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 $A B$ is $y-7=(x+5) \Rightarrow x-y+12=0$