Class 11: The Straight Line – Exercise 23.4 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Note: When slope $(m)$ and a point $(x_1, y_1)$ is given, the equation of the line is $y - y_1 = m ( x-x_1)$

Question 1: Find the equation of the straight line passing through the point $(6,2)$ and having slope $-3$ .

Answer:

Here $m = -3$ and $(x_1, y_1) = ( 6, 2)$

Substituting in $y - y_1 = m ( x-x_1)$ we get

$y - 2 = - 3( x-6)$

$\Rightarrow y - 2 = - 3x + 18$

$\Rightarrow 3x + y - 20 = 0$

Hence the equation of the straight line is $3x + y - 20 = 0$

$\\$

Question 2: Find the equation of the straight line passing through $(-2, 3)$ and inclined at an angle of $45^{\circ}$ with the x-axis.

Answer:

Here $m = \tan 45^{\circ} = 1$ and $(x_1, y_1) = ( -2, 3)$

Substituting in $y - y_1 = m ( x-x_1)$ we get

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

$\Rightarrow y-3 = x + 2$

$\Rightarrow x-y+5=0$

Hence the equation of the straight line is $x-y+5=0$

$\\$

Question 3: Find the equation of the line passing through $(0, 0)$ with slope $m$ .

Answer:

Here $m = m$ and $(x_1, y_1) = ( 0,0)$

Substituting in $y - y_1 = m ( x-x_1)$ we get

$y - 0 = m( x-0)$

$\Rightarrow y = mx$

Hence the equation of the straight line is $y=mx$

$\\$

Question 4: Find the equation of the line passing through $(2, 2\sqrt{3})$ and inclined with x-axis at an angle of $75^{\circ}$ .

Answer:

Here $\displaystyle \ m = \tan 75^{\circ} = \tan ( 45^{\circ} + 30^{\circ})$

$\displaystyle = \frac{\tan 45^{\circ} + \tan 30^{\circ}}{1 - \tan 45^{\circ} \tan 30^{\circ}}$

$\displaystyle = \frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}} = \frac{\sqrt{3}+1}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} = \frac{3+1 + 2\sqrt{3}}{3-1} = \frac{4+2\sqrt{3}}{2} = 2 + \sqrt{3}$

Also $(x_1, y_1) = ( 2, 2\sqrt{3})$

Substituting in $y - y_1 = m ( x-x_1)$ we get

$y - 2\sqrt{3} = (2+\sqrt{3})( x-2)$

$\Rightarrow y-2\sqrt{3} = (2+\sqrt{3})x - 4 - 2\sqrt{3}$

$\Rightarrow (2+\sqrt{3})x-y-4=0$

Hence the equation of the straight line is $(2+\sqrt{3})x-y-4=0$

$\\$

Question 5: Find the equation of the straight line which passes through the point $(1,2)$ and makes such an angle with the positive direction of x-axis whose sine is $\frac{3}{5}$

Answer:

$\displaystyle \text{Given } \sin \theta = \frac{3}{5}$

We know,

$\displaystyle \tan \theta= \frac{\sin \theta}{1 - \sin^2 \theta } = \frac{3/5}{\sqrt{1 - 9/25}} = \frac{3}{\sqrt{16}} = \frac{3}{4}$

$\displaystyle \text{Therefore } m = \frac{3}{4} \text{ and } (x_1, y_1) = ( 1, 2)$

Substituting in $y - y_1 = m ( x-x_1)$ we get

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

$\Rightarrow 3x-4y+5=0$

Hence the equation of the straight line is $3x-4y+5=0$

$\\$

Question 6: Find the equation of the straight line passing through $(3, -2)$ and making an angle of $60^{\circ}$ with the positive direction of y-axis

Answer:

$\displaystyle \text{ Slope of line } = m = \tan 30^{\circ} = \frac{1}{\sqrt{3}} \text{ and } (x_1, y_1) = ( 3, -2)$

Substituting in $y - y_1 = m ( x-x_1)$ we get

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

$\Rightarrow \sqrt{3} y + 2\sqrt{3}= x-3$

$\Rightarrow x-\sqrt{3} y - 3 - 2\sqrt{3} = 0$

Hence the equation of the straight line is $x-\sqrt{3} y - 3 - 2\sqrt{3}=0$

$\\$

Question 7: Find the lines through the point $(0,2)$ making angle $\frac{\pi}{3}$ and $\frac{2\pi}{3}$ with the x-axis. Also, find the lines parallel to them cutting the y-axis at a distance of $2$ units below the origin.

Answer:

$\displaystyle \text{Given } m_1 = \tan \frac{\pi}{3} = \sqrt{3}$

$\displaystyle m_2 = \tan \frac{2\pi}{3} = - \tan \frac{\pi}{3} = -\sqrt{3}$

Also $(x_1, y_2) = ( 0, 2)$

Therefore the equation passing through $( 0,2)$ and having $m_1$ and $m_2$ slope respectively are

$y - 2 = \sqrt{3}(x-0) \Rightarrow y - 2 = \sqrt{3}x \Rightarrow \sqrt{3}x - y + 2 = 0$

$y - 2 = -\sqrt{3}(x-0) \Rightarrow y - 2 = -\sqrt{3}x \Rightarrow \sqrt{3}x + y - 2 = 0$

Now let us find lines parallel to these lines but passing through $( 0, -2)$

Therefore the equation passing through $( 0,-2)$ and having $m_1$ and $m_2$ slope respectively are

$y + 2 = \sqrt{3}(x-0) \Rightarrow y + 2 = \sqrt{3}x \Rightarrow \sqrt{3}x - y - 2 = 0$

$y + 2 = -\sqrt{3}(x-0) \Rightarrow y + 2 = -\sqrt{3}x \Rightarrow \sqrt{3}x + y + 2 = 0$

$\\$

Question 8: Find the equations of the straight lines which cut off an intercept $5$ from the y-axis and are equally inclined to the axes.

Answer:

Given the lines are equally inclined to the axes.

Hence, their inclination with positive x-axis are $45^{\circ}$ and $135^{\circ}$

Therefore for Line 1: $m_1 = \tan 45^{\circ} = 1$

and for Line 2: $m_2 = \tan 135^{\circ} = - \tan 45^{\circ} = -1$

Hence the equation of Line 1: $y - 5 = 1 ( x - 0) \Rightarrow x - y + 5 = 0$

And the equation for Line 2: $y - 5 = -1 ( x - 0) \Rightarrow x + y -5 = 0$

$\\$

Question 9: Find the equation of the line which intercepts a length $2$ on the positive direction of the x-axis and is inclined at an angle of $135^{\circ}$ with the positive direction of y-axis.

Answer:

Here $m = \tan 45^{\circ} = 1$ and $(x_1, y_1) = ( 2, 0)$

Therefore equation of the line is: $y-0 = 1 ( x-2) \Rightarrow x-y - 2 = 0$

$\\$

Question 10: Find the equation of the straight line which divides the join of the points $(2, 3)$ and $(- 5, 8)$ in the ratio $3 : 4$ and is also perpendicular to it.

Answer:

Given points $A(2, 3)$ and $B(- 5, 8)$

$\displaystyle \text{ Therefore slope of } AB = \frac{8-3}{-5-2} = \frac{-5}{7}$

$\displaystyle \text{ Therefore Slope of line perpendicular to } AB = \frac{-1}{(-5/7)} = \frac{7}{5}$

The coordinate of the point that divides AB in the ratio of $3:4$ is

$\displaystyle (x_1, y_1) = \Big( \frac{3(-5) + 4(2)}{3+4} , \frac{3(8)+4(3)}{3+4} \Big) = \Big( \frac{-7}{7} , \frac{36}{7} \Big) =\Big( -1, \frac{36}{9} \Big)$

Therefore the equation of the required line:

$\displaystyle y - \frac{36}{7} = \frac{7}{5} ( x+1)$

$\Rightarrow 7y - 36 =$ $\frac{7}{5}$ $( 7x + 7)$

$\Rightarrow 35y - 180 = 49x + 49$

$\Rightarrow 49x - 35y + 229 = 0$

$\\$

Question 11: Prove that the perpendicular drawn from the point $(4, 1)$ on the join of $(2, -1)$ and $(6,5)$ divides it in the ratio $5 : 8$ .

Answer:

Given points $A(2, -1)$ and $B(6,5 )$

$\displaystyle \text{ Slope of } AB = \frac{5-(-1)}{6-2} = \frac{6}{4} = \frac{3}{2}$

$\displaystyle PQ \text{ is } \perp \text{ to } AB . \text{ Therefore slope of } PQ = \frac{-1}{3/2} = \frac{-2}{3}$

Therefore equation of $PQ$ is:

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

$\Rightarrow 3y - 3 = - 2x + 8$

$\Rightarrow 2x + 3y = 11$ … … … … … i)

Equation of $AB$ :

$\displaystyle y - 5 = \frac{3}{7} (x-6)$

$\Rightarrow 2y - 10 = 3x - 18$

$\Rightarrow 3x - 2y = 8$ … … … … … ii)

Point of intersection will be found by solving the two equations.

Multiply i) by $3$ and ii) by $2$ and then subtract ii) from i) we get

${ \hspace{1.0cm} 3 \times (2x + 3y = 11)} \\ \underline {(-1) \ 2 \times ( 3x - 2y = 8)} \\ {\hspace{2.5cm}13y = 17}$

$\displaystyle \Rightarrow y = \frac{17}{13}$

Substituting in i) we get

$\displaystyle 2x + 3 \Big( \frac{17}{13} \Big ) = 11$

$\displaystyle \Rightarrow x = \frac{46}{13}$

Hence the point of intersection is

$\displaystyle \Big($ $\frac{46}{13}$ $,$ $\frac{17}{13}$ $\Big)$ $= \Big($ $\frac{5(6) + 8 ( 2) }{5+8}$ $,$ $\frac{5(5) + 8 ( -1) }{5+8}$ $\Big)$

$\displaystyle \text{Therefore we see that } Q \Big( \frac{46}{13} , \frac{17}{13} \Big) \text{ divides AB in the ratio of } 5:8$

$\\$

Question 12: Find the equations to the altitudes of the triangle whose angular points ate $A(2,-2), B (1, 1)$ and $C (- 1, 0)$ .

Answer:

Given $A(2,-2), B (1, 1)$ and $C (- 1, 0)$ .

Let $AD$ , $BE$ and $CF$ are the altitudes as shown in the figure.

$\displaystyle \text{ Slope of } BC = \frac{0-1}{-1-1} = \frac{1}{2}$

$\displaystyle \text{ Therefore slope of } AD = \frac{-1}{(1/2)} = -2$

Therefore equation of $AD$ :

$y - ( -2) = -2 ( x-2)$

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

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

$\displaystyle \text{ Slope of } AC = \frac{0-(-2)}{-1-2} = \frac{2}{-3} = \frac{-2}{3}$

$\displaystyle \text{ Therefore slope of } BE = \frac{-1}{(-2/3)} = \frac{3}{2}$

Therefore equation of $BE$ :

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

$\Rightarrow 2y - 2 = 3x - 3$

$\Rightarrow 3x-2y-1=0$

$\displaystyle \text{ Slope of }AB = \frac{1-(-2)}{1-2} = \frac{3}{-1} = -3$

$\displaystyle \text{ Therefore slope of } CF = \frac{-1}{-3} = \frac{1}{3}$

Therefore equation of $CF$ :

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

$\Rightarrow 3y = x + 1$

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

$\\$

Question 13: Find the equation of the right bisector of the line segment joining the points $(3, 4)$ and $(-1,2)$ .

Answer:

Given points $A(3, 4)$ and $B ( -1, 2)$

$\displaystyle \text{Therefore the midpoint of } AB = ( \frac{-1+3}{2} , \frac{2+4}{2} ) = ( 1, 3)$

$\displaystyle \text{Slope of } AB = \frac{2-4}{-1-3} = \frac{-2}{-4} \frac{1}{2}$

$\displaystyle \text{Therefore the slope of the bisector } - \frac{-1}{1/2} = -2$

Hence the equation of the bisector:

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

$\Rightarrow y - 3 = -2x + 2$

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

Hence the equation of the right bisector is $2x + y - 5 = 0$

$\\$

Question 14: Find the equation of the line passing through the point $(- 3, 5)$ and perpendicular to the line joining $(2,5)$ and $(-3, 6)$ .

Answer:

Given points $A( 2, 5) \text{ and } B ( -3, 6)$

$\displaystyle \text{ Slope of } AB = \frac{6-5}{-3-2} = \frac{1}{-5} = \frac{=1}{5}$

$\displaystyle \text{Therefore the slope of the perpendicular line } = \frac{-1}{(-1/5)} = 5$

Hence the equation of the perpendicular which is passing through $( -3, 5)$ :

$y - 5 = 5 ( x+3)$

$\Rightarrow 5x - y + 20 = 0$

Hence the equation of the line is $5x - y + 20 = 0$

$\\$

Question 15: Find the equation of the right bisector of the line segment joining the points $A(1, 0)$ and $B (2, 3)$ .

Answer:

Given points $A(1,0) \text{ and } B( 2, 3)$

$\displaystyle \text{Mid point of } AB = \Big( \frac{2+1}{2} , \frac{3+0}{2} \Big) = \Big( \frac{3}{2} , \frac{3}{2} \Big)$

$\displaystyle \text{Slope of } AB = \frac{3-0}{2-1} = \frac{3}{1} = 3$

Therefore slope of bisector $\displaystyle = \frac{-1}{3}$

Therefore the equation of the right bisector:

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

$\displaystyle \Rightarrow 6y - 9 = - 2 \Big( x - \frac{3}{2} \Big)$

$\Rightarrow 6y - 9 = - 2x + 3$

$\Rightarrow 2x + 6y - 12 = 0$

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

Hence the equation of the right bisector is $x + 3y - 6 = 0$

ICSE / ISC / CBSE Mathematics Portal for K12 Students

Mathematics Made Easy for School Students

Class 11: The Straight Line – Exercise 23.4

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