Note: The equation of a line having slope of $m$ and $\text{y-intercept}$ as $c$ is given by $y= mx + c$

Question 1: Find the equation of a line making an angle of  $150^{\circ}$ with the x-axis and cutting off an intercept $2$ from y-axis

Given: $m = \tan 150^{\circ} = - \tan 30^{\circ} = -$ $\frac{1}{\sqrt{3}}$

$c = \text{y-intercept} = 2$

Substituting in $y = mx+c$ we get $y = -$ $\frac{1}{\sqrt{3}}$ $x + 2$

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

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

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Question 2: Find the equation of a straight line:

(i) with slope $2$ and y-intercept $3$

(ii) with slope $-1/3$ and y-intercept $- 4$

(iii) with slope $-2$ and intersecting the x-axis at a distance of $3$ units to the left of origin.

i)        Given $m = 2 , \hspace{0.5cm} c=3$

Substituting in $y = mx+c$ we get  $y = 2x + 3$

Hence the equation of a straight line with slope $2$ and y-intercept $3$ is $y = 2x + 3$

ii)     Given $m = -$ $\frac{1}{3}$ $, \hspace{0.5cm} c=-4$

Substituting in $y = mx+c$ we get  $y = -$ $\frac{1}{3}$ $x -4$

Hence the equation of a straight line with slope $-$ $\frac{1}{3}$ and y-intercept $- 4$ is $y = -$ $\frac{1}{3}$ $x -4$

iii)     Given $m = -2 ,$ and passes through $(-3,0)$

Substituting in $y = mx+c$ we get  $0 = -2(-3) + c \Rightarrow c = -6$

Substituting in $y = mx+c$ we get  the equation of the line is $y = -2x - 6$

Hence the equation of a straight line with slope $-2$ and passes through $(-3,0)$ is $y = -2x - 6$

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Question 3: Find the equations of the bisectors of the angles, between the coordinate axes.

For line 1, $m_1 = \tan 45^{\circ} = 1, c = \text{ y-intercept } = 0$

Substituting in $y = mx+c$ we get  the equation of line 1 is $y = x$

For line 2, $m_2 = \tan 135^{\circ} = -1, c = \text{ y-intercept } = 0$

Substituting in $y = mx+c$ we get  the equation of line 2 is $y = -x$

Hence the equations of the bisectors of the angles, between the coordinate axes $x \pm y = 0$

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Question 4: Find the equation of a line which makes an angle of  $\tan^{-1} (3)$ with the x-axis and cuts off an intercept of $4$ units on negative direction of y-axis.

Here $m = \tan^{-1} (3) = 3, \hspace{0.5cm} c = \text{y-intercept } = -4$

Substituting in $y = mx+c$ we get the equation as $y = 3x - 4$

Hence the equation of a line which makes an angle of  $\tan^{-1} (3)$ with the x-axis and cuts off an intercept of $4$ units on negative direction of y-axis is $y = 3x - 4$

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Question 5: Find the equation of a line that has y-intercept $-4$ and is parallel to the line joining $(2, -5)$ and $(1,2)$.

Here slope $(m) =$ $\frac{2-(-5)}{1-2}$ $=$ $\frac{7}{-1}$ $= -7$

$c = \text{y-intercept } = -4$

Substituting in $y = mx+c$ we get the equation as $y = -7x - 4 \Rightarrow 7x +y+4=0$

Hence equation of a line that has y-intercept $-4$ and is parallel to the line joining $(2, -5)$ and $(1,2)$ is $7x +y+4=0$

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Question 6: Find the equation of a line which is perpendicular to the line joining $(4,2)$ and $(3, 5)$ and cuts off an intercept of length $3$ on y-axis.

Slope of line joining $(4,2)$ and $(3, 5) =$ $\frac{5-2}{3-4}$ $=$ $\frac{3}{-1}$ $= -3$

The slope of the required line, which is $\perp$ to the above line $=$ $\frac{-1}{-3}$ $=$ $\frac{1}{3}$

$c = \text{y-intercept } = 3$

Substituting in $y = mx+c$ we get the equation as $y =$ $\frac{1}{3}$ $x + 3 \Rightarrow x-3y+ 9 = 0$

Hence the equation of the required equation is $x-3y+ 9 = 0$

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Question 7: Find the equation of the perpendicular to the line segment joining $(4, 3)$ and $(-1, 1)$ if it cuts off an intercept $-3$ from y-axis.

Slope of line joining $(4,3)$ and $(-1, 1) =$ $\frac{1-3}{-1-4}$ $=$ $\frac{-2}{-5}$ $=$ $\frac{2}{5}$

The slope of the required line, which is $\perp$ to the above line $=$ $\frac{-1}{2/5}$ $=$ $\frac{-5}{2}$

$c = \text{y-intercept } = -3$

Substituting in $y = mx+c$ we get the equation as

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

Hence the equation of the required equation is $5x+2y+6 = 0$

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Question 8: Find the equation of the straight line intersecting y-axis at a distance of $2$ units above the origin and making an angle of $30^{\circ}$ with the positive direction of the x-axis

Here $m = \tan 30^{\circ} =$ $\frac{1}{\sqrt{3}}$
$c = \text{y-intercept } = -3$
Substituting in $y = mx+c$ we get the equation as
$y =$ $\frac{1}{\sqrt{3}}$ $x + 2 \Rightarrow x - \sqrt{3}y + 2 \sqrt{3}= 0$
Hence the equation of the required equation is $x - \sqrt{3}y + 2 \sqrt{3}= 0$