Question

If the equation of base of an equilateral triangle is $2x-y=1$and the vertex $(-1,2)$, then the length of the side of the triangle is

• A. $\sqrt{\frac{20}{3}}$
• B. $\frac{2}{\sqrt{15}}$
• C. $5$
• D. $\sqrt{5}$

Answer 1

The correct option is A

$\sqrt{\frac{20}{3}}$

Step 1. Find the height of the equilateral triangle

We know that the distance of the point $\left(x_1,y_1\right)$ from the straight line $ax+by+c=0$ is $\frac{\left|a{x}_1+b{y}_1+c\right|}{\sqrt{a^2+b^2}}$

Given equation of the base is $2x-y=1$

Point $A$$(-1,2)$

Therefore $AD=\frac{\left|a{x}_1+b{y}_1+c\right|}{\sqrt{a^2+b^2}}$

$$\begin{gathered} =\frac{\left|2\left(-1\right)-1\left(2\right)-1\right|}{\sqrt{{\left(2\right)}^2+{\left(-1\right)}^2}} \\ =\frac{\left|5\right|}{\sqrt{5}} \\ =\sqrt{5} \end{gathered}$$

Step 2. Find the length of each side

Consider $∆ABD$

$∆ABD$ is a right angle triangle.

$$\begin{gathered} \therefore \sin \left(60°\right)=\frac{AD}{AB} \\ \Rightarrow \frac{\sqrt{3}}{2}=\frac{\sqrt{5}}{AB}\left[\because AD=\sqrt{5}\right] \\ \Rightarrow AB=\frac{2\sqrt{5}}{\sqrt{3}} \\ \Rightarrow AB=\sqrt{\frac{20}{3}} \end{gathered}$$

Hence Option (A) is correct i.e. $\sqrt{\frac{20}{3}}$