Straight Lines and Pair of Straight Lines — Page 12 | JEE Mathematics

Q111

If

x^2-y^2+2 h x y+2 g x+2 f y+c=0

is the locus of a point, which moves such that it is always equidistant from the lines

x+2 y+7=0

and

2 x-y+8=0

, then the value of

g+c+h-f

equals

A 8

B 14

C 29

D 6

Correct Answer

Option B

Solution

Cocus of point

\mathrm{P}(\mathrm{x}, \mathrm{y})

whose distance from Gives

X+2 y+7=0

&

2 x-y+8=0

are equal is

\frac{x+2 y+7}{\sqrt{5}}= \pm \frac{2 x-y+8}{\sqrt{5}}

(x+2 y+7)^2-(2 x-y+8)^2=0

Combined equation of lines

\begin{aligned} & (x-3 y+1)(3 x+y+15)=0 \\ & 3 x^2-3 y^2-8 x y+18 x-44 y+15=0 \\ & x^2-y^2-\frac{8}{3} x y+6 x-\frac{44}{3} y+5=0 \\ & x^2-y^2+2 h x y+2 g x 2+2 f y+c=0 \\ & h=\frac{4}{3}, g=3, f=-\frac{22}{3}, c=5 \\ & g+c+h-f=3+5-\frac{4}{3}+\frac{22}{3}=8+6=14 \end{aligned}

Q112

A line passing through the point

\mathrm{A}(9,0)

makes an angle of

30^{\circ}

with the positive direction of

x

-axis. If this line is rotated about A through an angle of

15^{\circ}

in the clockwise direction, then its equation in the new position is :

A

\dfrac{y}{\sqrt{3}+2}+x=9

B

\dfrac{x}{\sqrt{3}+2}+y=9

C

\dfrac{x}{\sqrt{3}-2}+y=9

D

\dfrac{y}{\sqrt{3}-2}+x=9

Correct Answer

Option D

Solution

\mathrm{Eq}^{\mathrm{n}}: y-0=\tan 15^{\circ}(x-9) \Rightarrow y=(2-\sqrt{3})(x-9)

Q113

A variable line

\mathrm{L}

passes through the point

(3,5)

and intersects the positive coordinate axes at the points

\mathrm{A}

and

\mathrm{B}

. The minimum area of the triangle

\mathrm{OAB}

, where

\mathrm{O}

is the origin, is :

A 35

B 25

C 30

D 40

Correct Answer

Option C

Solution

\begin{aligned} & \frac{x}{a}+\frac{y}{b}=1 \\ & \frac{3}{a}+\frac{5}{b}=1 \\ & 3 b+5 a=a b \\ & 5 a-a b=-3 b \\ & a(5-b)=-3 b \\ & a=\frac{3 b}{b-5} \end{aligned}

\begin{aligned} & \text{ Area of triangle }=\left|\frac{1}{2} \times a \times b\right| \\ & =\frac{1}{2} \times \frac{3 b}{b-5} \times b \\ & \Rightarrow f(b)=\frac{3 b^2}{2 b-10} \\ & \Rightarrow f^{\prime}(b)=0,(2 b-10) 6 b-2\left(3 b^2\right)=0 \\ & 12 b^2-60 b-6 b^2=0 \\ & 6 b^2-60 b=0 \\ & b^2-10 b=0 \\ & b(b-10)=0 \\ & b=0 \text{ or } b=10 \end{aligned}

So for minimum area,

b=10

then

\frac{1}{2} \times a \times b=30

Q114

A ray of light coming from the point

\mathrm{P}(1,2)

gets reflected from the point

\mathrm{Q}

on the

x

-axis and then passes through the point

R(4,3)

. If the point

S(h, k)

is such that

P Q R S

is a parallelogram, then

hk^2

is equal to:

A 60

B 70

C 80

D 90

Correct Answer

Option B

Solution

\begin{aligned} & P^{\prime} R: y+2=\frac{5}{3}(x-1) \\ & \text{ For Point } Q \Rightarrow y=0 \\ & \frac{6}{5}=a-1 \Rightarrow a=\frac{11}{5} \end{aligned}

Now,

P Q R S

is parallelogram

\begin{aligned} & \therefore \frac{h+a}{2}=\frac{4+1}{2} \Rightarrow h=5-\frac{11}{5}=\frac{14}{5} \\ & \text{ and } \frac{2+3}{2}=\frac{k}{2} \Rightarrow K=5 \end{aligned}

Now

h k^2=25 \times \frac{14}{5}=14 \times 5=70

Q115

The vertices of a triangle are

\mathrm{A}(-1,3), \mathrm{B}(-2,2)

and

\mathrm{C}(3,-1)

. A new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to origin is :

A

-x+y-(2-\sqrt{2})=0

B

x+y-(2-\sqrt{2})=0

C

x+y+(2-\sqrt{2})=0

D

x-y-(2+\sqrt{2})=0

Correct Answer

Option B

Solution

Equation of

A C: x+y=2

Equation of

A B: x-y+4=0

Equation of

B C: 3 x+5 y=4

The line nearest to origin is parallel to

A C

and inward. Let its equation is

x+y=C

.

\therefore\left|\frac{C-2}{\sqrt{2}}\right|=1

\therefore \quad C=2-\sqrt{2}

$\therefore$ required equation line is :

x+y-(2-\sqrt{2})=0

Q116

If the line segment joining the points

(5,2)

and

(2, a)

subtends an angle

\dfrac{\pi}{4}

at the origin, then the absolute value of the product of all possible values of

a

is :

A 4

B 8

C 6

D 2

Correct Answer

Option A

Solution

To solve this problem, we will use the concept of the slope and tangent of the angle subtended by the line segments at the origin.

Given points:

(5,2)

and

(2,a)

The slope of the line joining the origin and

(5,2)

is:

m_1 = \frac{2-0}{5-0} = \frac{2}{5}

The slope of the line joining the origin and

(2,a)

is:

m_2 = \frac{a-0}{2-0} = \frac{a}{2}

The line segment joining these points subtends an angle

\theta = \frac{\pi}{4}

at the origin. According to the tangent formula for the angle between two lines:

\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|

Here,

\theta = \frac{\pi}{4}

so,

\tan \left( \frac{\pi}{4} \right) = 1

. Therefore,

1 = \left| \frac{\frac{a}{2} - \frac{2}{5}}{1 + \left(\frac{2}{5} \cdot \frac{a}{2}\right)} \right|

This simplifies to:

1 = \left| \frac{\frac{5a - 4}{10}}{1 + \frac{2a}{10}} \right|

Multiplying both the numerator and denominator by 10 to simplify:

1 = \left| \frac{5a - 4}{10 + 2a} \right|

This leads to two equations due to the absolute value: 1.

5a - 4 = 10 + 2a

3a = 14

a = \frac{14}{3}

2.

5a - 4 = -(10 + 2a)

5a - 4 = -10 - 2a

7a = -6

a = -\frac{6}{7}

Thus, the possible values of

a

are

\frac{14}{3}

and

-\frac{6}{7}

. The product of these values is:

\left| \frac{14}{3} \times -\frac{6}{7} \right|

= \left| -\frac{84}{21} \right|

= \left| -4 \right|

= 4

Therefore, the absolute value of the product of all possible values of

a

is 4. Answer: Option A

Q117

The equations of two sides

\mathrm{AB}

and

\mathrm{AC}

of a triangle

\mathrm{ABC}

are

4 x+y=14

and

3 x-2 y=5

, respectively. The point

\left(2,-\dfrac{4}{3}\right)

divides the third side

\mathrm{BC}

internally in the ratio

2: 1

, the equation of the side

\mathrm{BC}

is

A

x+6 y+6=0

B

x-3 y-6=0

C

x+3 y+2=0

D

x-6 y-10=0

Correct Answer

Option C

Solution

\begin{aligned} & 2=\frac{2 x_2+x_1}{3}, \frac{-4}{3}=\frac{2 y_2+y_1}{3} \\ & 2 x_2+x_1=6,2 y_2+y_1=-4 \end{aligned}

\begin{aligned} & x_1=6-2 x_2 \quad \text{.... (1)}\\ & y_1=-4-2 y_2 \quad \text{.... (2)}\\ & 4 x_1+y_1=14 \quad \text{.... (3)}\\ & 3 x_2-2 y_2=5 \quad \text{.... (4)} \end{aligned}

From here,

x_2=1, y_2=-1, x_1=4, y_1=-2

\begin{aligned} & B(4,-2) C(1,-1) \\ & y+2=\frac{-1+2}{1-4}(x-4) \\ & -3 y-6=x-4 \\ & x+3 y+2=0 \end{aligned}

Q118

Let two straight lines drawn from the origin

\mathrm{O}

intersect the line

3 x+4 y=12

at the points

\mathrm{P}

and

\mathrm{Q}

such that

\triangle \mathrm{OPQ}

is an isosceles triangle and

\angle \mathrm{POQ}=90^{\circ}

. If

l=\mathrm{OP}^2+\mathrm{PQ}^2+\mathrm{QO}^2

, then the greatest integer less than or equal to

l

is :

A 42

B 46

C 48

D 44

Correct Answer

Option B

Solution

O P=O Q

(

\triangle P Q R

is isosceles triangle) Let slope of line

O P \rightarrow m_1

So, equation

\rightarrow y=m_1 x

\begin{aligned} &\begin{aligned} & \tan 45^{\circ}=\left|\frac{m_1-m_2}{m_1 m_2}\right| \\ & 1=\left|\frac{m_1+\frac{3}{4}}{1-\frac{3}{4} m_1}\right| \\ & \Rightarrow 1-\frac{3}{4} m_1=m_1+\frac{3}{4} \\ & \frac{1}{4}=\frac{7}{4} m_1 \Rightarrow m_1=\frac{1}{7} \end{aligned}\\ &\text{ Equation } O P \rightarrow y=\frac{1}{7} x \end{aligned}

Point of intersection of

O P

& line

3 x+4 y=12

is

P\left(\frac{84}{25}, \frac{12}{25}\right)

\begin{aligned} & \Rightarrow \quad O P^2=a^2=\left(\frac{84}{25}\right)^2+\left(\frac{12}{25}\right)^2=\frac{288}{25} \\ & \therefore \quad I=O P^2+P Q^2+Q O^2 \\ &=a^2+a^2+2 a^2 \\ &=4 a^2 \\ &=4 \times \frac{288}{25} \\ & I=46.08 \\ & {[I] }=46 \end{aligned}

Q119

Let

\mathrm{A}(-1,1)

and

\mathrm{B}(2,3)

be two points and

\mathrm{P}

be a variable point above the line

\mathrm{AB}

such that the area of

\triangle \mathrm{PAB}

is 10. If the locus of

\mathrm{P}

is

\mathrm{a} x+\mathrm{by}=15

, then

5 \mathrm{a}+2 \mathrm{~b}

is :

A

-\dfrac{12}{5}

B

-\dfrac{6}{5}

C 6

D 4

Correct Answer

Option A

Solution

$\begin{aligned} & \dfrac{1}{2}\left|\begin{array}{ccc}h & k & 1 \\ -1 & 1 & 1 \\ 2 & 3 & 1\end{array}\right|=10 \\\\ & -2 x+3 y=25 \\\\ & -\dfrac{6}{5} x+\dfrac{9}{5} y=15\end{aligned}$ $\begin{aligned} & a=-\dfrac{6}{5}, b=\dfrac{9}{5} \\\\ & 5 a=-6,2 b=\dfrac{18}{5}\end{aligned}$ $\therefore$

5 \mathrm{a}+2 \mathrm{~b}

=

-\frac{12}{5}

Q120

If the locus of the point, whose distances from the point

(2,1)

and

(1,3)

are in the ratio

5: 4

, is

a x^2+b y^2+c x y+d x+e y+170=0

, then the value of

a^2+2 b+3 c+4 d+e

is equal to :

A 37

B

-27

C 437

D 5

Correct Answer

Option A

Solution

\begin{aligned} & \therefore \frac{(h-2)^2+(k-1)^2}{(h-1)^2+(k-3)^2}=\frac{25}{16} \\ & \frac{h^2+k^2-4 h-2 k+5}{h^2+k^2-2 h-6 k+10}=\frac{25}{16} \\ \end{aligned}

Replacing

h \rightarrow x

and

k \rightarrow y

.

\begin{aligned} & 16 x^2+16 y^2-64 x-32 y+80 \\ & =25 x^2+25 y^2-50 x-150 y+250 \\ & 9 x^2+9 y^2+14 x-118 y+170=0 \\ & \Rightarrow a=9, b=9, c=0, d=14, e=-118 \\ & a^2+2 b+3 c+4 d+e \\ & 81+18+56-118 \\ & \Rightarrow 37 \end{aligned}