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The equations of the lines which pass through the point (3, –2) and are inclined at 60° to the line $\sqrt{3}x+y=1$ is
A. y + 2 = 0, $\sqrt{3}x-y-2-3\sqrt{3}=0$
B. x – 2 = 0, $\sqrt{3}x-y+2+3\sqrt{3}=0$
C. $\sqrt{3}x-y-2-3\sqrt{3}=0$
D. None of these

Answers (1)

Given equation is $\sqrt{3}x+y=1$ and θ= 60⁰

Slope of the equation $\sqrt{3}x+y=1$

$y=1-\sqrt{3}x$

Slope of the equation $m_{1}=-\sqrt{3}$

Let slope of the required line be m₂

Then $\tan \theta = \left | \frac{\left ( m_{1}-m_{2} \right )}{1+m_{1}m_{2}} \right |$

Putting the values in the above equation we gettan $60^{\circ}=\left | \frac{-\sqrt{3}-m_{2}}{1+\left ( -\sqrt{3} \right )*m_{2}} \right |$

$\sqrt{3}=\left | \frac{-\sqrt{3}-m_{2}}{1+\left ( -\sqrt{3} \right )*m_{2}} \right |$

$\sqrt{3}= \pm\left ( \frac{-\sqrt{3}-m_{2}}{1+\left ( -\sqrt{3} \right )*m_{2}} \right )$

Taking+sign we get $-\sqrt{3}-m_{2}=\sqrt{3}\left ( 1-\sqrt{3}m_{2} \right )$

$-\sqrt{3}-m_{2}=\sqrt{3}-3m_{2}$

$3m_{2}-m_{2}=\sqrt{3}+\sqrt{3}$

$2m_{2}=2\sqrt{3}$

$m_{2}=\sqrt{3}$

Taking -ve sign we get $\sqrt{3}+m_{2}=\sqrt{3}\left ( 1-\sqrt{3}m_{2} \right )$

$\sqrt{3}+m_{2}=\sqrt{3}-3m_{2}$

$3m_{2}+m_{2} =0$

$4m_{2} =0$

m₂=0

Equation of line passing through (3,-2) with slope $\sqrt{3}$ is y-y₁=m(x-x₁)

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

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

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

and equation of line passing through (3, -2) with slope 0 is y-y₁=m(x-x₁)

y-(-2)=0(x-3)

y+2=0

Hence, the required equations are $\sqrt{3}x-y-\left ( 3\sqrt{3}+2 \right )=0$ and y+2=0

Hence, the correct option is (a)

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infoexpert22

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