Question

Let $A\left(6,7\right),B\left(2,3\right)$ and $C\left(-2,1\right)$ be the vertices of a triangle. The point $P$ nearer to the point $A$ such that $\Delta PBC$ is an equilateral triangle is

• A. $\left(-\sqrt{3},2+2\sqrt{3}\right)$
• B. $\left(\sqrt{3},2+2\sqrt{3}\right)$
• C. $\left(\sqrt{3},2-2\sqrt{3}\right)$
• D. $\left(-\sqrt{3},2-2\sqrt{3}\right)$

Answer 1

Answer: (A)

We have, $BC=2\sqrt{5}.$
Since, $\Delta PBC$ is an equilateral triangle and the point $P$ lies in the interior of $\Delta ABC.$
Therefore, $P$ lies on the perpendicular bisector of $BC$ at a distance of $\frac{\sqrt{3}}{2}\left(BC\right)=\sqrt{15}$ from the mid-point of $BC$ such that $P$ and $A$ are on the same side of $BC.$
The equation of $BC$ is $x-2y+4=0$
The coordinates of the mid-point of $BC$ are $D\left(0,2\right)$
The slope of a line perpendicular to $BC$ is $-2.$
So, if it makes an angle $\theta$ with the $x$ -axis. Then,
$tan\theta =-2\Rightarrow sin\theta =\frac{2}{\sqrt{5}}$ and $cos\theta =-\frac{1}{\sqrt{5}}$
The parametric form of the perpendicular bisectors of $BC$ is
$\frac{x-0}{-\frac{1}{\sqrt{5}}}=\frac{y-2}{\frac{2}{\sqrt{5}}}$
So, the coordinates of $P$ are given by
$\frac{x-0}{-\frac{1}{\sqrt{5}}}=\frac{y-2}{\frac{2}{\sqrt{5}}}=\pm \sqrt{15}$
or, $x=\pm \sqrt{3},y=2\pm 2\sqrt{3}$
Thus, the coordinates of $P$ are $\left(\sqrt{3},2-2\sqrt{3}\right)$ or $\left(-\sqrt{3},2+2\sqrt{3}\right).$
Clearly, $A\left(6,7\right)$ and $\left(-\sqrt{3},2+2\sqrt{3}\right)$ lie on the same side of $BC.$ Hence, the coordinates of $P$ are $\left(-\sqrt{3},2+2\sqrt{3}\right).$