Question

The equation of two straight lines through $(7,9)$ and making an angle of ${60}^o$ with the line $x-\sqrt{3y}-2\sqrt{3}=0$ is

• A. $x=7,x+\sqrt{3y}=7+9\sqrt{3}$
• B. $x=\sqrt{3},x+\sqrt{3y}=7+9\sqrt{3}$
• C. $x=7,x-\sqrt{3y}=7+9\sqrt{3}$
• D. $x=\sqrt{3},x-\sqrt{3y}=7+9\sqrt{3}$

Answer 1

Answer: (A)

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

Given line equation is $x-\sqrt{3}y-2\sqrt{3}=0⟹y=\frac{1}{\sqrt{3}}x-2$

Therefore, the slope of the given line is $\frac{1}{\sqrt{3}}$

Let the slope of the required line be $m$

Given that angle between the two lines is ${60}^0$

Therefore $tan{60}^0=|\frac{(m-\frac{1}{\sqrt{3}})}{1+\frac{1}{\sqrt{3}}m}|$

$⟹\sqrt{3}=|\frac{(\sqrt{3}m-1)}{\sqrt{3}+m}|$

$⟹\pm \sqrt{3}=\frac{(\sqrt{3}m-1)}{\sqrt{3}+m}$

$⟹\sqrt{3}m+3=\sqrt{3}m-1$ or $-\sqrt{3}m-3=\sqrt{3}m-1$

$⟹\sqrt{3}m+3=\sqrt{3}m-1$ or $-\sqrt{3}m-3=\sqrt{3}m-1$

$⟹m=undefined$ or $2\sqrt{3}m=-2$

$⟹m=undefined$ or $m=\frac{-1}{\sqrt{3}}$

If the slope is undefined then the line is parallel to the y-axis. Therefore a line which is parallel to y-axis and passing through the point $(7,9)$ is $x=7$

The equation of the line with slope $\frac{-1}{\sqrt{3}}$ and passing through the point $(7,9)$ is $x=7$ is given by

$y-9=\frac{-1}{\sqrt{3}}(x-7)$

$⟹-x+7=\sqrt{3}y-9\sqrt{3}$

$⟹x+\sqrt{3}y=7+9\sqrt{3}$

Therefore,the two lines are $x=7$ and $x+\sqrt{3}y=7+9\sqrt{3}$