Question

If 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. $\sqrt{\frac{3}{2}}$ unit
• B. $\sqrt{\frac{2}{3}}$ unit
• C. $\sqrt{2}$ unit
• D. $\sqrt{3}$ unit

Answer 1

Answer: (B)

Let $ABC$ be an equilateral triangle. Draw $AD\perp BC$.
Length of $AD=$ Perpendicular distance from $A(2,-1)$ to the line $x+y-2=0$
$=\left|\frac{2-1-2}{\sqrt{1+1}}\right|=\frac{1}{\sqrt{2}}$
Since, $△ABC$ is an equilateral triangle, so $\angle ABC=60^{\circ }$.

In $△ABD$,
$\Rightarrow \sin 60^{\circ }=\frac{AD}{AB}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{\frac{1}{\sqrt{2}}}{AB}$
$\Rightarrow AB=\frac{\sqrt{3}}{2}=\frac{1}{\sqrt{2}}AB$
$\Rightarrow AB=\sqrt{\frac{2}{3}}$ unit
Since, $△ABC$ is an equilateral triangle, so all sides are equal length.
$\therefore$ Length of $AB=$ Length of $BC=$ Length of $CA=\sqrt{\frac{2}{3}}$