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

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

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

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:

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

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

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

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

$y - 2 =$ $\frac{3}{4}$  $( x-1)$

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

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:

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

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

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

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

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

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:

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

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

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 $B(- 5, 8)$

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

$\therefore$ Slope of line $\perp$ to $AB =$ $\frac{-1}{(-5/7)}$ $=$ $\frac{7}{5}$

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

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

Therefore the equation of the required line:

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

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

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

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

$\\$

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 $B(6,5 )$

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

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

Therefore equation of $PQ$ is:

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

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

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

Equation of $AB$:

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

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

$\Rightarrow 3x - 2y = 8$  … … … … … 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

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

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

Substituting in i) we get

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

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

Hence the point of intersection is $\Big($ $\frac{46}{13}$ $,$ $\frac{17}{13}$ $\Big)$ $= \Big($ $\frac{5(6) + 8 ( 2) }{5+8}$ $,$ $\frac{5(5) + 8 ( -1) }{5+8}$ $\Big)$

Therefore we see that  $Q \Big($ $\frac{46}{13}$ $,$ $\frac{17}{13}$ $\Big)$ 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.

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

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

Therefore equation of $AD$:

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

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

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

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

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

Therefore equation of $BE$:

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

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

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

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

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

Therefore equation of $CF$:

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

$\Rightarrow 3y = x + 1$

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

$\\$

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 $B ( -1, 2)$

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

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

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 $A( 2, 5)$ and $B ( -3, 6)$

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

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 $A(1,0)$ and $B( 2, 3)$

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

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

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

Therefore the equation of the right bisector:

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

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

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

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

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

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