question_answer

If the base of an equilateral triangle is\[x+y=2\]and vertex,\[(2,-1)\]is, then length of each side is equal to

A)  \[\sqrt{\frac{2}{3}}\]

B)  \[\sqrt{\frac{3}{2}}\]

C)  \[\frac{3}{\sqrt{2}}\]

D)  \[\frac{2}{3}\]

Correct Answer: A

Solution :

 The length of perpendicular drawn from the vertex\[A(2,-1)\]to the base\[x+y=2\]is \[p=\left| \frac{2+(-1)-2}{\sqrt{{{1}^{2}}+{{1}^{2}}}} \right|\] \[=\left| -\frac{1}{\sqrt{2}} \right|=\frac{1}{\sqrt{2}}\] Let length of the side is a. In \[\Delta ABM,,\frac{p}{a}=\sin 60{}^\circ \] \[\Rightarrow \] \[a=\frac{p}{\sin 60{}^\circ }\] \[\Rightarrow \] \[a=\frac{\frac{1}{\sqrt{2}}}{\frac{\sqrt{3}}{2}}=\frac{2}{\sqrt{2}\sqrt{3}}\] \[\Rightarrow \] \[a=\sqrt{\frac{2}{3}}\]