Question 1: Reduce the equation $\sqrt{3}x +y +2 = 0$  to:

(i) slope-intercept form and find slope and y-intercept     (ii) intercept form and find intercept on the axes;     (iii) the normal form and find $p$ and $\alpha$.

i)       Given $\sqrt{3}x +y +2 = 0$

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

Therefore Slope $= -\sqrt{3}$ and y-intercept $= -2$

ii)      Given $\sqrt{3}x +y +2 = 0$

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

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

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

Therefore x-intercept $(a) =$ $\frac{-2}{\sqrt{3}}$ and y-intercept $(b) = -2$

iii)     Given $\sqrt{3}x +y +2 = 0$

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

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

$\Rightarrow$ $\frac{-\sqrt{3}x}{2}$ $-$ $\frac{y}{2}$ $= 1$

Therefore $p = 1, \hspace{0.5cm} \cos \alpha =$ $\frac{-\sqrt{3}}{2}$ $, \hspace{0.5cm} \sin \alpha =$ $\frac{-1}{2}$   and   $\alpha = 210^{\circ}$

Since the coefficient of both $x$ and $y$ are negative, $\alpha$ lies in the $3^{rd}$ Quadrant.

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Question 2: Reduce the following equations to the normal form and find $p$ and $\alpha$ in each case:

i) $x+\sqrt{3}y - 4 = 0$     ii) $x+y + \sqrt{2} = 0$     iii) $x-y + 2\sqrt{2} = 0$

iv) $x-3=0$     v) $y-2=0$

i)       Given $x+\sqrt{3}y - 4 = 0$

$\Rightarrow x+\sqrt{3}y = 4$

$\Rightarrow \frac{x}{\sqrt{(1)^2 + ( \sqrt{3} )^2}} + \frac{\sqrt{3} y}{\sqrt{(1)^2 + ( \sqrt{3} )^2}} = \frac{4}{\sqrt{(1)^2 + ( \sqrt{3} )^2}}$

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

Therefore $p = 2, \hspace{0.5cm} \cos \alpha =$ $\frac{1}{2}$ $, \hspace{0.5cm} \sin \alpha =$ $\frac{\sqrt{3}}{2}$   and   $\alpha = 60^{\circ}$

Since the coefficient of both $x$ and $y$ are positive, $\alpha$ lies in the $1^{st}$ Quadrant.

ii)     Given $x+y + \sqrt{2} = 0$

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

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

$\Rightarrow -$ $\frac{x}{\sqrt{2} }$ $-$ $\frac{y}{\sqrt{2}}$ $= 1$

Therefore $p = 1, \hspace{0.5cm} \cos \alpha =$ $\frac{-1}{\sqrt{2}}$ $, \hspace{0.5cm} \sin \alpha =$ $\frac{-1}{\sqrt{2}}$   and   $\alpha = 225^{\circ}$

Since the coefficient of both $x$ and $y$ are negative, $\alpha$ lies in the $3^{rd}$ Quadrant.

iii)    Given $x-y + 2\sqrt{2} = 0$

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

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

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

Therefore $p = 2, \hspace{0.5cm} \cos \alpha =$ $\frac{-1}{2}$ $, \hspace{0.5cm} \sin \alpha =$ $\frac{1}{\sqrt{2}}$   and   $\alpha = 135^{\circ}$

Since the coefficient of both $x$ is negative and $y$ y is positive, $\alpha$ lies in the $2^{nd}$ Quadrant.

iv)    Given $x-3=0$

$\Rightarrow x + 0y = 3$

$\Rightarrow \frac{x}{\sqrt{(1)^2 + ( 0 )^2}} + \frac{ 0 \cdot y}{\sqrt{(1)^2 + ( 0 )^2}} = \frac{3}{\sqrt{(1)^2 + ( 0 )^2}}$

$\Rightarrow$ $\frac{x}{1}$ $+$ $\frac{0 \cdot y}{1}$ $= 3$

Therefore $p = 3, \hspace{0.5cm} \cos \alpha = 1$  $, \hspace{0.5cm} \sin \alpha = 0$    and   $\alpha = 0^{\circ}$

v)      Given $y-2=0$

$\Rightarrow 0x + 1y = 2$

$\Rightarrow \frac{0 \cdot x}{\sqrt{(0)^2 + ( 1 )^2}} + \frac{ y}{\sqrt{(0)^2 + ( 1 )^2}} = \frac{2}{\sqrt{(0)^2 + ( 1 )^2}}$

$\Rightarrow$ $\frac{0 \cdot x}{1}$ $+$ $\frac{y}{1}$ $= 2$

Therefore $p = 2, \hspace{0.5cm} \cos \alpha = 0$  $, \hspace{0.5cm} \sin \alpha = 1$    and   $\alpha = 90^{\circ}$

$\\$

Question 3: Put the equation  $\frac{x}{a}$ $+$ $\frac{y}{b}$ $=1$ to the slope intercept form and find its slope and y-intercept.

Given $\frac{x}{a}$ $+$ $\frac{y}{b}$ $=1$

$\Rightarrow bx + ay = ab$

$\Rightarrow ay = - bx + ab$

$\Rightarrow y =$ $\frac{-b}{a}$ $x + b$

Comparing with $y = mx + c$ where slope $= m$ and y-intercept $= c$

Therefore $\text{Slope} =$ $\frac{-b}{a}$ and y-intercept $= b$

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Question 4: Reduce the lines $3 x - 4 y + 4 =0$ and $2 x + 4 y -5 = 0$ to the normal form and hence find which line is nearer to the origin.

Line 1: Given $3x-4y+4=0$

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

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

$\Rightarrow -$ $\frac{3x}{5 }$ $+$ $\frac{4y}{5}$ $=$ $\frac{4}{5}$

Therefore $p_1 =$ $\frac{4}{5}$

Line 2: Given $2x+4y-5=0$

$\Rightarrow 2x+4y=5$

$\Rightarrow \frac{2x}{\sqrt{(2)^2 + ( 4 )^2}} + \frac{ 4y}{\sqrt{(2)^2 + ( 4 )^2}} = \frac{5}{\sqrt{(-3)^2 + ( 4 )^2}}$

$\Rightarrow$ $\frac{2x}{2\sqrt{5} }$ $+$ $\frac{2y}{2\sqrt{5}}$ $=$ $\frac{\sqrt{5}}{2}$

Therefore $p_2 =$ $\frac{\sqrt{5}}{2}$

Since $\frac{\sqrt{5}}{2}$ $>$ $\frac{4}{5}$, we can say that $3x-4y+4 = 0$ is nearer to the origin as compared to $2x+ 4y -5=0$

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Question 5: Show that the origin is equidistant from the lines $4x + 3y + 10 = 0$; $5x -12y + 26 =0$ and $7x+24y=50$.

Line 1: Given $4x + 3y + 10 = 0$

$\Rightarrow -4x-3y=10$

$\Rightarrow \frac{-4x}{\sqrt{(-4)^2 + ( -3 )^2}} - \frac{ 3y}{\sqrt{(-4)^2 + ( -3 )^2}} = \frac{10}{\sqrt{(-4)^2 + ( -3 )^2}}$

$\Rightarrow -$ $\frac{4x}{5 }$ $-$ $\frac{3y}{5}$ $= 2$

Therefore $p_1 = 2$

Line 2: Given $5x -12y + 26 =0$

$\Rightarrow -5x+12y=26$

$\Rightarrow \frac{-5x}{\sqrt{(-5)^2 + ( 12 )^2}} + \frac{ 12y}{\sqrt{(-5)^2 + ( 12 )^2}} = \frac{26}{\sqrt{(-5)^2 + ( 12 )^2}}$

$\Rightarrow -$ $\frac{5x}{13 }$ $+$ $\frac{12y}{13}$ $= 2$

Therefore $p_2 = 2$

Line 3: Given $7x+24y=50$

$\Rightarrow \frac{7x}{\sqrt{(7)^2 + ( 24 )^2}} + \frac{ 24y}{\sqrt{(7)^2 + ( 24 )^2}} = \frac{50}{\sqrt{(7)^2 + ( 24 )^2}}$

$\Rightarrow$ $\frac{7x}{25 }$ $+$ $\frac{24y}{25}$ $= 2$

Therefore $p_3 = 2$

Since $p_1=p_2=p_3$, the given lines $4x + 3y + 10 = 0$; $5x -12y + 26 =0$ and $7x+24y=50$ are equidistance from the origin.

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Question 6: Find the values of $0$ and $p$, if the equation $x \cos \theta + y \sin \theta =p$ is the normal form of the line $\sqrt{3}x+y+2=0$.

Given $\sqrt{3}x +y +2 = 0$

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

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

$\Rightarrow$ $\frac{-\sqrt{3}x}{2}$ $-$ $\frac{y}{2}$ $= 1$

Therefore $p = 1, \hspace{0.5cm} \cos \theta =$ $\frac{-\sqrt{3}}{2}$ $, \hspace{0.5cm} \sin \theta =$ $\frac{-1}{2}$   and   $\theta = 210^{\circ}$

$\\$

Question 7: Reduce the equation $3 x -2y + 5 =0$ to the intercept form and find the $x$ and $y$ intercepts.

Given equation $3 x -2y + 6 =0$

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

$\Rightarrow$ $\frac{3x}{-6}$ $-$ $\frac{2y}{-6}$ $= 1$

$\Rightarrow -$ $\frac{1}{2}$ $x+$ $\frac{1}{3}$ $y =1$

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

Hence, x-intercept $\Rightarrow = -2$ and y-intercept $\Rightarrow = 3$

$\\$

Question 8: The perpendicular distance of a line from the origin is $5$ units and its slope is $- 1$. Find the equation of the line.

Given $p = 5$

Therefore equation of line:

$x \cos \alpha + y \sin \alpha = 5$

$\Rightarrow y \sin \alpha = - x \cos \alpha + 5$

$\Rightarrow y = - x \cot \alpha +$ $\frac{5}{\sin \alpha}$

Comparing with $y = mx + c$ where slope $= m$ and y-intercept $= c$

Slope $= - \cot \alpha = - 1$

$\Rightarrow \cot \alpha = 1 \Rightarrow \alpha = 45^{\circ}$

Therefore the equation of line is

$x \cos 45^{\circ} + \sin 45^{\circ} = 5$

$\Rightarrow$ $\frac{x}{\sqrt{2}}$ $+$ $\frac{y}{\sqrt{2}}$ $= 5$

$\Rightarrow x + y = 5\sqrt{2}$