Class 11: The Straight Line – Exercise 23.9 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Reduce the equation $\displaystyle \sqrt{3}x +y +2 = 0$ 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 $\displaystyle p \text{ and } \alpha$ .

Answer:

i) $\displaystyle \text{Given } \sqrt{3}x +y +2 = 0$

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

$\displaystyle \text{Therefore slope } = -\sqrt{3} \text{ and y-intercept } = -2$

ii) $\displaystyle \text{Given } \sqrt{3}x +y +2 = 0$

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

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

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

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

iii) $\displaystyle \text{Given } \sqrt{3}x +y +2 = 0$

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

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

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

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

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

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

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

iv) $\displaystyle x-3=0$ v) $\displaystyle y-2=0$

Answer:

i) $\displaystyle \text{Given } x+\sqrt{3}y - 4 = 0$

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

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

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

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

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

ii) $\displaystyle \text{Given } x+y + \sqrt{2} = 0$

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

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

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

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

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

iii) $\displaystyle \text{Given } x-y + 2\sqrt{2} = 0$

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

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

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

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

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

iv) $\displaystyle \text{Given } x-3=0$

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

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

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

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

v) $\displaystyle \text{Given } y-2=0$

$\displaystyle \Rightarrow 0x + 1y = 2$

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

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

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

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

Answer:

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

$\displaystyle \Rightarrow bx + ay = ab$

$\displaystyle \Rightarrow ay = - bx + ab$

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

Comparing with $\displaystyle y = mx + c \text{ where slope } = m \text{ and y-intercept } = c$

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

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

Answer:

$\displaystyle \text{ Line 1: } \text{Given } 3x-4y+4=0$

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

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

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

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

$\displaystyle \text{ Line 2: } \text{Given } 2x+4y-5=0$

$\displaystyle \Rightarrow 2x+4y=5$

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

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

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

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

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

Answer:

$\displaystyle \text{ Line 1: } \text{Given } 4x + 3y + 10 = 0$

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

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

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

$\displaystyle \text{Therefore } p_1 = 2$

$\displaystyle \text{ Line 2: } \text{Given } 5x -12y + 26 =0$

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

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

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

$\displaystyle \text{Therefore } p_2 = 2$

$\displaystyle \text{ Line 3: } \text{Given } 7x+24y=50$

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

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

$\displaystyle \text{Therefore } p_3 = 2$

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

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

Answer:

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

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

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

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

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

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

Answer:

$\displaystyle \text{Given equation } 3 x -2y + 6 =0$

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

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

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

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

$\displaystyle \text{Hence, x-intercept } \Rightarrow = -2 \text{ and y-intercept } \Rightarrow = 3$

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