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Straight Lines and Pair of Straight Lines

JEE Mathematics · 144 questions · Page 4 of 15 · Click an option or "Show Solution" to reveal answer

Q31
If (a,a2)\left( {a,{a^2}} \right) falls inside the angle made by the lines y=x2,y = {x \over 2}, x>0x > 0 and y=3x,y = 3x, x>0,x > 0, then a belong to :
A (0,12)\left( {0,{1 \over 2}} \right)
B (3,∞)\left( {3,\infty } \right)
C (12,3)\left( {{1 \over 2},3} \right)
D (−3,−12)\left( {-3,-{1 \over 2}} \right)
Correct Answer
Option C
Solution

Clearly for point

P,P,
a2−3a<0{a^2} - 3a < 0

and

a2−a2>0⇒12<a<3{a^2} - {a \over 2} > 0 \Rightarrow {1 \over 2} < a < 3
Q32
A straight line through the point A(3,4)A (3, 4) is such that its intercept between the axes is bisected at AA. Its equation is :
A x+y=7x + y = 7
B 3x−4y+7=03x - 4y + 7 = 0
C 4x+3y=244x + 3y = 24
D 3x+4y=253x + 4y = 25
Correct Answer
Option C
Solution

As is the mid point of

PQ,PQ,

therefore

a+02=3,0+b2=4⇒a=6,b=8{{a + 0} \over 2} = 3,{{0 + b} \over 2} = 4 \Rightarrow a = 6,b = 8

∴\therefore Equation of line is

x6+y8=1{x \over 6} + {y \over 8} = 1

or

4x+3y=244x + 3y = 24
Q33
Let P=(−1,0), Q=(0,0)P = \left( { - 1,0} \right),\,Q = \left( {0,0} \right) and R=(3,33)R = \left( {3,3\sqrt 3 } \right) be three point. The equation of the bisector of the angle PQRPQR is :
A 32x+y=0{{\sqrt 3 } \over 2}x + y = 0
B x+3y=0x + \sqrt {3y} = 0
C 3x+y=0\sqrt 3 x + y = 0
D x+32y=0x + {{\sqrt 3 } \over 2}y = 0
Correct Answer
Option C
Solution

Given : The coordinates of points

P,Q,RP,Q,R

are

(−1,0),(-1,0),
(0,0), (3,33)\left( {0,0} \right),\,\left( {3,3\sqrt 3 } \right)

respectively Slope of

QRQR
=y2−y1x2−x1=333= {{{y_2} - {y_1}} \over {{x_2} - {x_1}}} = {{3\sqrt 3 } \over 3}
⇒tan⁡θ=3⇒θ=π3\Rightarrow \tan \theta = \sqrt 3 \Rightarrow \theta = {\pi \over 3}
⇒∠RQX=π3\Rightarrow \angle RQX = {\pi \over 3}

∴\therefore

∠RQP=π−π3=2π3\angle RQP = \pi - {\pi \over 3} = {{2\pi } \over 3}

Let

QMQM

bisects the

∠PQR,\angle PQR,

∴\therefore Slope of the line

QM=tanQM=tan
2π3=−3{{2\pi } \over 3} = - \sqrt 3

∴\therefore Equation of line

QMQM

is

(y−0)=−3(x−0)\left( {y - 0} \right) = - \sqrt 3 \left( {x - 0} \right)
⇒y=−3 x⇒3x+y=0\Rightarrow y = - \sqrt 3 \,x \Rightarrow \sqrt 3 x + y = 0
Q34
If one of the lines of my2+(1−m2)xy−mx2=0m{y^2} + \left( {1 - {m^2}} \right)xy - m{x^2} = 0 is a bisector of angle between the lines xy=0,xy = 0, then mm is :
A 11
B 22
C −1/2-1/2
D −2-2
Correct Answer
Option A
Solution

Equation of bisectors of lines,

xy=0xy=0

are

y=±xy = \pm x

∴\therefore Put

y=± xy = \pm \,x

in the given equation

my2+(1−m2)xy−mx2=0m{y^2} + \left( {1 - {m^2}} \right)xy - m{x^2} = 0

∴\therefore

mx2+(1−m2)x2−mx2=0m{x^2} + \left( {1 - {m^2}} \right){x^2} - m{x^2} = 0
⇒1−m2=0⇒m=±1\Rightarrow 1 - {m^2} = 0 \Rightarrow m = \pm 1
Q35
Let A (h,k)\left( {h,k} \right), B(1,1)\left( {1,1} \right) and C (2,1)(2, 1) be the vertices of a right angled triangle with AC as its hypotenuse. If the area of the triangle is 11 square unit, then the set of values which ′k′'k' can take is given by :
A {−1,3}\left\{ { - 1,3} \right\}
B {−3,−2}\left\{ { - 3, - 2} \right\}
C {1,3}\left\{ { 1,3} \right\}
D {0,2}\left\{ {0,2} \right\}
Correct Answer
Option A
Solution

Given : The vertices of a right angled triangle

A(1,k),A\left( {1,k} \right),
B(1,1)B\left( {1,1} \right)

and

C(2,1)C\left( {2,1} \right)

and area of

ΔABC=1\Delta ABC = 1

square unit We know that, area of night angled triangle

=12×BC×AB=1=12(1)∣(k−1)∣= {1 \over 2} \times BC \times AB = 1 = {1 \over 2}\left( 1 \right)\left| {\left( {k - 1} \right)} \right|
⇒±(k−1)=2⇒k=−1,3\Rightarrow \pm \left( {k - 1} \right) = 2 \Rightarrow k = - 1,3
Q36
The lines p(p2+1)x−y+q=0p\left( {{p^2} + 1} \right)x - y + q = 0 and (p2+1)2x+(p2+1)y+2q\left( {{p^2} + 1} \right){}^2x + \left( {{p^2} + 1} \right)y + 2q =0=0 are perpendicular to a common line for :
A exactly one values of pp
B exactly two values of pp
C more than two values of pp
D no value of pp
Correct Answer
Option A
Solution

If the lines

p(p2+1)x−y+q=0p\left( {{p^2} + 1} \right)x - y + q = 0

and

(p2+1)2x+(p2+1)y+2q=0{\left( {{p^2} + 1} \right)^2}x + \left( {{p^2} + 1} \right)y + 2q = 0

are perpendicular to a common line then these lines - must be parallel to each other, ∴\therefore

m1=m2⇒−p(p2+1)−1=−(p2+1)2p2+1{m_1} = {m_2} \Rightarrow - {{p\left( {{p^2} + 1} \right)} \over { - 1}} = - {{{{\left( {{p^2} + 1} \right)}^2}} \over {{p^2} + 1}}
⇒(p2+1)(p+1)=0\Rightarrow \left( {{p^2} + 1} \right)\left( {p + 1} \right) = 0
⇒p=−1\Rightarrow p = - 1

∴\therefore

pp

can have exactly one value.

Q37
The line LL given by x5+yb=1{x \over 5} + {y \over b} = 1 passes through the point (13,32)\left( {13,32} \right). The line K is parrallel to LL and has the equation xc+y3=1.{x \over c} + {y \over 3} = 1. Then the distance between LL and KK is :
A 17\sqrt {17}
B 1715{{17} \over {\sqrt {15} }}
C 2317{{23} \over {\sqrt {17} }}
D 2315{{23} \over {\sqrt {15} }}
Correct Answer
Option C
Solution

Slope of line

L=−b5L = - {b \over 5}

Slope of line

K=−3cK = - {3 \over c}

Line

LL

is parallel to line

k.k.
⇒b5=3c⇒bc=15\Rightarrow {b \over 5} = {3 \over c} \Rightarrow bc = 15
(13,32)(13,32)

is a point on

L.L.

∴\therefore

135+32b=1⇒32b=−85{{13} \over 5} + {{32} \over b} = 1 \Rightarrow {{32} \over b} = - {8 \over 5}
⇒b=−20⇒c=−34\Rightarrow b = - 20 \Rightarrow c = - {3 \over 4}

Equation of

K:K:
y−4x=3y - 4x = 3
           \,\,\,\,\,\,\,\,\,\,\,
⇒4x−y+3=0\Rightarrow 4x - y + 3 = 0

Distance between

LL

and

KK
=∣52−32+3∣17=2317= {{\left| {52 - 32 + 3} \right|} \over {\sqrt {17} }} = {{23} \over {\sqrt {17} }}
Q38
The lines L1:y−x=0{L_1}:y - x = 0 and L2:2x+y=0{L_2}:2x + y = 0 intersect the line L3:y+2=0{L_3}:y + 2 = 0 at PP and QQ respectively. The bisector of the acute angle between L1{L_1} and L2{L_2} intersects L3{L_3} at RR. Statement-1: The ratio PRPR : RQRQ equals 22:52\sqrt 2 :\sqrt 5 Statement-2: In any triangle, bisector of an angle divide the triangle into two similar triangles.
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.
Correct Answer
Option B
Solution
L1:y−x=0{L_1}:y - x = 0
L2:2x+y=0{L_2}:2x + y = 0
L3:y+2=0{L_3}:y + 2 = 0

On solving the equation of line

L1{L_1}

and

L2{L_2}

we get their point of intersection

(0,0)(0, 0)

i.e., origin

O.O.

On solving the equation of line

L1{L_1}

and

L3,{L_3},

we get

P=(−2,−2).P=(-2, -2).

Similarly, we get

Q=(−1,−2)Q = \left( { - 1, - 2} \right)

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 ] ∴\therefore

PRRQ=OPOQ=(−2)2+(−2)2(−1)2+(−2)2{{PR} \over {RQ}} = {{OP} \over {OQ}} = {{\sqrt {{{\left( { - 2} \right)}^2} + {{\left( { - 2} \right)}^2}} } \over {\sqrt {{{\left( { - 1} \right)}^2} + {{\left( { - 2} \right)}^2}} }}
=225= {{2\sqrt 2 } \over {\sqrt 5 }}
Q39
The xx-coordinate of the incentre of the triangle that has the coordinates of mid points of its sides as (0,1)(1,1)(0, 1) (1, 1) and (1,0)(1, 0) is :
A 2+22 + \sqrt 2
B 2−22 - \sqrt 2
C 1+21 + \sqrt 2
D 1−21 - \sqrt 2
Correct Answer
Option B
Solution

From the figure, we have

a=2,b=22,c=2a = 2,b = 2\sqrt 2 ,c = 2
x1=0, x2=0, x3=2{x_1} = 0,\,{x^2} = 0,\,{x_3} = 2

Now,

xx

-co-ordinate of incenter is given as

ax1+bx2+cx3a+b+c{{a{x_1} + b{x_2} + c{x_3}} \over {a + b + c}}
⇒x\Rightarrow x

-coordinate of incentre

=2×0+22.0+2.22+2+22= {{2 \times 0 + 2\sqrt 2 .0 + 2.2} \over {2 + 2 + 2\sqrt 2 }}

==

22+2=2−2{2 \over {2 + \sqrt 2 }} = 2 - \sqrt 2
Q40
A ray of light along x+3y=3x + \sqrt 3 y = \sqrt 3 gets reflected upon reaching XX-axis, the equation of the reflected ray is :
A y=x+3y = x + \sqrt 3
B 3y=x−3\sqrt 3 y = x - \sqrt 3
C y=3x−3y = \sqrt 3 x - \sqrt 3
D 3y=x−1\sqrt 3 y = x - 1
Correct Answer
Option B
Solution
x+3y=3x + \sqrt 3 y = \sqrt 3

or

y=−13x+1y = - {1 \over {\sqrt 3 }}x + 1

Let θ\theta be the angle which the line makes with the positive x-axis. ∴\therefore

tan⁡θ=−13=tan⁡(π−π6)\tan \theta = - {1 \over {\sqrt 3 }} = \tan \left( {\pi - {\pi \over 6}} \right)

or

θ=π−π6\theta = \pi - {\pi \over 6}

∴\therefore

∠ABC=π6\angle ABC = {\pi \over 6}

; ∴\therefore

∠DBE=π6\angle DBE = {\pi \over 6}

∴\therefore the equation of the line BD is,

y=tan⁡π6x+cy = \tan {\pi \over 6}x + c

or

y=x3+cy = {x \over {\sqrt 3 }} + c

..... (1) The line

x+3y=3x + \sqrt 3 y = \sqrt 3

intersects the x-axis at

B(3,0)B(\sqrt 3 ,0)

and, the line (1) passes through

B(3,0)B(\sqrt 3 ,0)

. ∴\therefore

0=33+c0 = {{\sqrt 3 } \over {\sqrt 3 }} + c

or, c = −-1 Hence, the equation of the reflected ray is,

y=x3−1y = {x \over {\sqrt 3 }} - 1

or

y3=x−3y\sqrt 3 = x - \sqrt 3
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