question_answer

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)  \[{{\left( \frac{3}{2} \right)}^{\frac{1}{2}}}\]

C)  \[{{\left( \frac{1}{2} \right)}^{\frac{1}{2}}}\]      

D)         \[\left( \frac{1}{\sqrt{2}} \right)\]

E)  \[{{\left( \frac{2}{3} \right)}^{\frac{1}{2}}}\]

Correct Answer: E

Solution :

Length of perpendicular from\[A(2,-1)\]to the line\[x+y-2=0\]is \[\left| \frac{2-1.2}{\sqrt{1+1}} \right|=\frac{1}{\sqrt{2}}\] In\[\Delta ABC,\] \[\frac{AD}{AB}=sin\text{ }60{}^\circ \] \[\Rightarrow \]               \[\frac{1}{\sqrt{2}AB}=\frac{\sqrt{3}}{2}\] \[\Rightarrow \]               \[AB=\sqrt{\frac{2}{3}}\]