Question

A straight line is drawn through the point $(2,3)$ and is inclined at an angle of $30^{\circ }$ with the $x$-axis . Find the coordinates of two points on it at a distance $4$ from $P$.

• A. $(2+2\sqrt{3},5)$ or $(2-2\sqrt{3},1)$
• B. $(2-2\sqrt{3},5)$ or $(2-2\sqrt{3},1)$
• C. $(2-2\sqrt{3},5)$ or $(2+2\sqrt{3},1)$
• D. $(2+2\sqrt{3},5)$ or $(2+2\sqrt{3},1)$

Answer 1

Answer: (A)

Here $\left(x_1,y_1\right)=(2,3),\theta =30^{\circ }$, the equation of the line is
$\frac{x-2}{\cos 30^{\circ }}=\frac{y-3}{\sin 30^{\circ }}$
$\Rightarrow \frac{x-2}{\frac{\sqrt{3}}{2}}=\frac{y-3}{\frac{1}{2}}$
$\Rightarrow x-2=\sqrt{3}(y-3)$
$\Rightarrow x-\sqrt{3}y=2-3\sqrt{3}$
Points on the line at a distance 4 from $P(2,3)$ are
$\left(x_1\pm r\cos \theta ,y_1\pm r\sin \theta \right)$
$\Rightarrow \left(2\pm 4\cos 30^{\circ },3\pm 4\sin 30^{\circ }\right)$
$\Rightarrow (2\pm 2\sqrt{3},3\pm 2)\Rightarrow (2+2\sqrt{3},5)\text{ or }(2-2\sqrt{3},1)$

Hence equation of $AB$ is $y-7=(x+5)\Rightarrow x-y+12=0$