Q. The equation of the base of an equilateral triangle is $x + y = 2$ and the vertex is $(2,\ - 1)$ , then the length of the side of the triangle is
A) $\sqrt{3/2}\ /\sqrt{2/3}$
B) $\sqrt{2}$
C) $\sqrt{2/3}$
D) $\sqrt{3/2}$
Explanation
(c)
Let $p\$be the length of the perpendicular from the vertex $(2,\ - 1)$ to the base
$x + y = 2$, then
$p = \left| \frac{2 - 1 - 2}{\sqrt{1^{2} + 1^{2}}} \right|$
$= \frac{1}{\sqrt{2}}$
If $a$ be the length of the side of triangle then,
$p = a\sin{60{^\circ}}$
$\Rightarrow \ \frac{1}{\sqrt{2}} = \frac{a\sqrt{3}}{2}$
$\Rightarrow \ a = \sqrt{\frac{2}{3}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.