Class 11: The Straight Line – Exercise 23.19 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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

Answer:

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

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

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

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

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

Since lines are parallel to each other,

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

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

$\displaystyle \Rightarrow 30 \lambda = - 32$

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

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

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

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

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

$\displaystyle \\$

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

Answer:

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

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

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

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

$\displaystyle \text{Slope of line } x-y+9 = 0 \text{ is } 1$

Since the two lines are perpendicular to each other,

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

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

$\displaystyle \Rightarrow \lambda = - 1$

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

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

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

$\displaystyle \Rightarrow x+y+2=0$

$\displaystyle \\$

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

Answer:

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

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

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

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

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

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

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

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

$\displaystyle \Rightarrow -13 y + 27 = 0$

$\displaystyle \Rightarrow 13 y - 27 = 0$

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

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

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

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

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

$\displaystyle \Rightarrow 13x-23= 0$

$\displaystyle \\$

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

Answer:

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

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

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

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

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

Case 1: Slope $\displaystyle = 1$

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

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

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

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

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

Case 2: Slope $\displaystyle = -1$

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

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

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

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

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

$\displaystyle \\$

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

Answer:

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

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

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

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

The equation of the line with intercepts of $\displaystyle 5 \text{ and } 6$ on axis is

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

Answer:

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

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

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

Therefore this line passes through the intersection of $\displaystyle x + 2y + 5 = 0 \text{ and } x - y = 0$

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

$\displaystyle \\$

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

Answer:

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

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

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

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

Therefore this line passes through the intersection of $\displaystyle 2x+y-5 = 0 \text{ and } x + y-7 = 0$

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

$\displaystyle \\$

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

Answer:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

Answer:

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

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

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

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

Since this line makes equal intercepts,

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

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

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

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

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

$\displaystyle \\$

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

Answer:

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

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

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

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

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

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

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

Squaring both sides we get

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

Answer:

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

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

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

Give, distance from origin $\displaystyle ( 3,2) \text{ is } 7/5 .$ Therefore

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

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