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)}^{\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}}} $

Answer 1

Answer: (E)

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}} $