Straight Lines and Pair of Straight Lines — Page 14 | JEE Mathematics

Q131

Let a be the length of a side of a square OABC with O being the origin. Its side OA makes an acute angle

\alpha

with the positive x-axis and the equations of its diagonals are

(\sqrt{3}+1)x+(\sqrt{3}-1)y=0

and

(\sqrt{3}-1)x-(\sqrt{3}+1)y+8\sqrt{3}=0

. Then

a

2 is equal to :

A 48

B 16

C 24

D 32

Correct Answer

Option A

Solution

\begin{aligned} &\begin{aligned} & \text{ Slope of diagonal } \mathrm{OB}=\frac{\sqrt{3}+1}{1-\sqrt{3}} \\ & =\tan 105^{\circ} \\ & \therefore \alpha=60^{\circ} \\ & \therefore \mathrm{A}\left(\operatorname{acos} 60^{\circ}, \mathrm{asin} 60^{\circ}\right) \\ & \therefore \mathrm{A}\left(\frac{\mathrm{a}}{2}, \frac{\sqrt{3} \mathrm{a}}{2}\right) \end{aligned}\\ &\text{ A Lies on other diagonal }\\ &\begin{aligned} & \therefore\left(\frac{\sqrt{3}-1}{2}\right) a-\left(\frac{\sqrt{3}+1}{2}\right) \cdot \sqrt{3} a+8 \sqrt{3}=0 \\ & a\left[\frac{\sqrt{3}-1-3-\sqrt{3}}{2}\right]=-8 \sqrt{3} \end{aligned} \end{aligned}

\begin{aligned} & \mathrm{a}=4 \sqrt{3} \\ & \therefore \mathrm{a}^2=48 \end{aligned}

Q132

A line passing through the point P(

a

, 0) makes an acute angle

\alpha

with the positive x-axis. Let this line be rotated about the point P through an angle

\dfrac{\alpha}{2}

in the clockwise direction. If in the new position, the slope of the line is

2 - \sqrt{3}

and its distance from the origin is

\dfrac{1}{\sqrt{2}}

, then the value of

3a^2 \tan^2 \alpha - 2\sqrt{3}

is :

A 8

B 4

C 5

D 6

Correct Answer

Option B

Solution

\begin{aligned} & \mathrm{m}_{\mathrm{PR}}=2-\sqrt{3}=\tan 15^{\circ} \\ & \therefore \frac{\alpha}{2}=15^{\circ} \end{aligned}

$\Rightarrow \alpha=30^{\circ}$ equation of PR:

\begin{aligned} & y=\tan 15^{\circ}(x-a) \\ & y=(2-\sqrt{3})(x-a) \end{aligned}

$\perp$ distance from origin $=\dfrac{1}{\sqrt{2}}$

\begin{aligned} & \left|\frac{\sqrt{3} a-2 a}{\sqrt{4+3-4 \sqrt{3}+1}}\right|=\frac{1}{\sqrt{2}} \\ & \frac{|a|(2-\sqrt{3})}{2 \sqrt{(2-\sqrt{3})}}=\frac{1}{\sqrt{2}} \\ & |a|=\frac{\sqrt{2}}{\sqrt{2-\sqrt{3}}}=\sqrt{2}(\sqrt{2+\sqrt{3}}) \\ & a^2=2(2+\sqrt{3}) \\ & 3 a^2 \tan ^2 \alpha-2 \sqrt{3} \\ & 3 \times(4+2 \sqrt{3}) \cdot \frac{1}{3}-2 \sqrt{3}=4 \end{aligned}

Q133

Let ABC be the triangle such that the equations of lines AB and AC be

3 y-x=2

and

x+y=2

, respectively, and the points B and C lie on

x

-axis. If P is the orthocentre of the triangle ABC , then the area of the triangle PBC is equal to

A 8

B 4

C 10

D 6

Correct Answer

Option D

Solution

Equation of line $A B$ is $3 y-x=2$ And $A C$ is $x+y=2$ In line $A B$ , When $y=0, x=-2$

\therefore \quad B(-2,0)

In line $A C$ , When $y=0, x=2$

\therefore \quad C(2 x, 0)

Equation of altitude of $B C$ ,

Y=x+2

Similarly, equation of altitude of $A B$ ,

y=-3 x+6

$\therefore$ On solving, orthocentre $P(1,3)$

\therefore \quad \operatorname{ar}(\triangle P B C)=6

Q134

If the orthocenter of the triangle formed by the lines y = x + 1, y = 4x - 8 and y = mx + c is at (3, -1), then m - c is :

A 0

B 2

C -2

D 4

Correct Answer

Option A

Solution

Solve line PQ & QR

\begin{aligned} & \text{ Point } \mathrm{Q}\left(\frac{1-\mathrm{c}}{\mathrm{~m}-1}, \frac{1-\mathrm{c}}{\mathrm{~m}-1}+1\right) \\ & \mathrm{m}_{2 \mathrm{H}}=\frac{\frac{1-\mathrm{c}}{\mathrm{~m}-1}+2}{\frac{1-\mathrm{c}}{\mathrm{~m}-1}-3}=\frac{1-\mathrm{c}+2 \mathrm{~m}-2}{1-\mathrm{c}-3 \mathrm{~m}+3}=-\frac{1}{4} \quad\text{..... (1)}\\ & \because \mathrm{~m}_{\mathrm{PH}}=\frac{5}{0} \rightarrow \infty \\ & \Rightarrow \text{ Slope of line } \mathrm{QR}(\mathrm{~m})=0 \end{aligned}

Put value of $m$ in equation (1)

\frac{1-c-2}{1-c+3}=-\frac{1}{4} \Rightarrow c=0

so $\mathrm{m}-\mathrm{c}=0$ Ans.

Q135

Let the area of the triangle formed by a straight line

\mathrm{L}: x+\mathrm{b} y+\mathrm{c}=0

with co-ordinate axes be 48 square units. If the perpendicular drawn from the origin to the line L makes an angle of

45^{\circ}

with the positive

x

-axis, then the value of

\mathrm{b}^2+\mathrm{c}^2

is :

A 90

B 83

C 93

D 97

Correct Answer

Option D

Solution

\begin{aligned} & L: x+b y+c=0 \\ & \because \frac{1}{2}\left|(-c) \cdot\left(\frac{-c}{b}\right)\right|=48 \\ & \therefore\left|\frac{c^2}{b}\right|=96\quad\text{..... (i)} \end{aligned}

Slope of line $L=-\dfrac{1}{b}$ $\therefore$ Slope of line perpendicular to $L$ is $b$ .

\begin{aligned} & \therefore b=1 \\ & \therefore c^2=96 \\ & \therefore b^2+c^2=97 \end{aligned}

Q136

In a triangle ABC, coordinates of A are (1, 2) and the equations of the medians through B and C are respectively, x + y = 5 and x = 4. Then area of

\Delta

ABC (in sq. units) is :

A 12

B 4

C 5

D 9

Correct Answer

Option D

Solution

Median through C is x = 4 So the 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.

$\therefore$

\,\,\,

D =

\left( {{{1 + 4} \over 2},{{2 + y} \over 2}} \right)

Now,

{{1 + 4 + 2 + y} \over 2}

= 5 $\Rightarrow$ y = 3.

So, C $\equiv$ (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

\Delta

ABC = 3 $\times$

\Delta

AGC = 3 $\times$

{1 \over 2}

[1(1 $-$ 3) + 4(3 $-$ 2) + 4(2 $-$ 1)] = 9.

Q137

A line passes through the origin and makes equal angles with the positive coordinate axes. It intersects the lines

\mathrm{L}_1: 2 x+y+6=0

and

\mathrm{L}_2: 4 x+2 y-p=0, p>0

, at the points A and B , respectively. If

A B=\dfrac{9}{\sqrt{2}}

and the foot of the perpendicular from the point

A

on the line

L_2

is

M

, then

\dfrac{A M}{B M}

is equal to

A 5

B 3

C 2

D 4

Correct Answer

Option B

Solution

\begin{aligned} & \tan \theta=\left|\frac{-2-1}{1+(-2)(1)}\right|=\frac{3}{1} \\ & \sin \theta=\frac{3}{\sqrt{10}} \\ & A M=\frac{3}{\sqrt{10}} \times \frac{9}{\sqrt{2}}=\frac{27}{2 \sqrt{5}} \end{aligned}

Dist. between $I_1 ~\& ~l_2, \quad A M=\left|\dfrac{\dfrac{P}{2}+6}{\sqrt{5}}\right|=\dfrac{27}{2 \sqrt{5}}$

\begin{aligned} & \frac{P}{2}+6= \pm \frac{27}{2} \\ & \frac{P}{2}=\frac{27}{2}-6 \Rightarrow P=15, \text{ As } P>0 \\ & B M=\sqrt{\frac{81}{2}-\left(\frac{27}{2 \sqrt{5}}\right)^2} \\ & =\sqrt{\frac{810-729}{20}}=\sqrt{\frac{81}{20}}=\frac{9}{2 \sqrt{5}} \\ & \text{ Now, } \frac{A M}{B M}=\frac{\frac{27}{\frac{2 \sqrt{5}}{9}}}{\frac{2 \sqrt{5}}{2 \sqrt{2}}}=3 \end{aligned}

Q138

Consider the lines

x(3 \lambda+1)+y(7 \lambda+2)=17 \lambda+5, \lambda

being a parameter, all passing through a point P. One of these lines (say

L

) is farthest from the origin. If the distance of

L

from the point

(3,6)

is

d

, then the value of

d^2

is

A 10

B 20

C 15

D 30

Correct Answer

Option B

Solution

\begin{aligned} & x(3 \lambda+1)+y(7 \lambda+2)=17 \lambda+5 \\ & (x+2 y-5)+\lambda(3 x+7 y-17)=0 \\ & L_1+\lambda L_2=0 \\ & \Rightarrow \quad P \text{ is intersection of } L_1 \& L_2 \text{ i.e. }(1,2) \\ & y-2=m(x-1) \\ & m x-y+2-m=0 \\ & \text{ Distance from origin }=\left|\frac{2-m}{\sqrt{1+m^2}}\right|=\max \end{aligned}

\begin{aligned} & \text{ For } m=-\frac{1}{2} \\ & \therefore L=y-2=\frac{-1}{2}(x-1) \\ & L: x+2 y-5=0 \\ & \text{ Now, } d=\left|\frac{3+12-5}{\sqrt{5}}\right|=\left|\frac{10}{\sqrt{5}}\right| \\ & d^2=\frac{100}{5}=20 \end{aligned}

Q139

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 :

A 41x − 38y + 38 = 0

B 41x + 25y − 25 = 0

C 41x + 38y − 38 = 0

D 41x − 25y + 25 = 0

Correct Answer

Option A

Solution

Let slope of incident ray be m. $\therefore$ angle of incidence = angle of reflection $\therefore$

\left| {{{m - 7} \over {1 + 7m}}} \right| = \left| {{{ - 2 - 7} \over {1 - 14}}} \right| = {9 \over {13}}

$\Rightarrow$

{{m - 7} \over {1 + 7m}} = {9 \over {13}}

or

{{m - 7} \over {1 + 7m}} = - {9 \over {13}}

$\Rightarrow$ 13m $-$ 91 $=$ 9 + 63m or 13m $-$ 91 $=$ $-$ 9 $-$ 63m $\Rightarrow$ 50m $=$ $-$ 100 or 76m $=$ 82 $\Rightarrow$ m $=$

- {1 \over 2}

or m $=$

{{41} \over {38}}

$\Rightarrow$ y $-$ 1 $=$ $-$

{1 \over 2}

(x $-$ 0) or y $-$ 1 $=$

{{41} \over {38}}

(x $-$ 0) i.e x + 2y $-$ 2 $=$ 0 or 38y $-$ 38 $-$ 41x $=$ 0 $\Rightarrow$ 41x $-$ 38y + 38 $=$ 0

Q140

The point (2, 1) is translated parallel to the line L : x− y = 4 by

2\sqrt 3

units. If the newpoint Q lies in the third quadrant, then the equation of the line passing through Q and perpendicular to L is :

A x + y = 2

-

\sqrt 6

B x + y = 3

-

3

\sqrt 6

C x + y = 3

-

2

\sqrt 6

D 2x + 2y = 1

-

\sqrt 6

Correct Answer

Option C

Solution

x $-$ y = 4 To find equation of R slope of L = 0 is 1 $\Rightarrow$ 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

\sqrt 3

2

\sqrt 3

=

{{\left| {2 + 1 - c} \right|} \over {\sqrt 2 }}

2

\sqrt 6

=

\left| {3 - c} \right|

c $-$ 3 =

\pm 2\sqrt 6

c = 3 $\pm$ 2

\sqrt 6

Line can be x + y = 3 $\pm$ 2

\sqrt 6

x + y = 3 $-$ 2

\sqrt 6