What are the equations of the two lines passi | vedclass

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(A) The slope of the given line $x - \sqrt{3}y - 2\sqrt{3} = 0$ is $m = \frac{1}{\sqrt{3}}$.Let the slope of the required lines be $m'$. The angle bet

Vedclass · Accepted Answer

$x = 7, x + \sqrt{3}y = 7 + 9\sqrt{3}$

Vedclass · Answer

(A) The slope of the given line $x - \sqrt{3}y - 2\sqrt{3} = 0$ is $m = \frac{1}{\sqrt{3}}$.Let the slope of the required lines be $m'$. The angle between the lines is $	heta = 60^{\circ}$.Using the formula $	an 	heta = \left| \frac{m' - m}{1 + m'm} ight|$,we have $	an 60^{\circ} = \left| \frac{m' - 1/\sqrt{3}}{1 + m'/\sqrt{3}} ight|$.$\sqrt{3} = \left| \frac{\sqrt{3}m' - 1}{\sqrt{3} + m'} ight|$.Case $1$: $\sqrt{3} = \frac{\sqrt{3}m' - 1}{\sqrt{3} + m'}$ $\Rightarrow 3 + \sqrt{3}m' = \sqrt{3}m' - 1$ $\Rightarrow 3 = -1$ (Impossible,implies vertical line).$A$ vertical line passing through $(7, 9)$ is $x = 7$.Case $2$: $-\sqrt{3} = \frac{\sqrt{3}m' - 1}{\sqrt{3} + m'}$ $\Rightarrow -3 - \sqrt{3}m' = \sqrt{3}m' - 1$ $\Rightarrow 2\sqrt{3}m' = -2$ $\Rightarrow m' = -\frac{1}{\sqrt{3}}$.The equation of the line is $y - 9 = -\frac{1}{\sqrt{3}}(x - 7)$ $\Rightarrow \sqrt{3}y - 9\sqrt{3} = -x + 7$ $\Rightarrow x + \sqrt{3}y = 7 + 9\sqrt{3}$.Thus,the equations are $x = 7$ and $x + \sqrt{3}y = 7 + 9\sqrt{3}$.

What are the equations of the two lines passing through the point $(7, 9)$ and making an angle of $60^{\circ}$ with the line $x - \sqrt{3}y - 2\sqrt{3} = 0$?

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