question_answer

The vertex of an equilateral triangle is \[(2,-1)\] and the equation of its base is\[x+2y=1\]. The length of its sides is:

A) \[4/\sqrt{15}\]                         

B)   \[2/\sqrt{15}\]

C)   \[4/3\sqrt{15}\]                       

D)   \[1/\sqrt{15}\]

Correct Answer: B

Solution :

Let ABC be an equilateral triangle with base BC. So,   \[AD\bot BC\] Now, length \[AD=\left| \frac{2+2(-1)-1}{\sqrt{{{(1)}^{2}}+{{(2)}^{2}}}} \right|=\left| \frac{2-2-1}{\sqrt{5}} \right|=\frac{1}{\sqrt{5}}\] Let \[AB=BC=AC=x\] So, in  \[\Delta ABD,\] \[A{{D}^{2}}+B{{D}^{2}}=A{{B}^{2}}\] \[\Rightarrow {{\left( \frac{1}{\sqrt{5}} \right)}^{2}}+\frac{{{x}^{2}}}{4}={{x}^{2}}\]          \[\left( \because \,\,BD=\frac{BC}{2} \right)\] \[\Rightarrow \,\,\,\frac{1}{5}={{x}^{2}}-\frac{{{x}^{2}}}{4}\Rightarrow \frac{3}{4}{{x}^{2}}=\frac{1}{5}\] \[\Rightarrow \,\,{{x}^{2}}=\frac{4}{15}\Rightarrow x=\frac{2}{\sqrt{15}}\]