Question 1: Find the equation of a straight line through the point of intersection of the lines $4 x-3y=0$ and $2x-5y+3 = 0$ and parallel to $4x+5y+6=0$.

The equation of a straight line passing through the point of intersection of $4 x-3y=0$ and $2x-5y+3 = 0$ is given by:

$( 4x-3y) + \lambda ( 2x-5y+3) = 0$     … … … … … i)

$\Rightarrow ( 4 + 2 \lambda) x - ( 3 +5\lambda) y + 3 \lambda = 0$

$\Rightarrow y = \Big($ $\frac{4+ 2 \lambda}{3+5 \lambda}$ $\Big) x+ \Big($ $\frac{3 \lambda}{3+5\lambda}$ $\Big)$     … … … … … ii)

Slope of line $4x+5y+6 = 0$  is $\frac{-4}{5}$

Since lines are parallel to each other,

$\frac{4+ 2 \lambda}{3+5 \lambda}$ $=$ $\frac{-4}{5}$

$\Rightarrow 20 + 10\lambda = - 12 - 20 \lambda$

$\Rightarrow 30 \lambda = - 32$

$\Rightarrow \lambda =$ $\frac{-16}{15}$

Substituting the value of  $\lambda$ in i) we get the equation of the required line.

$(4x-3y) -$ $\frac{16}{15}$ $( 2x-5y+3) = 0$

$\Rightarrow 60x-45y-32x+80y-48=0$

$\Rightarrow 28x+35y-48=0$

$\\$

Question 2: Find the equation of a straight line passing through the point of intersection of $x +2y + 3 =0$ and $3 x + 4y + 7 =0$ and perpendicular to the straight line $x -y +9=0$.

The equation of a straight line passing through the point of intersection of $x +2y + 3 =0$ and $3 x + 4y + 7 =0$ is given by:

$( x +2y + 3) + \lambda ( 3 x + 4y + 7) = 0$     … … … … … i)

$\Rightarrow ( 1 + 3 \lambda) x + ( 2 +4\lambda) y + (3 + 7 \lambda) = 0$

$\Rightarrow y = -\Big($ $\frac{1+ 3 \lambda}{2+4 \lambda}$ $\Big) x- \Big($ $\frac{3+7 \lambda}{2+4\lambda}$ $\Big)$     … … … … … ii)

Slope of line $x-y+9 = 0$  is $1$

Since the two lines are perpendicular to each other,

$- \Big($ $\frac{1+ 3 \lambda}{2+4 \lambda}$ $\Big) \times 1 = - 1$

$\Rightarrow 1 + 3\lambda = 2+4 \lambda$

$\Rightarrow \lambda = - 1$

Substituting the value of  $\lambda$ in i) we get the equation of the required line.

$( x +2y + 3) + (-1) ( 3 x + 4y + 7) = 0$

$\Rightarrow -2x-2y-4=0$

$\Rightarrow x+y+2=0$

$\\$

Question 3: Find the equation of the line passing through the point of intersection of $2x-7y+11=0$ and $x+ 3y -8=0$ and is parallel to (i) x-axis (ii) y-axis

The equation of a straight line passing through the point of intersection of $2x-7y+11=0$ and $x+ 3y -8=0$ is given by:

$( 2x-7y+11) + \lambda ( x+ 3y -8) = 0$     … … … … … i)

$\Rightarrow ( 2 + \lambda) x - ( 7-3\lambda) y + ( 11 - 8 \lambda) = 0$

$\Rightarrow y = \Big($ $\frac{2+ \lambda}{7-3 \lambda}$ $\Big) x+ \Big($ $\frac{11-8\lambda}{7-3\lambda}$ $\Big)$     … … … … … ii)

i)       When line is parallel to x-axis, then its slope is 0. Therefore

$\frac{2+ \lambda}{7-3 \lambda}$ $= 0$      $\Rightarrow \lambda = -2$

Substituting the value of  $\lambda$ in i) we get the equation of the required line.

$( 2x-7y+11) + (-2) ( x+ 3y -8) = 0$

$\Rightarrow -13 y + 27 = 0$

$\Rightarrow 13 y - 27 = 0$

ii)     When line is parallel to y-axis, then $\frac{-1}{\text{Slope}}$ $= 0$. Therefore

$\frac{3\lambda -7}{2+ \lambda}$ $= 0$      $\Rightarrow \lambda =$ $\frac{7}{3}$

Substituting the value of  $\lambda$ in i) we get the equation of the required line.

$( 2x-7y+11) + \Big($ $\frac{7}{3}$ $\Big) ( x+ 3y -8=0) = 0$

$\Rightarrow 6x-21y+33+7x+21y-56=0$

$\Rightarrow 13x-23= 0$

$\\$

Question 4: Find the equation of the straight line passing through the point of intersection of $2x+ 3y + 1=0$ and $3 x -5y -5 =0$ and equally inclined to the axes.

The equation of a straight line passing through the point of intersection of $2x+ 3y + 1=0$ and $3 x -5y -5 =0$ is given by:

$( 2x+ 3y + 1) + \lambda ( 3 x -5y -5 ) = 0$     … … … … … i)

$\Rightarrow ( 2 + 3\lambda) x + ( 3-5\lambda) y + ( 1-5 \lambda) = 0$

$\Rightarrow y = - \Big($ $\frac{2+ 3\lambda}{3-5 \lambda}$ $\Big) x+ \Big($ $\frac{1-5\lambda}{3-5\lambda}$ $\Big)$     … … … … … ii)

Since the line is equally inclined to both axis, the slope of the line would be either $1 \text{ or } -1$

Case 1: Slope $= 1$

$-$ $\frac{2+ 3\lambda}{3-5 \lambda}$ $= 1$     $\Rightarrow -2-3\lambda = 3 - 5 \lambda$      $\Rightarrow 2\lambda = 5$      $\Rightarrow \lambda =$ $\frac{5}{2}$

Substituting the value of  $\lambda$ in i) we get the equation of the required line.

$( 2x+ 3y + 1) + \Big($ $\frac{5}{2}$ $\Big) ( 3 x -5y -5 ) = 0$

$\Rightarrow 4x+6y+2+15x-25y-25=0$

$\Rightarrow 19x-19y-23= 0$

Case 2: Slope $= -1$

$-$ $\frac{2+ 3\lambda}{3-5 \lambda}$ $= -1$     $\Rightarrow 2+3\lambda = 3 - 5 \lambda$      $\Rightarrow 8\lambda = 1$     $\Rightarrow \lambda =$ $\frac{1}{8}$

Substituting the value of  $\lambda$ in i) we get the equation of the required line.

$( 2x+ 3y + 1) + \Big($ $\frac{1}{8}$ $\Big) ( 3 x -5y -5 ) = 0$

$\Rightarrow 16x+24y+8+3x-5y-5=0$

$\Rightarrow 19x+19y+3= 0$

$\\$

Question 5: Find the equation of the straight line drawn through the point of intersection of the lines $x+y=4$ and $2x-3y=1$ and perpendicular to the line cutting off intercepts $5,6$ on the axes.

The equation of a straight line passing through the point of intersection of $x+y=4$ and $2x-3y=1$ is given by:

$( x + y - 4) + \lambda ( 2x - 3y -1) = 0$     … … … … … i)

$\Rightarrow ( 1 + 2 \lambda) x + ( 1 - 3\lambda) y - ( 4 + \lambda) = 0$

$\Rightarrow y = - \Big($ $\frac{1+ 2 \lambda}{1- 3 \lambda}$ $\Big) x + \Big($ $\frac{4 + \lambda}{1- 3\lambda}$ $\Big)$     … … … … … ii)

The equation of the line with intercepts of $5$ and $6$ on axis is

$\frac{x}{5}$ $+$ $\frac{y}{6}$ $= 1$     … … … … … iii)

Therefore the slope  iii) is $\frac{-6}{5}$

Since lines i) and iii) are perpendicular to each other, therefore

$-\Big($ $\frac{1+ 2 \lambda}{1- 3 \lambda}$ $\Big) \times$ $\frac{-6}{5}$ $= -1 = -1$

$\Rightarrow 6 ( 1 + 2 \lambda) = - 5 ( 1 - 3 \lambda)$

$\Rightarrow 6 + 12 \lambda = - 5 + 15 \lambda$     $\Rightarrow 3 \lambda = 11$      $\Rightarrow \lambda =$ $\frac{11}{3}$

Substituting the value of  $\lambda$ in i) we get the equation of the required line.

$( x + y - 4) +$ $\frac{11}{3}$ $( 2x - 3y -1) = 0$

$\Rightarrow 3x + 3 y - 12 + 22x - 33y - 11 = 0$

$\Rightarrow 25x - 30y-23=0$

$\\$

Question 6: Prove that the family of lines represented by $x (1 + \lambda) + y (2 - \lambda) + 5 = 0$, being arbitrary, pass through a fixed point. Also, find the fixed point.

Given: $x (1 + \lambda) + y (2 - \lambda) + 5 = 0$

$\Rightarrow ( x + 2y + 5) + \lambda ( x - y) = 0$

This line is of the form $L_1 + \lambda L_2 = 0$

Therefore this line passes through the intersection of $x + 2y + 5 = 0$ and $x - y = 0$

Solving the above two equations, we get the point of intersection as                  $\Big($ $\frac{-5}{3}$ $,$ $\frac{-5}{3}$ $\Big)$ which is a fixed point  through which the given family of lines passes for any value of $\lambda$.

$\\$

Question 7: Show that the straight lines given by $(2 + k) x + (1 + k) y = 5 +7k$ for different values of $k$ pass through a fixed point. Also, find that point.

Given: $(2 + k) x + (1 + k) y = 5 +7k$

$\Rightarrow ( 2x+y) + k ( x + y) = 5+7k$

$\Rightarrow ( 2x+y-5) + k ( x + y-7) = 0$

This line is of the form $L_1 + k L_2 = 0$

Therefore this line passes through the intersection of $2x+y-5 = 0$ and $x + y-7 = 0$

Solving the above two equations, we get the point of intersection as                  $(-2, 9)$ which is a fixed point  through which the given family of lines passes for any value of $k$.

$\\$

Question 8: Find the equation of the straight line passing through the point of intersection of $2x + y -1 =0$ and $x + 3 y -2= 0$ and making with the coordinate axes a triangle of area $3/8$ sq. units.

The equation of a straight line passing through the point of intersection of $2x + y -1 =0$ and $x + 3 y -2= 0$ is given by:

$( 2x + y -1) + \lambda ( x + 3 y -2) = 0$     … … … … … i)

$\Rightarrow ( 2 + \lambda) x + ( 1 + 3\lambda) y - ( 1 + 2\lambda) = 0$

$\frac{x}{( \frac{1+2\lambda}{2+\lambda} ) } + \frac{y}{( \frac{1+2\lambda}{1+3\lambda} )} = 1$   … … … … … ii)

Therefore the point of intersection of this line with the axis are                             $\Big($ $\frac{1+2\lambda}{2+\lambda}$ $, 0 \Big)$ and $\Big( 0,$ $\frac{1+2\lambda}{1+3\lambda}$ $\Big)$

It is given that the required line makes an area $\frac{3}{8}$ sq. units with the coordinate axis. Therefore,

$\frac{1}{2}$ $\Big($ $\frac{1+2\lambda}{2+\lambda}$ $\Big) \times \Big($ $\frac{1+2\lambda}{1+3\lambda}$ $\Big) =$ $\frac{3}{8}$

$\Rightarrow 4 ( 1 + 4 \lambda^2 + 4 \lambda) = 3 ( 2 + 6\lambda + \lambda + 3 \lambda^2)$

$\Rightarrow 4 ( 1 + 4 \lambda^2 + 4 \lambda) = ( 6 + 21 \lambda + 9 \lambda^2)$

$\Rightarrow 7 \lambda^2 - 5 \lambda - 2 = 0$

$\Rightarrow ( 7 \lambda + 2) ( \lambda - 1) = 0$

$\lambda = 1$ or $\lambda =$ $\frac{-2}{7}$

Substituting the value of  $\lambda = 1$ in i) we get the equation of the required line.

$( 2x + y -1) + (1) ( x + 3 y -2) = 0$

$\Rightarrow 3x+4y-3 = 0$

Substituting the value of  $\lambda =$ $\frac{-2}{7}$ in i) we get the equation of the required line.

$( 2x + y -1) + (\frac{-2}{7}) ( x + 3 y -2) = 0$

$\Rightarrow 14x+7y-7-2x-6y+4 = 0$

$\Rightarrow 12x+y - 3 = 0$

$\\$

Question 9: Find the equation of the straight line which passes through the point of intersection of the lines $3x - y =5$ and $x + 3y =1$ and makes equal and positive intercepts on the axes.

The equation of a straight line passing through the point of intersection of $3x - y =5$ and $x + 3y =1$ is given by:

$( 3x-y-5 ) + \lambda (x+3y-1 ) = 0$     … … … … … i)

$\Rightarrow ( 3 + \lambda) x + ( -1 + 3\lambda) y - ( 5 + \lambda) = 0$

$\frac{x}{( \frac{5+\lambda}{3+\lambda} ) } + \frac{y}{( \frac{5+\lambda}{-1+3\lambda} )} = 1$   … … … … … ii)

Since this line makes equal intercepts,

$\frac{5+\lambda}{3+\lambda}$ $=$ $\frac{5+\lambda}{-1+3\lambda}$

$\Rightarrow 3 \lambda - 1 = 3 + \lambda$      $\Rightarrow 2 \lambda = 4$     $\Rightarrow \lambda = 2$

Substituting the value of  $\lambda = 2$  in i) we get the equation of the required line.

$( 3x-y-5 ) + (2) (x+3y-1 ) = 0$

$\Rightarrow 5x+5y-7=0$

$\\$

Question 10: Find the equations of the lines through the point of intersection of the lines $x - 3y + 1 = 0$ and $2x +5y -9 =0$ and whose distance from the origin is $\sqrt{5}$.

The equation of a straight line passing through the point of intersection of $x - 3y + 1 = 0$ and $2x +5y -9 =0$ is given by:

$( x - 3y + 1 ) + \lambda (2x +5y -9 ) = 0$     … … … … … i)

$\Rightarrow ( 1 + 2\lambda) x + ( -3 + 5\lambda) y + ( 1 -9 \lambda) = 0$

Give, distance from origin $( 0,0)$ is  $\sqrt{5}$. Therefore

$\Bigg|$ $\frac{( 1 + 2\lambda) (0) + ( -3 + 5\lambda) (0) + ( 1 -9 \lambda)}{\sqrt{ (1 + 2\lambda)^2 + (-3 + 5\lambda)^2}}$ $\Bigg | = \sqrt{5}$

$\Rightarrow \Bigg|$ $\frac{ ( 1 -9 \lambda)}{\sqrt{ 1+ 4\lambda^2 + 4 \lambda + 25 \lambda^2 + 9 - 30 \lambda }}$ $\Bigg | = \sqrt{5}$

$\Rightarrow \Bigg|$ $\frac{ ( 1 -9 \lambda)}{\sqrt{ 29 \lambda^2 - 26 \lambda+10 }}$ $\Bigg | = \sqrt{5}$

Squaring both sides we get

$\Rightarrow 5 ( 29 \lambda^2 - 26 \lambda+10) = ( 1 -9 \lambda)^2$

$\Rightarrow 145 \lambda^2 - 130 \lambda + 50 = 1 + 81 \lambda^2 - 18\lambda$

$\Rightarrow 64 \lambda^2 - 112 \lambda + 49 = 0$

$\Rightarrow ( 8 \lambda-7)^2 = 0$

$\Rightarrow \lambda = \frac{7}{8}$

Substituting the value of  $\lambda = 2$  in i) we get the equation of the required line.

$( x - 3y + 1 ) + \Big($ $\frac{7}{8}$ $\Big) (2x +5y -9 ) = 0$

$\Rightarrow 8x-24y+8+14x+35y-63=0$

$\Rightarrow 22x+11y-55=0$

$\Rightarrow 2x+y-5=0$

$\\$

Question 11: Find the equations of the lines through the point of intersection of the lines $x -y + 1 = 0$ and $2x - 3y +5 = 0$ whose distance from the point $(3, 2)$ is $7 / 5$.

The equation of a straight line passing through the point of intersection of $x -y + 1 = 0$ and $2x - 3y +5 = 0$ is given by:

$( x -y + 1 ) + \lambda (2x - 3y +5) = 0$     … … … … … i)

$\Rightarrow ( 1 + 2\lambda) x + ( -1 -3\lambda) y + ( 1 +5 \lambda) = 0$

Give, distance from origin $( 3,2)$ is  $7/5$. Therefore

$\Bigg|$ $\frac{( 1 + 2\lambda) (3) + ( -1 -3\lambda) (2) + ( 1 +5 \lambda)}{\sqrt{ (1 + 2\lambda)^2 + ( -1 -3\lambda)^2}}$ $\Bigg | =$ $\frac{7}{5}$

$\Rightarrow \Bigg|$ $\frac{ 3+6\lambda-2-6\lambda + 1 + 5 \lambda}{\sqrt{ 1+4 \lambda^2+4\lambda+1+9\lambda^2+6\lambda }}$ $\Bigg | =$ $\frac{7}{5}$

$\Rightarrow \Bigg|$ $\frac{ 2+5\lambda}{\sqrt{ 13 \lambda^2 +10 \lambda+2 }}$ $\Bigg | =$ $\frac{7}{5}$

Squaring both sides we get

$\Rightarrow 25 ( 2 + 5 \lambda) = 49(13 \lambda^2 +10 \lambda+2)$

$\Rightarrow 6\lambda^2-5\lambda-1=0$

$\Rightarrow (\lambda-1)(6\lambda+1)=0$

$\Rightarrow \lambda = 1 \text {or } \lambda=$ $\frac{-1}{6}$

Substituting the value of  $\lambda = 1$  in i) we get the equation of the required line.

$( x -y + 1 ) + (1) (2x - 3y +5) = 0$

$\Rightarrow 3x-4y+6=0$

Substituting the value of  $\lambda =$ $\frac{-1}{6}$  in i) we get the equation of the required line.

$( x -y + 1 ) + \Big($ $\frac{-1}{6}$ $\Big) (2x - 3y +5) = 0$

$\Rightarrow 6x-6y+6-2x+3y-5=0$

$\Rightarrow 4x-3y+1=0$