Question

The equation of the base $BC$ of an equilateral triangle $ABC$ is $x+y=2$ and $A$ is $(2,-1)$.

The length of the side of the triangle is:

• A. $\sqrt{2}$
• B. ${\left(\frac{3}{2}\right)}^{1/2}$
• C. ${\left(\frac{1}{2}\right)}^{1/2}$
• D. ${\left(\frac{2}{3}\right)}^{1/2}$

Answer 1

Answer: (D)

${\left(\frac{2}{3}\right)}^{1/2}$

The vertex $A$ is $(2,-1)$. Draw perpendicular $AM$ with length $d$. Now, as $△ABC$ is equilateral, $BM=MC=x$.

$\therefore d=\left|\frac{2-1-2}{\sqrt{2}}\right|=\frac{1}{\sqrt{2}}$

Now, in $△AMC$, $\angle C={90}^{\circ }$

$\therefore \frac{d}{x}=\tan {60}^{\circ }$

$\therefore x=\frac{1}{\sqrt{6}}$

$\therefore BC=2x=\frac{2}{\sqrt{6}}={\left(\frac{2}{3}\right)}^{0.5}$ which is the required side length.