis2x +y−3=0andtheequationof
anglebisectorofCis7x −4y −1=0,thentan θisequalto
[2021,26Aug.Shift-I]
Options:
A.1 ∕2
B.3 ∕4
C.4/3
D.2
Answer:C
Solution:
Solution:
Letmid-pointofM Cis(h,k).
Then,(h,k) = t
2,3−t2
6
⇒h=t
2,k=3−t2
6
Eliminatingt,weget
2h2=3(3−k)
Locusof(h,k)
2x2=3(3−y)
⇒2x2+3y −9=0
( )
Given,theequationofmedianthroughB
i.e.BE :2x +y−3=0
Equation,ofanglebisectorofCi.e.
CD :7x −4y =1
Since,EsatisfiestheequationofBE .
r2 a−3
2+b+1
2−3=0
2a −6+b+1−6=0
2a +b=11 ⋅⋅⋅⋅⋅⋅ ⋅ (i)
Since,CsatisfiesC(d)
∴7a −4b =1⋅⋅⋅⋅⋅⋅ ⋅ (ii)
OnsolvingEqs.(i)and(ii),weget
a=3,b=5
SlopeofAC =2
3SlopeofCD =7
4
( ) ( )
-------------------------------------------------------------------------------------------------
Question77
Ifpandqarethelengthsoftheperpendicularsfromtheoriginonthe
lines,
xcosecα −y sec α =k cot 2 αandx sin α +y cos α =k sin 2 α
respectively,thenk2isequalto
[2021,31Aug.Shift-1]
Options:
A.4p2+q2
B.2p2+q2
C.p2+2q2
D.2p2+q2p2+4q2
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question78
LetA(1,0), B(6,2)andC 3
2,6 betheverticesofatriangleABC.IfPis
( )
∴tan θ
2=
2
3−7
4
1+14
12∣
=1
2
Now,tan θ =
2 tan θ
2
1−tan2θ
2
=
2⋅1
2
1−1
4
=4
3
( ) | |
( )
( )
p=k cot 2 α
cosec2α+sec2α
⇒q=k sin 2 α
sin2α+cos2α
⇒p=
kcos 2 α
sin 2 α
sin2α+cos2α
sin2αcos2α
=k cos 2 α
sin 2 α
⇒p=k
2cos 2 α
⇒q=k sin 2 α
⇒cos 2 α = (2p ∕k)
⇒sin 2 α = (q∕k)
⇒sin22α +cos22α =1
⇒4p2
k2+q2
k2=1
⇒4p2+q2=k2
√
√
( )
√
( )
apointinsidethetriangleABCsuchthatthetrianglesAPC,APBand
BPChaveequalareas,thenthelengthofthelinesegmentPQ,whereQ
isthepoint −7
6, − 1
3,is________.
[NAJan.7,2020(I)]
Answer:5
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question79
Thelocusofthemid-pointsoftheperpendicularsdrawnfrompointson
theline,x =2ytothelinex =yis:
[Jan.7,2020(II)]
Options:
A.2x −3y =0
B.5x −7y =0
C.3x −2y =0
D.7x −5y =0
Answer:B
Solution:
Solution:
( )
Pwillbecentroidof△ABC
P17
6,8
3⇒PQ =(4)2+ (3)2=5
( ) √
Since,slopeofPQ =k−α
h−2α = −1
⇒k−α= −h+2α
⇒α=h+k
3
Also,2h =2α +βand
2k =α+β
⇒2h =α+2k
-------------------------------------------------------------------------------------------------
Question80
LetCbethecentroidofthetrianglewithvertices(3,-1)(1,3)and(2,4).
LetPbethepointofintersectionofthelinesx +3y −1=0and
3x −y+1=0.ThenthelinepassingthroughthepointsCandPalso
passesthroughthepoint:
[Jan.9,2020(I)]
Options:
A.(-9,-6)
B.(9,7)
C.(7,6)
D.(-9,-7)
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question81
Ifa∆ABChasverticesA(−1,7), B(−7,1)andC(5, −5),thenits
orthocentrehascoordinates:
[Sep.03,2020(II)]
Options:
A. −3
5,3
5
B.(-3,3)
( )
⇒α=2h −2k
From(i)and(ii),wehave
h+k
3=2(h−k)
So,locusis6x −6y =x+y
⇒5x =7y ⇒5x −7y =0
Coordinatesofcentroides
C=x1+x2+x3
3,y1+y2+y3
3
=3+1+2
3,−1+3+4
3= (2,2)
Thegivenequationoflinesare
x+3y −1=0...(i)
3x −y+1=0...(ii)
Then,from(i)and(ii)
pointofintersectionP−1
5,2
5
equationoflineDP
8x −11y +6=0
( )
( )
( )
C. 3
5, − 3
5
D.(3,-3)
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question82
AtriangleABClyinginthefirstquadranthastwoverticesasA(1,2)and
B(3,1).If∠BAC =90∘,andar(∆ABC) = 5√5sq.units,thentheabscissa
ofthevertexCis:
[Sep.04,2020(I)]
Options:
A.1 + √5
B.1 +2√5
C.2 + √5
D.2√5−1ṁ
Answer:B
Solution:
Solution:
( )
mBC =6
−12 = − 1
2
∴EquationofASisy−7=2(x+1)
y=2x +9...(i)
mAC =12
−6= −2
∴EquationofBPisy−1=1
2(x+7)
y=x
2+9
2...(ii)
Fromequs.(i)and(ii),
2x +9=x+9
2
⇒4x +18 =x+9
⇒3x =9⇒x= −3
∴y=3
-------------------------------------------------------------------------------------------------
Question83
Iftheperpendicularbisectorofthelinesegmentjoiningthepoints
P(1,4)andQ(k,3)hasy-interceptequalto-4thenavalueofkis:
[Sep.04,2020(II)]
Options:
A.-2
B.-4
C.√14
D.√15
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Let∆ABCbeinthefirstquadrant
SlopeoflineAB = − 1
2
SlopeoflineAC =2
LengthofAB = √5
Itisgiventhatar(∆ABC) = 5√5
∴1
2AB ⋅AC =5√5⇒AC =10
∴CoordinateofvertexC= (1+10 cos θ,2+10 sin θ)
∵tan θ =2⇒cos θ =1
√5,sin θ =2
√5
∴CoordinateofC= (1+2√5,2+4√5)
∴AbscissaofvertexCis1+2√5.
MidpointoflinesegmentPQbe k+1
2,7
2.
∴Slopeofperpendicularlinepassingthrough
(0,-4)and k+1
2,7
2=
7
2+4
k+1
2−0
=15
k+1
SlopeofPQ =4−3
1−k=1
1−k
∴15
1+k×1
1−k= −1
1−k2= −15 ⇒k= ±4
( )
( )
Question84
Iftheline,2x −y+3=0isatadistance 1
√5and 2
√5fromthelines
4x −2y +α=0and6x −3y +β=0,respectively,thenthesumofall
possiblevalueofαandβis_______.
[NASep.05,2020(I)]
Answer:30
Solution:
-------------------------------------------------------------------------------------------------
Question85
Letf :R→Rbedefinedas
f(x) =
x5sin 1
x+5x2x<0
0 x =0
x5cos 1
x+λx2x>0.
Thevalueoflambdaforwhichf "(0)exists,is______.
[NASep.06,2020(I)]
Answer:5
Solution:
{( )
( )
L1:2x −y+3=0
L1:4x −2y +α=0⇒2x −y+α
2=0
L1:6x −3y +β=0⇒2x −y+β
3=0
DistancebetweenL1andL2
α−6
2√5=1
√5⇒α−6=2
⇒α=4,8
DistancebetweenL1andL3:
β−9
3√5=2
√5⇒β−9=6
⇒β=15,3
Sumofallvalues = 4+8+15 +3=30
| | | |
| | | |
-------------------------------------------------------------------------------------------------
Question86
LetLdenotethelineinthexy-planewithxandyinterceptsas3and1
respectively.Thentheimageofthepoint(-1,-4)inthislineis:
[Sep.06,2020(II)]
Options:
A. 11
5,28
5
B. 29
5,8
5
C. 8
5,29
5
D. 29
5,11
5
Answer:A
Solution:
Solution:
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Question87
Considerthesetofalllinespx +qy +r=0suchthat3p +2q +4r =0.
Whichoneofthefollowingstatementsistrue?
[Jan.9,2019(I)]
( )
( )
( )
( )
f′(x) =
5x4⋅sin 1
x−x3cos 1
x+10x x <0
0x=0
5x4cos 1
x+x3sin 1
x+2λx x >0.
f′′(x) =
(20x3−x)sin 1
x−8x2cos 1
x+10 x <0
0x=0
(20x3−x)cos 1
x+8x2sin 1
x+2λ x >0.
Now,f′′(0+) = f′′(0−) ⇒ 2λ =10 ⇒λ=5
{( ) ( )
( ) ( )
{( ) ( )
( ) ( )
Thelineinxy-planeis,
x
3+y=1⇒x+3y −3=0
Letimageofthepoint(-1,-4)be(α,β),then
α+1
1=β+y
3= − 2(−1−12 −3)
10
⇒α+1=β+4
3=16
5
⇒α=11
5,β=28
5
Options:
A.Thelinesareconcurrentatthepoint 3
4,1
2.
B.Eachlinepassesthroughtheorigin.
C.Thelinesareallparallel.
D.Thelinesarenotconcurrent.
Answer:A
Solution:
Solution:
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Question88
Lettheequationsoftwosidesofatrianglebe3x −2y +6=0and
4x +5y −20 =0.Iftheorthocentreofthistriangleisat(1,1),thenthe
equationofitsthirdsideis:
[Jan.09,2019(II)]
Options:
A.122y −26x −1675 =0
B.122y +26x +1675 =0
C.26x +61y +1675 =0
D.26x −122y −1675 =0
Answer:D
Solution:
Solution:
( )
Thegivenequationsofthesetofalllines
px +qy +r=0...(i)
andgivenconditionis:
3p +2q +4r =0
⇒3
4p+2
4q+r=0...(ii)
From(i)&(ii)weget:
∴x=3
4,y=1
2
Hencethesetoflinesareconcurrentandpassingthrough
thefixedpoint
3
4,1
2
( )
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Question89
ApointPmovesontheline2x −3y +4=0.IfQ(1,4)andR(3, −2)are
fixedpoints,thenthelocusofthecentroidof∆PQRisaline:
[Jan.10,2019(I)]
Options:
A.withslope 3
2
B.paralleltox-axis
C.withslope 2
3
D.paralleltoy-axis
x1,3x1+6
2
Since,AH isperpendiculartoBC
Hence,mAH ⋅mBC = −1
20 −4x2
5−1
x2−1×3
2= −1
15 −4x2
5(x2−1)= − 2
3
45 −12x2= −10x2+10
2x2=35 ⇒x2=35
2
⇒A35
2, −10
Since,BH isperpendiculartoCA.
Hence,mBH ×mCA = −1
3x1
2+3−1
x1−1−4
5= −1
(3x1+4)
2(x1−1)×4=5
⇒6x1+8=5x1−5⇒x1= −13 ⇒ −13,−33
2
⇒EquationoflineABis
y+10 =
−33
2+10
−13 −35 x−35
2
⇒−61y −610 = −13x +455
2
⇒−122y −1220 = −26x +455
⇒26x −122y −1675 =0
( )
( )
( )
( ) ( )
( )
( ) ( )
Answer:C
Solution:
Solution:
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Question90
Iftheline3x +4y −24 =0intersectsthex-axisatthepointAandthey-
axisatthepointB,thentheincentreofthetriangleOAB,whereOisthe
origin,is:
[Jan.10,2019(I)]
Options:
A.(3,4)
B.(2,2)
C.(4,3)
D.(4,4)
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
LetcentroidCbe(α,β)
wehaveα=1+3+h
3⇒h=3α −4
β=4−2+k
3⇒k=3β −2
butP(h,k)lieson2x −3y +4=0
⇒2(3α −4) − 3(3β −2) + 4=0
⇒6α −9β −8+6+4=0
⇒6α −9β +2=0
Locus:6x −9y +2=0
⇒y=6
9x+2
9∴itsslope = 6
9=2
3
Equationofthelineis:3x +4y =24or x
8+y
6=1
∴coordinatesofA,B&Oare(8,0), (0,6) & (0,0)respectively.
⇒OA =8,OB =6&AB =10
∴Incentreof∆OABisgivenas:
I≡8×0+6×8+10 ×0
8+6+10 ,8×6+6×0+10 ×0
8+6+10 ≡ (2,2).
( )
Question91
Twoverticesofatriangleare(0,2)and(4,3).Ifitsorthocentreisatthe
origin,thenitsthirdvertexliesinwhichquadrant?
[Jan.10,2019(II)]
Options:
A.third
B.second
C.first
D.fourth
Answer:B
Solution:
Solution:
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Question92
Twosidesofaparallelogramarealongthelines,x +y=3and
x−y+3=0.Ifitsdiagonalsintersectat(2,4),thenoneofitsvertexis:
[Jan.10,2019(II)]
Options:
A.(3,5)
B.(2,1)
C.(2,6)
D.(3,6)
Answer:D
Solution:
Since,mQR ×mPH = −1
⇒mQR = − 1
mPH
⇒mQR =y−3
x−4=0
⇒y=3
mPQ ×mRH = −1
⇒1
4×y
x= −1
⇒y= −4x
⇒x= − 3
4
VertexRis −3
4,3
Hence,vertexRliesinsecondquadrant.
( )
Solution:
-------------------------------------------------------------------------------------------------
Question93
IfinaparallelogramABDC,thecoordinatesofA,BandCare
respectively(1,2),(3,4)and(2,5),thentheequationofthediagonalAD
is:
[Jan.11,2019(II)]
Options:
A.5x −3y +1=0
B.5x +3y −11 =0
C.3x −5y +7=0
D.3x +5y −13 =0
Answer:A
Solution:
Solution:
Since,x−y+3=0andx+y=3areperpendicularlinesandintersectionpointofx−y+3=0andx+y=3isP(0,3).
⇒Mismid-pointofPR ⇒R(4,5)
LetS(x1,x1+3)andQ(x2,3−x2)
Mismid-pointofSQ
⇒x1+x2=4,x1+3+3−x2=8
⇒x1=3,x2=1
Then,thevertexDis(3,6)
Since,inparallelogrammidpointsofbothdiagonalsconsiders.
∴mid-pointofAD = mid-pointofBC
x1+1
2,y1+2
2=3+2
2,4+5
2
∴ (x1,y1) = (4,7)
Then,equationofADis,
y−7=2−7
1−4(x−4)
y−7=5
3(x−4)
3y −21 =5x −20
5x −3y +1=0
( ) ( )
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Question94
IfastraightlinepassingthroughthepointP(−3,4)issuchthatits
interceptedportionbetweenthecoordinateaxesisbisectedatP,then
itsequationis:
[Jan.12,2019(II)]
Options:
A.3x −4y +25 =0
B.4x −3y +24 =0
C.x −y+7=0
D.4x +3y =0
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question95
Ifthestraightline,2x −3y +17 =0isperpendiculartothelinepassing
throughthepoints(7,17)and(15,β),thenβequals:
[Jan.12,2019(I)]
Options:
A. 35
3
B.-5
C.−35
3
Since,PismidpointofMN
Then, 0+x
2= −3
⇒x= −3×2⇒x= −6
and y+0
2=4⇒y+0=2×4⇒y=8
HencerequiredequationofstraightlineM N is
x
−6+y
8=1⇒4x −3y +24 =0
D.5
Answer:D
Solution:
Solution:
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Question96
LetO(0,0)andA(0,1)betwofixedpoints.ThenthelocusofapointP
suchthattheperimeterof△AOPis4,is:
[April8,2019(I)]
Options:
A.8x2−9y2+9y =18
B.9x2−8y2+8y =16
C.9x2+8y2−8y =16
D.8x2+9y2−9y =18
Answer:C
Solution:
Solution:
∵Equationofstraightlinecanberewrittenas,
y=2
3x+17
3
∴Slopeofstraightline = 2
3
Slopeoflinepassingthroughthepoints(7,17)and(15,β)
=β−17
15 −7=β−17
8
Since,linesareperpendiculartoeachother.
Hence,m1m2= −1
⇒2
3
β−17
8= −1⇒β=5
( ) ( )
LetpointP(h,k)
∵OA =1
So,OP +AP =3
⇒h2+k2+h2+ (k−1)2=3
√ √
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Question97
SlopeofalinepassingthroughP(2,3)andintersectingtheline
x+y=7atadistanceof4unitsfromP,is:
[April9,2019(I)]
Options:
A. 1− √5
1+ √5
B. 1− √7
1+ √7
C. √7−1
√7+1
D. √5−1
√5+1
Answer:B
Solution:
Solution:
⇒h2+ (k−1)2=9+h2+k2−6 h2+k2
⇒6 h2+k2=2k +8
⇒9h2+8k2−8k −16 =0
Hence,locusofpointPis
9x2+8y2−8y −16 =0
√
√
Sincepointat4unitsfromP(2,3)willbe
A(4 cos θ +2,4 sin θ +3)andthispointwillsatisfytheequationoflinex+y=7
-------------------------------------------------------------------------------------------------
Question98
Apointonthestraightline,3x +5y =15whichisequidistantfromthe
coordinateaxeswilllieonlyin:
[April8,2019(I)]
Options:
A.4 th quadrant
B.1 st quadrant
C.1 st and2 nd quadrants
D.1 st ,2nd and4 th quadrants
Answer:C
Solution:
Solution:
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Question99
Twoverticalpolesofheights,20mand80mstandapartonahorizontal
plane.Theheight(inmeters)ofthepointofintersectionofthelines
joiningthetopofeachpoletothefootoftheother,fromthishorizontal
planeis:
⇒cos θ +sin θ =1
2
Onsquaring
⇒sin 2 θ −3
4⇒2 tan θ
1+tan2θ= − 3
4
⇒3tan2θ+8 tan θ +3=0
⇒tan θ =−8±2√7
6(ignoring-vesign)
⇒tan θ =−8±2√7
6=1− √7
1+ √7
Apointwhichisequidistantfromboththeaxesliesoneithery=xandy= −x.
Since,pointliesontheline3x +5y =15
Thentherequiredpoint
3x +5y =15
x+y=0
x= − 15
2
y=15
2⇒ (x,y) = − 15
2,15
2{2nd quadrant }
3x +5y =15
or x−y=0
15
x=15
8
y=15
8⇒ (x,y) = 15
8,15
8{1st quadrant }
Hence,therequiredpointliesin1st and2nd quadrant.
( )
( )
[April08,2019(II)]
Options:
A.15
B.18
C.12
D.16
Answer:D
Solution:
Solution:
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Question100
Supposethatthepoints(h,k), (1,2)and(-3,4)lieonthelineL1.Ifa
lineL2passingthroughthepoints(h,k)and(4,3)isperpendicularon
L1,thenk
hequals:
[April08,2019(II)]
Options:
A.1
3
B.0
C.3
D.−1
7
Answer:A
EquationsoflinesOBandACarerespectively
y=80
x1
x...(i)
x
x1
+y
20 =1...(ii)
∵equations(i)and(ii)intersecteachother
∴substitutethevalueofxfromequation(i)toequation(ii),weget
y
80 +y
20 =1
⇒y+4y =80 ⇒y=16m
Hence,heightofintersectionpointis16m.
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question101
Ifthetwolinesx + (a−1)y=1and2x +a2y=1(a∈R− {0,1})are
perpendicular,thenthedistanceoftheirpointofintersectionfromthe
originis:
[April09,2019(II)]
Options:
A. 2
5
B. 2
5
C. 2
√5
D. √2
5
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
√
∵(h,k), (1,2)and(-3,4)arecollinear
∴
h k 1
1 2 1
−341
=0⇒ −2h −4k +10 =0
⇒h+2k =5...(i)
Now,mL1=4−2
−3−1= − 1
2⇒mL2=2[∵L1⟂L2]
Bythegivenpoints(h,k)and(4,3),
mL2=k−3
h−4⇒k−3
h−4=2⇒k−3=2h −8
2h −k=5...(ii)
From(i)and(ii)
h=3,k=1⇒k
h=1
3
| |
∵twolinesareperpendicular⇒m1m2= −1
⇒−1
a−1
−2
a2= −1
⇒2=a2(1−a) ⇒ a3−a2+2=0
⇒(a+1)(a2+2a +2) = 0⇒a= −1
Henceequationsoflinesarex−2y =1and2x +y=1
∴intersectionpointis 3
5,−1
5
Now,distancefromorigin = 9
25 +1
25 =10
25 =2
5
( ) ( )
( )
√ √ √
Question102
Arectangleisinscribedinacirclewithadiameterlyingalongtheline
3y =x+7.Ifthetwoadjacentverticesoftherectangleare(-8,5)and
(6,5),thentheareaoftherectangle(insq.units)is:
[April09,2019(II)]
Options:
A.84
B.98
C.72
D.56
Answer:A
Solution:
Solution:
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Question103
Linesaredrawnparalleltotheline4x −3y +2=0,atadistance 3
5from
theorigin.Thenwhichoneofthefollowingpointsliesonanyofthese
lines?
[April10,2019(II)]
Options:
A. −1
4,2
3
( )
Givensituation
∴perpendicularbisectorofABwillpassfromcentre.
∴equationofperpendicularbisectorx= −1
Hencecentreofthecircleis(-1,2)
Letco-ordinateofDis(α,β)
⇒α+6
2= −1and β+5
2=2
⇒α= −8andβ= −1, ∴D≡ (−8, −1)
|AD| = 6and|AB| = 14
Area = 6×14 =84
B. 1
4, − 1
3
C. 1
4,1
3
D. −1
4, − 2
3
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question104
Atrianglehasavertexat(1,2)andthemidpointsofthetwosides
throughitare(-1,1)and(2,3).Thenthecentroidofthistriangleis:
[April12,2019(II)]
Options:
A. 1,7
3
B. 1
3,2
C. 1
3,1
D. 1
3,5
3
Answer:B
Solution:
Solution:
( )
( )
( )
( )
( )
( )
( )
Letstraightlinebe4x −3y +α=0
∵distancefromorigin = 3
5
∴3
5=α
5⇒α= ±3
Hence,lineis4x −3y +3=0or4x −3y −3=0
Clearly − 1
4,2
3satisfies4x −3y +3=0
| |
( )
Fromthemid-pointformulaco-ordinatesofvertexBandCareB(−3,0)andC(3,4)
Now,centroidofthetriangle
G≡3−3+1
3,0+4+2
3⇒G≡1
3,2
( ) ( )
-------------------------------------------------------------------------------------------------
Question105
AstraightlineLatadistanceof4unitsfromtheoriginmakespositive
interceptsonthecoordinateaxesandtheperpendicularfromtheorigin
tothislinemakesanangleof60∘withthelinex +y=0.Thenan
equationofthelineLis:
[April12,2019(II)]
Options:
A.x + √3y=8
B.(√3+1)x+ (√3−1)y=8√2
C.√3x+y=8
D.Noneofthese
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question106
Theequationy =sin x sin(x+2) − sin2(x+1)representsastraightline
lyingin:
∵perpendicularmakesanangleof60∘withthelinex+y=0
∴theperpendicularmakesanangleof15∘or75∘withx-axis
Hence,theequationoflinewillbe
x cos 75∘+y sin 75∘=4
or x cos 15∘+y sin 15∘=4
(√3−1)x+ (√3+1)y=8√2
or (√3+1)x+ (√3−1)y=8√2
[April12,2019(I)]
Options:
A.secondandthirdquadrantsonly
B.first,secondandfourthquadrant
C.first,thirdandfourthquadrants
D.thirdandfourthquadrantsonly
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question107
LettheorthocentreandcentroidofatrianglebeA(−3,5)andB(3,3)
respectively.IfCisthecircumcentreofthistriangle,thentheradiusof
thecirclehavinglinesegmentACasdiameter,is:
[2018]
Options:
A.2√10
B.3 5
2
C. 3√5
2
D.√10
Answer:B
√
Considertheequation,y=sin x ⋅sin(x+2) − sin2(x+1)
=1
2cos(−2) − cos(2x +2)
2−1−cos(2x +2)
2
=(cos 2) − 1
2= −sin21
BythegraphyliesinIIIandIVquadrant.
[ ]
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question108
Astraightlinethroughafixedpoint(2,3)intersectsthecoordinateaxes
atdistinctpointsPandQ.IfOistheoriginandtherectangleOPRQis
completed,thenthelocusofRis:
[2018]
Options:
A.2x +3y =xy
B.3x +2y =xy
C.3x +2y =6xy
D.3x +2y =6
Answer:B
Solution:
Solution:
SinceOrthocentreofthetriangleisA(−3,5)andcentriodofthetriangleisB(3,3),then
AB = √40 =2√10
Centroiddividesorthocentreandcircumcentreofthetriangleinratio2:1
∴AB :BC =2:1
N ow,AB =2
3AC
⇒AC =3
2AB =3
2(2√10) ⇒ AC =3√10
∴RadiusofcirclewithACasdiametre
=AC
2=3
2√10 =35
2
√
EquationofPQis
x
h+y
k=1...(i)
Since,(i)passesthroughthefixedpoint(2,3)Then,
2
h+3
k=1
-------------------------------------------------------------------------------------------------
Question109
InatriangleABC,coordianatesofAare(1,2)andtheequationsofthe
mediansthroughBandCarex +y=5andx =4respectively.Thenarea
of∆ABC(insq.units)is
[OnlineApril15,2018]
Options:
A.5
B.9
C.12
D.4
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question110
Thefootoftheperpendiculardrawnfromtheorigin,ontheline,
3x +y=λ(λ≠0)isP.Ifthelinemeetsx-axisatAandy-axisatB,then
theratioBP :PAis
[OnlineApril15,2018]
Options:
Then,thelocusofRis 2
x+3
y=1or3x +2y =xy.
MedianthroughCisx=4
SothexcoordinateofCis4.letC≡ (4,y),thenthemidpointofA(1,2)andC(4,y)isDwhichliesonthemedian
throughB.
∴D≡1+4
2,2+y
2
Now, 1+4+2+y
2=5⇒y=3
So,C≡ (4,3)
Thecentroidofthetriangleistheintersectionofthemesians.Herethemediansx=4andx+4andx+y=5intersect
atG(4,1)
Theareaoftriangle∆ABC =3× ∆ AGC
=3×1
2[1(1−3) + 4(3−2) + 4(2−1)] = 9
( )
A.9:1
B.1:3
C.1:9
D.3:1
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question111
ThesidesofarhombusABCDareparalleltothelines,x −y+2=0and
7x −y+3=0.IfthediagonalsoftherhombusintersectatP(1,2)and
thevertexA(differentfromtheorigin)isonthey-axis,thenthe
ordinateofAis
[OnlineApril15,2018]
Options:
A.2
B. 7
4
C. 7
2
Let(x,y)befootofperpendiculardrawntothepoint(x1,y1)onthelineax+by +c=0
Relation: x−x1
a=y−y1
b=−(ax1+by1+cz1)
a2+b2
Here(x1,y1) = (o,0)
givenlineis:3x +y−λ=0
x−0
3=y−0
1=−((3×0) + (1×0) − λ)
32+12
x=3λ
10andy=λ
10
HencefootofperpendicularP=3λ
10,λ
10
LinemeetsX-axisatA=λ
3,0
andmeetsY-axisatB= (0,λ)
BP =3λ
10
2+λ
10 −λ2
⇒BP =9λ2
100 +81λ2
100
∴BP =90λ2
100
AP =λ
3−3λ
10
2+0−λ
10
2
⇒AP =λ2
900 +λ2
100
∴AP =10λ2
900
∴AP :AP =9:1
( )
( )
√( ) ( )
√
√
√( ) ( )
√
√
D. 5
2
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question112
Asquare,ofeachside2,liesabovethex-axisandhasonevertexatthe
origin.Ifoneofthesidespassingthroughtheoriginmakesanangle30∘
withthepositivedirectionofthex-axis,thenthesumofthex-
coordinatesoftheverticesofthesquareis:
[OnlineApril9,2017]
Options:
A.2√3−1
B.2√3−2
C.√3−2
D.√3−1
Answer:B
Solution:
Solution:
LetthecoordinateAbe(0,c)
Equationsofthegivenlinesare
x−y+2=0and
7x −y+3=0
Weknowthatthediagonalsoftherhombuswillbeparalleltotheanglebisectorsofthetwogivenlines;
y=x+2andy=7x +3
∴equationofanglebisectorsisgivenas:
x−y+2
√2= ± 7x −y+3
5√2
5x −5y +10 = ±(7x −y+3)
∴Parallelequationsofthediagonalsare2x +4y −7=0and12x −6y +13 =0
∴slopesofdiagonalsare −1
2and2.
Now,slopeofthediagonalfromA(0,c)andpassingthroughP(1,2)is(2−c)
∴2−c=2⇒c=0(notpossible)
∴2−c=−1
2⇒c=5
2
∴ordinateofAis 5
2
-------------------------------------------------------------------------------------------------
Question113
Arayoflightisincidentalongalinewhichmeetsanotherline,
7x −y+1=0,atthepoint(0,1).Therayisthenreflectedfromthis
pointalongtheline,y +2x =1.Thentheequationofthelineof
incidenceoftherayoflightis:
[OnlineApril10,2016]
Options:
A.41x −25y +25 =0
B.41x +25y −25 =0
C.41x −38y +38 =0
D.41x +38y −38 =0
Answer:C
Solution:
Solution:
ForA;
x
cos 30∘=y
sin 30∘=2
⇒x= √3andy=1
ForC
x
cos 120∘=y
sin 120∘=2
⇒x= −1,y= √3
ForB
x
cos 75∘=y
sin 75∘=2√2
⇒x= √3−1andy= √3+1
∴Sum =2√3−2
Letslopeofincidentraybem
∴angleofincidence=angleofreflection
-------------------------------------------------------------------------------------------------
Question114
Twosidesofarhombusarealongthelines,x −y+1=0and
7x −y−5=0.Ifitsdiagonalsintersectat(−1, −2),thenwhichoneof
thefollowingisavertexofthisrhombus?
[2016]
Options:
A. 1
3, − 8
3
B. −10
3, − 7
3
C.(-3,-9)
D.(-3,-8)
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question115
( )
( )
∴m−7
1+7m =−2−7
1−14 =9
13
⇒m−7
1+7m =9
13or m−7
1+7m = − 9
13
⇒13m −91 =9+63mor13m −91 = −9−63m
⇒50m = −100or76m =82
⇒m= − 1
2or m=41
38
⇒y−1= − 1
2(x−0) or y−1=41
38(x−0)
i.ex+2y −2=0or 38y −38 −41x =0
⇒41x −38y +38 =0
| | | |
Letothertwosidesofrhombusare
x−y+λ=0
and7x −y+µ=0
thenOisequidistantfromABandDCandfromADandBC
∴ | −1+2+1| = | −1+2+λ| ⇒λ= −3and|−7+2−5| = | −7+2+µ| ⇒µ=15
∴Othertwosidesarex−y−3=0and7x −y+15 =0
∴Onsolvingtheeqnsofsidespairwise,wegettheverticesas
,(1,2), −7
3,−4
3, (−3, −6) 1
3,−8
3
( ) ( )
Ifavariablelinedrawnthroughtheintersectionofthelines x
3+y
4=1
and x
4+y
3=1,meetsthecoordinateaxesatAandB, (A≠B),thenthe
locusofthemidpointofABis:
[OnlineApril9,2016]
Options:
A.7xy =6(x+y)
B.4(x+y)2−28(x+y) + 49 =0
C.6xy =7(x+y)
D.14(x+y)2−97(x+y) + 168 =0
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question116
Thepoint(2,1)istranslatedparalleltothelineL :x−y=4by2√3
units.IfthenewpointsQliesinthethirdquadrant,thentheequation
ofthelinepassingthroughQandperpendiculartoLis:
[OnlineApril9,2016]
Options:
A.x +y=2− √6
B.2x +2y =1− √6
C.x +y=3−3√6
D.x +y=3−2√6
Answer:D
Solution:
L1:4x +3y −12 =0
L2:3x +4y −12 =0
L1+λL2=0
(4x +3y −12) + λ(3x +4y −12) = 0
x(4+3λ) + y(3+4λ) − 12(1+λ) = 0
PointA12(1+λ)
4+3λ ,0
PointB 0,12(1+λ)
3+4λ
midpoint⇒h=6(1+λ)
4+3λ ... . (i)
k=6(1+λ)
3+4λ ... . (ii)
Eliminateλfrom(i)and(ii),then
6(h+k) = 7hk
6(x+y) = 7xy
( )
( )
Solution:
-------------------------------------------------------------------------------------------------
Question117
AstraightlinethroughoriginOmeetsthelines3y =10 −4xand
8x +6y +5=0atpointsAandBrespectively.ThenOdividesthe
segmentABintheratio:
[OnlineApril10,2016]
Options:
A.2:3
B.1:2
C.4:1
D.3:4
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question118
x−y=4
TofindequationofR
slopeofL=0is1
⇒slopeofQR = −1
LetQRisy=mx +c
y= −x+c
x+y−c=0
distanceofQRfrom(2,1)is2√3
2√3=|2+1−c|
√2
2√6= | 3−c|
c−3= ±2√6c=3±2√6
Linecanbex+y=3±2√6
x+y=3−2√6
Lengthof⟂to4x +3y =10fromorigin(0,0)
P1=10
5=2
Lengthof⟂to8x +6y +5=0fromorigin(0,0)
P2=5
10 =1
2
∵Linesareparalleltoeachother⇒ratiowillbe4:1or1:4
LetLbethelinepassingthroughthepointP(1,2)suchthatits
interceptedsegmentbetweentheco-ordinateaxesisbisectedatP.IfL1
isthelineperpendiculartoLandpassingthroughthepoint(−2,1),
thenthepointofintersectionofLandL1is:
[OnlineApril10,2015]
Options:
A. 4
5,12
5
B. 3
5,23
10
C. 11
20,29
10
D. 3
10,17
5
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question119
Thepoints 0,8
3, (1,3)and(82,30):
[OnlineApril10,2015]
Options:
A.formanacuteangledtriangle.
B.formarightangledtriangle.
C.lieonastraightline.
D.formanobtuseangledtriangle.
( )
( )
( )
( )
( )
EquationoflineL
x
2+y
4=1
2x +y=4...(i)
Forline
x−2y = −4...(ii)
solvingequation(i)and(ii);wegetpointofintersection
4∕5,12
5
( )
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question120
AstraightlineLthroughthepoint(3,-2)isinclinedatanangleof60∘to
theline√3x+y=1.IfLalsointersectsthex-axis,thentheequationof
Lis:
[OnlineApril11,2015]
Options:
A.y + √3x+2−3√3=0
B.√3y+x−3+2√3=0
C.y − √3x+2+3√3=0
D.√3y−x+3+2√3=0
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question121
Thecircumcentreofatriangleliesattheoriginanditscentroidisthe
midpointofthelinesegmentjoiningthepoints(a2+1,a2+1)and
(2a, −2a), a≠0.Thenforanya,theorthocentreofthistrianglelieson
A 0,8
3B(1,3)C(89,30)
SlopeofAB =1
3
SlopeofBC =1
3
So,liesonsameline
( )
Giveneqnoflineisy+ √3x −1=0
⇒y= −√3x +1
⇒(slope)m2= −√3
Lettheotherslopebem1
∴tan 60∘= ∣ m1− (−√3)
1+ (−√3m1)
⇒m1=0,m2= √3
SincelineLispassingthrough(3,-2)
∴y− (−2) = +√3(x−3)
⇒y+2= √3(x−3)
y− √3x +2+3√3=0
theline:
[OnlineApril11,2015]
Options:
A.y −2ax =0
B.y − (a2+1)x=0
C.y +x=0
D.(a−1)2x− (a+1)2y=0
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question122
GiventhreepointsP,Q,RwithP(5,3)andRliesonthex-axis.If
equationofRQisx −2y =2andPQisparalleltothex-axis,thenthe
centroidof∆PQRliesontheline:
[OnlineApril9,2014]
Options:
A.2x +y−9=0
B.x −2y +1=0
C.5x −2y =0
D.2x −5y =0
Answer:D
Solution:
Solution:
Circumcentre = (0,0)
Centroid = (a+1)2
2,(a−1)2
2
Weknowthecircumcentre(O),
Centroid(G)andorthocentre(H)ofatrianglelieonthelinejoiningtheOandG.
Also, H G
GO =2
1
⇒Coordinateoforthocentre = 3(a+1)2
2,3(a−1)2
2
Now,thesecoordinatessatisfieseqngiveninoption(d)Hence,requiredeqnoflineis
(a−1)2x− (a+1)2y=0
( )
-------------------------------------------------------------------------------------------------
Question123
Ifalineinterceptedbetweenthecoordinateaxesistrisectedatapoint
A(4,3),whichisnearertox-axis,thenitsequationis:
[OnlineApril12,2014]
Options:
A.4x −3y =7
B.3x +2y =18
C.3x +8y =36
D.x +3y =13
Answer:B
Solution:
Solution:
EquationofRQisx−2y =2...(i)
aty=0,x=2[R(2,0)]
asPQisparalleltox,y-coordinatesofQisalso3Puttingvalueofyinequation(i),weget
Q(8,3)
Centroidof∆PQR =8+5+2
3,3+3
3= (5,2)
Only(2x −5y =0)satisfythegivenco-ordinates.
( )
AdividesCBin2:1
⇒4=1×0+2×a
1+2=2a
3
⇒a=6⇒coordinateofBisB(6,0)
3=1×b+2×0
1+2=b
3
⇒b=9andC(0,9)
Slopeoflinepassingthrough(6,0),(0,9)
slope,m=9
−6= − 3
2
Equationofliney−0=−3
2(x−6)
2y = −3x +18
( )
( )
-------------------------------------------------------------------------------------------------
Question124
Leta,b,candd benon-zeronumbers.Ifthepointofintersectionofthe
lines4ax +2ay +c=0and5bx +2by +d =0liesinthefourthquadrant
andisequidistantfromthetwoaxesthen
[2014]
Options:
A.3bc −2ad =0
B.3bc +2ad =0
C.2bc −3ad =0
D.2bc +3ad =0
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question125
LetPSbethemedianofthetriangleverticesP(2,2), Q(6, −1)and
R(7,3).Theequationofthelinepassingthrough(1,-1)andparallelto
PSis:
[2014]
Options:
A.4x +7y +3=0
B.2x −9y −11 =0
C.4x −7y −11 =0
3x +2y =18
Givenlinesare
4ax +2ay +c=0
5bx +2by +d=0
Thepointofintersectionwillbe
x
2ad −2bc =−y
4ad −5bc =1
8ab −10ab
⇒x=2(ad −bc)
−2ab =bc −ad
ab
⇒y=5bc −4ad
−2ab =4ad −5bc
2ab
∵Pointofintersectionisinfourthquadrantsoxispositiveandyisnegative.
Alsodistancefromaxesissame
Sox= −y
( ∵distancefromx-axisis−yasyisnegative)
bc −ad
ab =5bc −4ad
2ab ⇒3bc −2ad =0
D.2x +9y +7=0
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question126
IfalineLisperpendiculartotheline5x −y=1,andtheareaofthe
triangleformedbythelineLandthecoordinateaxesis5,thenthe
distanceoflineLfromthelinex +5y =0is:
[OnlineApril19,2014]
Options:
A. 7
√5
B. 5
√13
C. 7
√13
D. 5
√7
Answer:B
Solution:
Solution:
LetP,Q,R,betheverticesof∆PQR
SincePSisthemedian
Sismid-pointofQR
So,S=7+6
2,3−1
2=13
2,1
Now,slopeofPS =2−1
2−13
2
= − 2
9
Since,requiredlineisparalleltoPSthereforeslopeofrequiredline=slopeofPS
Now,eqn.oflinepassingthrough(1,-1)andhavingslope− 2
9is
y− (−1) = − 2
9(x−1)
9y +9= −2x +2⇒2x +9y +7=0
( ) ( )
LetequationoflineL,perpendicularto5x −y=1bex+5y =c
-------------------------------------------------------------------------------------------------
Question127
Ifthethreedistinctlinesx +2ay +a=0,x+3by +b=0and
x+4ay +a=0areconcurrent,thenthepoint(a,b)liesona:
[OnlineApril12,2014]
Options:
A.circle
B.hyperbola
C.straightline
D.parabola
Answer:C
Solution:
Solution:
Giventhatareaof∆AOBis5.
Weknow
{area,A=1
2[x1(y2−y3) + x2(y3−y1) + x3(y1−y2)]
⇒5=1
2cc
5
∵(x1,y1) = (10,0) (x3,y3) = 0,c
5
(x2,y2) = (c,0)
⇒c= ±√50
DistancebetweenLandlinex+5y =0is
d=±√50 −0
12+52=√50
√26 =5
√13
}
[ ( ) ]
(( ) )
|√|
x+2ay +a=0.. . (i)
x+3by +b=0.. . (ii)
x+4ay +a=0.. . (iii)
Subtractingequation(iii)from(i)
−2ay =0
ay =0=y=0
Puttingvalueofyinequation(i),weget
x+0+a=0
x= −a
Puttingvalueofxandyinequation(ii),weget
−a+b=0⇒a=b
Thus,(a,b)liesonastraightline
-------------------------------------------------------------------------------------------------
Question128
Thebaseofanequilateraltriangleisalongthelinegivenby3x +4y =9.
Ifavertexofthetriangleis(1,2),thenthelengthofasideofthe
triangleis:
[OnlineApril11,2014]
Options:
A. 2√3
15
B. 4√3
15
C. 4√3
5
D. 2√3
5
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question129
Thex-coordinateoftheincentreofthetrianglethathasthe
coordinatesofmidpointsofitssidesas(0,1)(1,1)and(1,0)is:
Shortestdistanceofapoint(x1,y1)fromlineax +by =cisd=ax1+by1−c
a2+b2
NowshortestdistanceofP(1,2)from3x +4y =9isPC =d=3(1) + 4(2) − 9
32+42=2
5
Giventhat∆APBisanequilateraltriangleLet'a'beitssidethenPB =a,CB =a
2
Now,In∆PCB, (PB)2= (PC)2+ (CB)2
(ByPythagorastheoresm)
a2=2
5
2+a2
4
a2−a4
4=4
25 ⇒3a2
4=4
25
a2=16
75 ⇒a=16
75 =4
5√3×√3
√3=4√3
15
∴LengthofEquilateraltriangle(a) = 4√3
15
|√|
|√|
( )
√
[2013]
Options:
A.2 + √2
B.2 − √2
C.1 + √2
D.1 − √2
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question130
AlightrayemergingfromthepointsourceplacedatP(1,3)isreflected
atapointQintheaxisofx.Ifthereflectedraypassesthroughthepoint
R(6,7),thentheabscissaofQis:
[OnlineApril9,2013]
Options:
A.1
B.3
C. 7
2
D. 5
2
Fromthefigure,wehave
a=2,b=2√2,c=2
x1=0,x2=0,x3=2
Now,x-co-ordinateofincentreisgivenas
ax1+bx2+cx3
a+b+c
⇒x-coordinateofincentre
=2×0+2√2.0 +2.2
2+2+2√2=2
2+ √2=2− √2
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question131
Ifthethreelinesx −3y =p,ax +2y =qandax +y=rformaright-
angledtrianglethen:
[OnlineApril9,2013]
Options:
A.a2−9a +18 =0
B.a2−6a −12 =0
C.a2−6a −18 =0
D.a2−9a +12 =0
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
LetabcissaofQ=x
∴Q= (x,0)
tan θ =0−7
x−6,tan(180∘−θ) = 0−3
x−1
Now,tan(180∘−θ) = −tan θ
∴−3
x−1=−7
x−6⇒x=5
2
Sincethreelinesx−3y =p,
ax +2y =qandax +y=r
formarightangledtriangle
∴productofslopesofanytwolines = −1
Supposeax +2y =qandx−3y =pare⟂toeachother.
∴−a
2×1
3= −1⇒a=6
Now,consideroptiononebyonea=6satisfiesonlyoption(a)
∴Requiredanswerisa2−9a +18 =0
Question132
Arayoflightalongx + √3y= √3getsreflecteduponreachingx-axis,
theequationofthereflectedrayis
[2013]
Options:
A.y =x+ √3
B.√3y=x− √3
C.y = √3x− √3
D.√3y=x−1
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question133
Ifthex-interceptofsomelineLisdoubleasthatoftheline,
3x +4y =12andthey-interceptofLishalfasthatofthesameline,
thentheslopeofLis:
[OnlineApril22,2013]
Options:
A.-3
B.−3
8
C.−3
2
D.−3
16
SupposeB(0,1)beanypointongivenlineandco-ordinateofAis(√3,0).So,equationof
Reflectedrayis −1−0
0− √3=y−0
x− √3
⇒√3y=x− √3
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question134
Iftheextremitiesofthebaseofanisoscelestrianglearethepoints
(2a,0)and(0,a)andtheequationofoneofthesidesisx =2a,thenthe
areaofthetriangle,insquareunits,is:
[OnlineApril23,2013]
Options:
A. 5
4a2
B. 5
2a2
C. 25a2
4
D.5a2
Answer:B
Solution:
Solution:
Givenline3x +4y =12canberewrittenas
3x
12 +4y
12 =1⇒x
4+y
3=1
⇒x-intercept = 4andy-intercept = 3
Lettherequiredlinebe
L:x
a+y
b=1where
a=x-interceptandb=y-intercept
Accordingtothequestion
a=4×2=8andb=3∕2
∴Requiredlineis x
8+2y
3=1
⇒3x +16y =24
⇒y=−3
16 x+24
16
Hence,requiredslope = −3
16 .
Lety-coordinateofC=b
∴C= (2a,b)
-------------------------------------------------------------------------------------------------
Question135
Letθ1betheanglebetweentwolines2x +3y +c1=0and
−x+5y +c2=0andθ2betheanglebetweentwolines2x +3y +c1=0
and−x+5y +c3=0,wherec1,c2,c3areanyrealnumbers:
Statement-1:Ifc2andc3areproportional,thenθ1=θ2.
Statement-2:θ1=θ2forallc2andc3
[OnlineApril23,2013]
Options:
A.Statement-1istrue,Statement-2istrue;Statement-2isacorrectexplanationofStatement-
1.
B.Statement-1istrue,Statement-2istrue;Statement-2isnotacorrectexplanationof
Statement-1.
C.Statement-1isfalse;Statement-2istrue.
D.Statement-1istrue;Statement-2isfalse.
Answer:A
Solution:
AB =4a2+a2= √5a
Now,AC =BC ⇒b=4a2+ (b−a)2
⇒b2=4a2+b2+a2−2ab
⇒2ab =5a2⇒b=5a
2
∴C=2a,5a
2
Henceareaofthetriangle
=1
2
2a 0 1
0 a 1
2a 5a
21
=1
2
2a 0 1
0 a 1
05a
20
=1
2×2a −5a
2= − 5a2
2
Sinceareaisalways+ve,hencearea
=5a2
2sq ⋅unit
√
√
( )
| | | |
( )
Solution:
-------------------------------------------------------------------------------------------------
Question136
LetA(−3,2)andB(−2,1)betheverticesofatriangleABC.Ifthe
centroidofthistriangleliesontheline3x +4y +2=0,thenthevertex
Cliesontheline:
[OnlineApril25,2013]
Options:
A.4x +3y +5=0
B.3x +4y +3=0
C.4x +3y +3=0
D.3x +4y +5=0
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question137
IftheimageofpointP(2,3)inalineLisQ(4,5),thentheimageof
pointR(0,0)inthesamelineis:
[OnlineApril25,2013]
Options:
A.(2,2)
Twolines−x+5y +c2=0and−x+5y +c3=0areparalleltoeachother.Hencestatement-1istrue,statement2is
trueandstatement-2isthecorrectexplanationofstatement-1.
LetC= (x1,y1)
Centroid,E=x1−5
3,y1+3
3
Sincecentroidliesontheline
3x +4y +2=0
∴3x1−5
3+4y1+3
3+2=0
⇒3x1+4y1+3=0
Hencevertex(x1,y1)liesontheline
3x +4y +3=0
( )
( ) ( )
B.(4,5)
C.(3,4)
D.(7,7)
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question138
Iftheline2x +y=kpassesthroughthepointwhichdividestheline
segmentjoiningthepoints(1,1)and(2,4)intheratio3 :2,thenk
equals:
[2012]
Options:
A. 29
5
B.5
C.6
D. 11
5
Answer:C
Solution:
Solution:
Mid-pointofP(2,3)andQ(4,5) = (3,4)SlopeofPQ =1
SlopeofthelineL= −1
Mid-point(3,4)liesonthelineL.
EquationoflineL
y−4= −1(x−3) ⇒ x+y−7=0
LetimageofpointR(0,0)beS(x1,y1)
Mid-pointofRS =x1
2,y1
2
Mid-point x1
2,y1
2
liesontheline(i)
∴x1+y1=14
SlopeofRS =y1
x1
SinceRS⟂lineL
∴y1
x1
× (−1) = −1
∴x1=y1
From(ii)and(iii),
x1=y1=7
HencetheimageofR= (7,7)
( )
( )
LetthepointsbeA(1,1)andB(2,4).LetpointCdivideslineABintheratio3:2.So,bysectionformulawehave
-------------------------------------------------------------------------------------------------
Question139
Ifthestraightlinesx +3y =4,3x +y=4andx +y=0formatriangle,
thenthetriangleis
[OnlineMay7,2012]
Options:
A.scalene
B.equilateraltriangle
C.isosceles
D.rightangledisosceles
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question140
Iftwoverticalpoles20mand80mhighstandapartonahorizontal
plane,thentheheight(inm)ofthepointofintersectionofthelines
joiningthetopofeachpoletothefootofotheris
[OnlineMay7,2012]
Options:
A.16
B.18
C.50
D.15
Answer:A
C=3×2+2×1
3+2,3×4+2×1
3+2=8
5,14
5
SinceLine2x +y=kpassesthroughC8
5,14
5
⇒2×8
5+14
5=k⇒k=6
( ) ( )
( )
LetequationofAB :x+3y =4
LetequationofBC :3x +y=4
LetequationofCA :x+y=0
Now,Bysolvingtheseequationsweget
A= (−2,2), B= (1,1)and C = (2, −2)
Now, AB = √9+1= √10
BC = √1+9= √10
and CA = √16 +16 = √32
Since,lengthofABandBCaresamethereforetriangleisisosceles.
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question141
Thepointofintersectionofthelines(a3+3)x+ay +a−3=0and
(a5+2)x+ (a+2)y+2a +3=0(areal)liesonthey-axisfor
[OnlineMay7,2012]
Options:
A.novalueofa
B.morethantwovaluesofa
C.exactlyonevalueofa
D.exactlytwovaluesofa
Answer:A
Solution:
Solution:
Weputonepoleatorigin.
BC =80m,OA =20m
LineOCandABintersectatM.
Tofind:LengthofM N .
EqnofOC :y=80 −0
a−0x
⇒y=80
ax...(i)
EqnofAB :y=20 −0
0−a(x−a)
⇒y=−20
a(x−a)...(ii)
AtM: (i) = (ii)
⇒80
ax=−20
a(x−a)
⇒80
ax=−20
ax+20 ⇒x=a
5
∴y=80
a×a
5=16
( )
( )
Givenequationoflinesare
(a3+3)x+ay +a−3=0and(a5+2)x+ (a+2)y+2a +3=0(areal)
Sincepointofintersectionoflinesliesony-axis.
∴Putx=0ineachequation,weget
ay +a−3=0and(a+2)y+2a +3=0
Onsolvingtheseweget
-------------------------------------------------------------------------------------------------
Question142
Ifthepoint(1,a)liesbetweenthestraightlinesx +y=1and
2(x+y) = 3thenaliesininterval
[OnlineMay12,2012]
Options:
A. 3
2,∞
B. 1,3
2
C.(−∞,0)
D. 0,1
2
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question143
( )
( )
( )
(a+2)(a−3) − a(2a +3) = 0
⇒a2−a−6−2a2−3a =0
⇒−a2−4a −6=0⇒a2+4a +6=0
⇒a=−4± √16 −24
2=−4± √−8
2
(notreal)
Thisshowsthatthepointofintersectionofthelinesliesonthey-axisfornovalueof'a'.
Since,(1,a)liesbetweenx+y=1and2(x+y) = 3
∴Putx=1in2(x+y) = 3
Wegettherangeofy.Thus,
2(1+y) = 3⇒y=3
2−1=1
2
Thus'a'liesin 0,1
2
( )
Iftwoverticesofatriangleare(5,-1)and(-2,3)anditsorthocentreisat
(0,0),thenthethirdvertexis
[OnlineMay12,2012]
Options:
A.(4,-7)
B.(-4,-7)
C.(-4,7)
D.(4,7)
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question144
LetLbetheliney =2x,inthetwodimensionalplane.
Statement1:Theimageofthepoint(0,1)inListhepoint 4
5,3
5
Statement2:Thepoints(0,1)and 4
5,3
5lieonoppositesidesofthe
lineLandareatequaldistancefromit.
[OnlineMay19,2012]
Options:
( )
( )
Letthethirdvertexof∆ABCbe(a,b).
Orthocentre = H(0,0)
LetA(5, −1)andB(−2,3)beothertwoverticesof∆ABC.Now,(SlopeofAH ) × (SlopeofBC ) = −1
⇒−1−0
5−0
b−3
a+2= −1
⇒b−3=5(a+2)...(i)
Similarly,
(SlopeofBH ) × (SlopeofAC ) = −1
⇒− 3
2×b+1
a−5= −1
⇒3b +3=2a −10
⇒3b −2a +13 =0...(ii)
Onsolvingequations(i)and(ii)weget
a= −4,b= −7
Hence,thirdvertexis(-4,-7)
( ) ( )
( ) ( )
A.Statement1istrue,Statement2isfalse.
B.Statement1istrue,Statement2istrue,Statement2isnotacorrectexplanationfor
Statement1.
C.Statement1istrue,Statement2istrue,Statement2isacorrectexplanationforStatement1
D.Statement1isfalse,Statement2istrue.
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question145
Thelineparalleltox-axisandpassingthroughthepointofintersection
oflinesax +2by +3b =0andbx −2ay −3a =0where(a,b) ≠ (0,0)is
[OnlineMay26,2012]
Options:
A.abovex-axisatadistance2 ∕3fromit
B.abovex-axisatadistance3 ∕2fromit
C.belowx-axisatadistance3 ∕2fromit
D.belowx-axisatadistance2 ∕3fromit
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question146
Statement-1
LetP′(x1,y1)betheimageof(0,1)withrespecttotheline2x −y=0then
x1
2=y1−1
−1=−4(0) + 2(1)
5
⇒x1=4
5,y1=3
5
Thus,statement-1istrue.
Also,statement-2istrueandcorrectexplanationforstatement-1
Givenlinesare
ax +2by +3b =0andbx −2ay −3a =0
Since,requiredlineis||tox-axis
∴x=0Weputx=0ingivenequation,weget
2by = −3b ⇒y= − 3
2
Thisshowsthattherequiredlineisbelowx-axisatadistanceof 3
2fromit.
Considerthestraightlines
L1:x−y=1
L2:x+y=1
L3:2x +2y =5
L4:2x −2y =7
Thecorrectstatementis
[OnlineMay26,2012]
Options:
A.L1|L4,L2|L3,L1intersectL4
B.L1⟂L2,L1|L3,L1intersectL2
C.L1⟂L2,L2|L3,L1intersectL4
D.L1⟂L2,L1⟂L3,L2intersectL4
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question147
Ifa,b,c∈Rand1isarootofequationax2+bx +c=0,thenthecurve
y=4ax2+3bx +2c,a≠0intersectx-axisat
[OnlineMay26,2012]
Options:
A.twodistinctpointswhosecoordinatesarealwaysrationalnumbers
B.nopoint
C.exactlytwodistinctpoints
Considerthelines
L1:x−y=1
L2:x+y=1
L3:2x +2y =5
L4:2x −2y =7
L1⟂L2iscorrectstatement
( ∵Productoftheirslopes = −1)
L1⟂L3isalsocorrectstatement
( ∵Productoftheirslopes = −1)
Now,L2:x+y=1
L4:2x −2y =7
⇒2x −2(1−x) = 7
⇒2x −2+2x =7
⇒x=9
4andy=−5
4
Hence,L2intersectsL4
D.exactlyonepoint
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question148
IfA(2, −3)andB(−2,1)aretwoverticesofatriangleandthirdvertex
movesontheline2x +3y =9,thenthelocusofthecentroidofthe
triangleis:
[2011RS]
Options:
A.x −y=1
B.2x +3y =1
C.2x +3y =3
D.2x −3y =1
Answer:B
Solution:
Solution:
Givenax2+bx +c=0
⇒ax2= −bx −c
Now,consider
y=4ax2+3bx +2c
=4[−bx −c] + 3bx +2c
= −4bx −4c +3bx +2c = −bx −2c
Since,thiscurveintersectsx-axis
∴puty=0,weget
−bx −2c =0⇒ −bx =2c
⇒x=−2c
b
Thus,givencurveintersectsx-axisatexactlyonepoint.
Centroid(h,k) = 2−2+α
3,−3+1+β
3
∴α=3h
β−2=3k
β=3k +2
Thirdvertex(α,β)liesontheline
2x +3y =9
2α +3β =9
2(3h) + 3(3k +2) = 9
( )
-------------------------------------------------------------------------------------------------
Question149
Thelinesx +y= | a|andax −y=1intersecteachotherinthefirst
quadrant.Thenthesetofallpossiblevaluesofaintheinterval:
[2011RS]
Options:
A.(0,∞)
B.[1,∞)
C.(−1,∞)
D.(-1,1)
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question150
2h +3k =1
2x +3y =1
Giventhatx+y= | a|andax −y=1
CaseI:Ifa>0
x+y=a...(i)
ax −y=1...(ii)
Onaddingequations(i)and(ii),weget
x(1+a) = 1+a⇒x=1
y=a-1
Sincegiventhatintersectionpointliesinfirstquadrant
So,a−1≥0
⇒a≥1
⇒a∈ [1,∞)
CaseII:Ifa<0
x+y= −a...(iii)
ax −y=1...(iv)
Onaddingequations(iii)and(iv),weget
x(1+a) = 1−a
x=1−a
1+a>0⇒a−1
a+1<0
Sincea−1<0
∴a+1>0
⇒a> −1...(v)
←0
−1>
y= −a−1−a
1+a>0=−a−a2−1+a
1+a>0
⇒− a2+1
a+1>0⇒a2+1
a+1<0
Sincea2+1>0
∴a+1<0
⇒a< −1...(vi)
From(v)and(vi),a∈φ
Hence,Case-IIisnotpossible.
So,correctanswerisa∈ [1,∞)
( )
ThelinesL1:y−x=0andL2:2x +y=0intersectthelineL3:y+2=0
atPandQrespectively.ThebisectoroftheacuteanglebetweenL1and
L2intersectsL3atR.
Statement-1:TheratioPR :RQequals2√2: √5
Statement-2:Inanytriangle,bisectorofanangledividesthetriangle
intotwosimilartriangles.
[2011]
Options:
A.Statement-1istrue,Statement-2istrue;Statement-2isnotacorrectexplanationfor
Statement-1.
B.Statement-1istrue,Statement-2isfalse.
C.Statement-1isfalse,Statement-2istrue.
D.Statement-1istrue,Statement-2istrue;Statement-2isacorrectexplanationfor
Statement-1.
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question151
Thelinesp(p2+1)x−y+q=0and(p2+1)2x+ (p2+1)y+2q =0are
perpendiculartoacommonlinefor:
L1:y−x=0
L2:2x +y=0
L3:y+2=0
OnsolvingtheequationoflineL1andL2wegettheirpointofintersection(0,0)i.e.,originO.
OnsolvingtheequationoflineL1andL3,wegetP= (−2, −2).
Similarly,solvingequationoflineL2andL3,wegetQ= (−1, −2)
Weknowthatbisectorofanangleofatriangle,dividetheoppositesidethetriangleintheratioofthesidesincludingthe
angle[AngleBisectorTheoremofaTriangle]
∴PR
RQ =OP
OQ =(−2)2+ (−2)2
(−1)2+ (−2)2=2√2
√5
∴Statement1istruebut∠OPR ≠ ∠OQR
So∆OPRand∆OQRnotsimilar
∴Statement2isfalse
√
√
[2009]
Options:
A.exactlyonevaluesofp
B.exactlytwovaluesofp
C.morethantwovaluesofp
D.novalueofp
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question152
Theshortestdistancebetweentheliney −x=1andthecurvex =y2is:
[2009]
Options:
A. 2√3
8
B. 3√2
5
C. √3
4
D. 3√2
8
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Giventhatthelinesp(p2+1)x−y+q=0and(p2+1)2x+ (p2+1)y+2q =0areperpendiculartoacommonlinethen
theselinesmustbeparalleltoeachother,
∴m1=m2
⇒− p(p2+1)
−1= − (p2+1)2
p2+1
⇒(p2+1)2(p+1) = 0
⇒p= −1
∴pcanhaveexactlyonevalue.
Let(a2,a)bethepointofshortestdistanceonx=y2Thendistancebetween(a2,a)andlinex−y+1=0isgivenby
D=ax1+by1+c
a2+b2=|a2−a+1|
√2=1
√2a−1
2
2+3
4
Itisminwhena=1
2andDmin =3
4√2=3√2
8
|√| | ( ) |
Question153
TheperpendicularbisectorofthelinesegmentjoiningP(1,4)and
Q(k,3)hasy-intercept−4.Thenapossiblevalueofkis
[2008]
Options:
A.1
B.2
C.-2
D.-4
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question154
LetA(h,k), B(1,1)andC(2,1)betheverticesofarightangledtriangle
withACasitshypotenuse.Iftheareaofthetriangleislsquareunit,
thenthesetofvalueswhich'k'cantakeisgivenby
[2007]
Options:
A.{−1,3}
B.{−3, −2}
C.{1,3}
D.{0,2}
Answer:A
Solution:
SlopeofPQ =3−4
k−1=−1
k−1
∴SlopeofperpendicularbisectorofPQ = (k−1)
Also,midpointofPQ k+1
2,7
2.
∴EquationofperpendicularbisectorofPQis
y−7
2= (k−1)x−k+1
2
⇒2y −7=2(k−1)x− (k2−1)
⇒2(k−1)x−2y + (8−k2) = 0
Giventhaty-intercept
=8−k2
2= −4
⇒8−k2= −8 or k2=16 ⇒k= ±4
( )
( )
Solution:
-------------------------------------------------------------------------------------------------
Question155
LetP = (−1,0), Q= (0,0)andR = (3,3√3)bethreepoint.Theequation
ofthebisectoroftheanglePQRis
[2007]
Options:
A. √3
2x+y=0
B.x + √3y =0
C.√3x+y=0
D.x +√3
2y=0
Answer:C
Solution:
Solution:
Given:A(1,k), B(1,1)andC(2,1)areverticesofarightangledtriangleandareaof∆ABC =1squareunit
Weknowthat,areaofrightangledtriangle
=1
2×BC ×AB =1=1
2(a) ∣ (k−1)
⇒±(k−1) = 2⇒k= −1,3
Given:ThecoordinatesofpointsP,Q,Rare(−1,0),(0,0), (3,3√3)respectively.
SlopeofQR =y2−y1
x2−x1
=3√3
3
⇒tan θ = √3⇒θ=π
3
⇒ ∠RQX =π
3
∴ ∠RQP =π−π
3=2π
3
LetQMbisectsthe∠PQR,
-------------------------------------------------------------------------------------------------
Question156
Ifoneofthelinesofmy2+ (1−m2)xy −mx2=0isabisectoroftheangle
betweenthelinesxy =0,thenmis
[2007]
Options:
A.1
B.2
C.−1
2
D.-2
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question157
If(a,a2)fallsinsidetheanglemadebythelinesy =x
2,x >0and
y=3x,x>0,thenabelongto
[2006]
Options:
A. 0,1
2
B.(3,∞)
( )
∴ ∠M QR =π
3⇒ ∠M QX =2π
3
∴SlopeofthelineQM =tan 2π
3= −√3
∴ EquationoflineQM is(y−0) = −√3(x−0)
⇒y= −√3x⇒ √3x+y=0
Fromfigureequationofbisectorsoflines,xy =0arey= ±x
∴Puty= ±xinthegivenequation
my2+ (1−m2)xy −mx2=0
∴mx2± (1−m2)x2−mx2=0
⇒1−m2=0⇒m= ±1
C. 1
2,3
D. −3, − 1
2
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question158
AstraightlinethroughthepointA(3,4)issuchthatitsintercept
betweentheaxesisbisectedatA.Itsequationis
[2006]
Options:
A.x +y=7
B.3x −4y +7=0
C.4x +3y =24
D.3x +4y =25
Answer:C
Solution:
Solution:
( )
( )
ClearlyforpointP,
a2−3a <0anda2−a
2>0
⇒1
2<a<3
-------------------------------------------------------------------------------------------------
Question159
Ifavertexofatriangleis(1,1)andthemidpointsoftwosidesthrough
thisvertexare(-1,2)and(3,2)thenthecentroidofthetriangleis
[2005]
Options:
A. −1,7
3
B. −1
3,7
3
C. 1,7
3
D. 1
3,7
3
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question160
Thelineparalleltothex-axisandpassingthroughtheintersectionof
thelinesax +2by +3b =0andbx −2ay −3a =0,where(a,b) ≠ (0,0)is
[2005]
( )
( )
( )
( )
∵AisthemidpointofPQ
∴a+0
2=3,0+b
2=4⇒a=6,b=8
∴Equationoflineis x
6+y
8=1
or4x +3y =24
Vertexoftriangleis(1,1)andmidpointofsidesthroughthisvertexis(-1,2)and(3,2)
Co-ordinateofBis(x,y)
co-ordinatesofBis 1+x
2= −1,1+y
2=1: : ((x,y) : (−3,3))
Similarly,co-ordinateofCcomesouttobe(5,3)
Thuscentroidis, 1−3+5
3,1+3+3
3⇒1,7
3...(hint:usingformulaofcentroid)
( )
( )
Options:
A.belowthex-axisatadistanceof 3
2fromit
B.belowthex-axisatadistanceof 2
3fromit
C.abovethex-axisatadistanceof 3
2fromit
D.abovethex-axisatadistanceof 2
3fromit
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question161
Theequationofthestraightlinepassingthroughthepoint(4,3)and
makinginterceptsontheco-ordinateaxeswhosesumis-1is
[2004]
Options:
A. x
2−y
3=1and x
−2+y
1=1
B. x
2−y
3= −1and x
−2+y
1= −1
C. x
2+y
3=1and x
2+y
1=1
D. x
2+y
3= −1and x
−2+y
1= −1
Answer:A
Solution:
Solution:
Theeqn.oflinepassingthroughtheintersectionoflinesax +2by +3b =0and
bx −2ay −3a =0isax +2by +3b +λ(bx −2ay −3a) = 0
⇒(a+bλ)x+ (2b −2aλ)y+3b −3λa =0
Requiredlineisparalleltox-axis.
∴a+bλ =0⇒λ= −a∕b
⇒ax +2by +3b −a
b(bx −2ay −3a) = 0
⇒ax +2by +3b −ax +2a2
by+3a2
b=0
y 2b +2a2
b+3b +3a2
b=0
y2b2+2a2
b= − 3b2+3a2
b
y=−3(a2+b2)
2(b2+a2)=−3
2
Soitis3∕2unitsbelowx-axis.
( )
( ) ( )
-------------------------------------------------------------------------------------------------
Question162
LetA(2, −3)andB(−2,3)beverticesofatriangleABC.
Ifthecentroidofthistrianglemovesontheline2x +3y =1,thenthe
locusofthevertexCistheline
[2004]
Options:
A.3x −2y =3
B.2x −3y =7
C.3x +2y =5
D.2x +3y =9
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question163
Ifoneofthelinesgivenby6x2−xy +4cy2=0is3x +4y =0,thenc
equals
[2004]
Options:
A.-3
B.1
Lettherequiredlinebe x
a+y
b=1....(i)
thena+b= −1⇒b= −a−1....(ii)
(i)passesthrough(4,3), ⇒ 4
a+3
b=1
⇒4b +3a =ab....(iii)
Puttingvalueofbfrom(ii)in(iii),weget
a2−4=0⇒a= ±2⇒b= −3
or1∴Equationsofstraightlinesare
x
2+y
−3=1or x
−2+y
1=1
LetthevertexCbe(h,k),thenthe
centroidof∆ABCis x1+x2+x3
3,y1+y2+y3
3
=2−2+h
3,−3+1+k
3
=h
3,−2+k
3⋅Itlieson2x +3y =1
⇒2h
3−2+k=1⇒2h +3k =9
⇒LocusofCis2x +3y =9
( )
( )
( )
C.3
D.1
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question164
Ifthesumoftheslopesofthelinesgivenbyx2−2cxy −7y2=0isfour
timestheirproductchasthevalue
[2004]
Options:
A.-2
B.-1
C.2
D.1
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question165
Iftheequationofthelocusofapointequidistantfromthepoint(a1,b1)
and(a2,b2)is(a1−b2)x+ (a1−b2)y+c=0,thenthevalueof′c′is
[2003]
Options:
3x +4y =0isoneofthelineofthepairequations.oflines
6x2−xy +4cy2=0, Puty= − 3
4x
weget,6x2+3
4x2+4c −3
4x2=0
⇒6+3
4+9c
4=0⇒c= −3
( )
Letthelinesbey=m1xandy=m2xthen
m1+m2= − 2c
7andm1m2= − 1
7
Giventhatm1+m2=4m1m2
⇒− 2c
7= − 4
7⇒c=2
A. a1
2+b1
2−a2
2−b2
2
B. 1
2a2
2+b2
2−a1
2−b1
2
C.a1
2−a2
2+b1
2−b2
2
D. 1
2(a1
2+a2
2+b1
2+b2
2)
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question166
Locusofcentroidofthetrianglewhoseverticesare
(a cos t,a sin t), (b sin t, −b cos t)and(1,0),wheretisaparameter,is
[2003]
Options:
A.(3x +1)2+ (3y)2=a2−b2
B.(3x −1)2+ (3y)2=a2−b2
C.(3x −1)2+ (3y)2=a2+b2
D.(3x +1)2+ (3y)2=a2+b2
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
√
(x−a1)2+ (y−b1)2= (x−a2)2+ (y−b2)2
(a1−a2)x+ (b1−b2)y+1
2(a2
2+b2
2−a1
2−b1
2) = 0
Comparingwithgiveneqn.weget
c=1
2(a2
2+b2
2−a1
2−b1
2)
Weknowthatcentroid
(x,y) = x1+x2+x3
3,y1+y2+y3
3
x=a cos t +b sin t +1
3
⇒a cos t +b sin t =3x −1
y=a sin t −b cos t
3
⇒a sin t −b cos t =3y
Squaringandadding,
(3x −1)2+ (3y)2=a2+b2
( )
Question167
Ifx1,x2,x3andy1,y2,y3arebothinG.P.withthesamecommonratio,
thenthepoints(x1,y1), (x2,y2)and(x3,y3)
[2003]
Options:
A.areverticesofatriangle
B.lieonastraightline
C.lieonanellipse
D.lieonacircle.
Answer:B
Solution:
Solution:
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Question168
Asquareofsidealiesabovethex-axisandhasonevertexattheorigin.
Thesidepassingthroughtheoriginmakesanangleα 0 <α<π
4with
thepositivedirectionofx-axis.Theequationofitsdiagonalnotpassing
throughtheoriginis
[2003]
Options:
A.y(cos α +sin α) + x(cos α −sin α) = a
B.y(cos α −sin α) − x(sin α −cos α) = a
C.y(cos α +sin α) + x(sin α −cos α) = a
D.y(cos α +sin α) + x(sin α +cos α) = a
Answer:A
( )
Takingco-ordinatesas
Ax
r,y
r;B(x,y)andC(xr,yr)
Thenslopeoflinejoining
Ax
r,y
r,B(x,y) =
y 1 −1
r
x 1 −1
r
=y
x
andslopeoflinejoiningB(x,y)andC(xr,yr)
=y(r−1)
x(r−1)=y
x
∴m1=m2
∵SlopeofABandBCaresameandonepointBcommon.
⇒Pointslieonthestraightline.
( )
( ) ( )
( )
Solution:
Solution:
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Question169
Ifthepairofstraightlinesx2−2pxy −y2=0andx2−2qxy −y2=0be
suchthateachpairbisectstheanglebetweentheotherpair,then
[2003]
Options:
A.pq = −1
B.p =q
C.p = −q
D.pq =1.
Answer:A
Solution:
Solution:
Co-ordinatesofA= (a cos α,a sin α)EquationofOB,
y=tan π
4+α x
CA⟂rtoOB
∴SlopeofCA = −cot π
4+α
EquationofCA
y−a sin α = −cot π
4+α(x−a cos α)
⇒(y−a sin α)tan π
4+α= (a cos α −x)
⇒(y−a sin α)
tan π
4+tan α
1−tan π
4tan α
= (a cos α −x)
⇒(y−a sin α)(1+tan α) = (a cos α −x)(1−tan α)
⇒(y−a sin α)(cos α +sin α)
= (a cos α −x)(cos α −sin α)
⇒y(cos +sin α) − a sin α cos α −asin2α
=acos2α−a cos α sin α −x(cos α −sin α)
⇒y(cos α +sin α) + x(cos α −sin α) = a
y(sin α +cos α) + x(cos α −sin α) = a
( )
( )
( )
( ( ) )
( )
Equationofbisectorsofsecondpairofstraightlinesis,
qx2+2xy −qy2=0...(i)
Itmustbeidenticaltothefirstpair....
x2−2pxy −y2=0...(ii)
from(i)and(ii)
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Question170
Atrianglewithvertices(4,0),(-1,-1),(3,5)is
[2002]
Options:
A.isoscelesandrightangled
B.isoscelesbutnotrightangled
C.rightangledbutnotisosceles
D.neitherrightanglednorisosceles
Answer:A
Solution:
Solution:
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Question171
Locusofmidpointoftheportionbetweentheaxesofx cos α +y sin α =p
wherepisconstantis
[2002]
Options:
A.x2+y2=4
p2
B.x2+y2=4p2
C. 1
x2+1
y2=2
p2
D. 1
x2+1
y2=4
p2
Answer:D
Solution:
q
1=2
−2p =−q
−1⇒pq = −1
AB =(4+1)2+ (0+1)2= √26
BC =(3+1)2+ (5+1)2= √52
CA =(4−3)2+ (0−5)2= √26
∵AB =CA
∴Isoscelestriangle
∵(√26)2+ (√26)2=52
BC2=AB2+AC2
∴rightangledtriangle,
So,thegiventriangleisisoscelesrightangled.
√
√
√
Solution:
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Question172
Thepairoflinesrepresentedby3ax2+5xy + (a2−2)y2=0are
perpendiculartoeachotherfor
[2002]
Options:
A.twovaluesofa
B.∀a
C.foronevalueofa
D.fornovaluesofa
Answer:A
Solution:
Solution:
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Question173
EquationofABis
x cos α +y sin α =p;
⇒x cos α
p+y sin α
p=1
⇒x
p∕cos α +y
p∕sin α =1
So,co-ordinatesofAandBare
p
cos α,0and 0,p
sin α
So,coordinatesofmidpointofABare
M(x1,y1) = p
2 cos α,p
2 sin α
x1=p
2 cos α&y1=p
2 sin α
⇒cos α =p∕2x1andsin α =p∕2y1
∵cos2α+sin2α=1
Locusof(x1,y1)is 1
x2+1
y2=4
p2.
( ) ( )
( )
Weknowthatpairofstraightylines
ax2+2hxy +by2=0areperpendicularwhena+b=0
3a +a2−2=0⇒a2+3a −2=0
⇒a=−3± √9+8
2=−3± √17
2
Ifthepairoflinesax2+2hxy +by2+2gx +2f y +c=0intersectonthey
-axisthen
[2002]
Options:
A.2f gh =bg2+ch2
B.bg2≠ch2
C.abc =2f gh
D.noneofthese
Answer:A
Solution:
Solution:
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Putx=0inthegivenequation
⇒by2+2f y +c=0
Foruniquepointofintersection,f2−bc =0
⇒af 2−abc =0
Weknowthatforpairofstraightline
abc +2f gh −af 2−bg2−ch2=0
⇒2f gh −bg2−ch2=0