Question

A straight line is drawn through the point $P(2,2)$ and is inclined at an angle of $30$ with the $X-$axis. Find the coordinates of two points on it at a distance $4$ from $P$ on either side on $P$.

Answer 1

Given that $\theta ={30}^0$

Therefore, $\Rightarrow m=\tan \theta =\tan \left({30}^0\right)=\frac{1}{\sqrt{3}}$

Given that the line has slope $\frac{1}{\sqrt{3}}$ and passes through $P(2,2)$

$\Rightarrow y-y_1=m(x-x_1)$

$\Rightarrow y-2=\frac{1}{\sqrt{3}}(x-2)$ ..$\left(1\right)$

Let the required points be $(x_3,y_3)$ and $(x_4,y_4)$

Given that this point is $4$ units from $P$ and lies on the line $\left(1\right)$

Thereore, $\Rightarrow \sqrt{(x_3-2{)}^2+(y_3-2{)}^2}=4$

$\Rightarrow (x_3-2{)}^2+(y_3-2{)}^2=16$

$\Rightarrow (x_3-2{)}^2+\frac{1}{3}(x_3-2{)}^2=16$ (from $\left(1\right)$)

$\Rightarrow \frac{4}{3}(x_3-2{)}^2=16$

$\Rightarrow (x_3-2{)}^2=12$

$\Rightarrow (x_3-2)=\pm 2\sqrt{3}$

$\Rightarrow x_3=2\pm 2\sqrt{3}$

Substituting these values in $\left(1\right)$ we get

$\Rightarrow (x_3,y_3)=(2+2\sqrt{3},4)$ and $(x_4,y_4)=(2-2\sqrt{3},0)$