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Question: The points \(\left( \frac{a}{\sqrt{3}},a \right),\mspace{6mu}\left( \frac{2a}{\sqrt{3}},2a \right),\...

The points (a3,a),6mu(2a3,2a),6mu(a3,3a)\left( \frac{a}{\sqrt{3}},a \right),\mspace{6mu}\left( \frac{2a}{\sqrt{3}},2a \right),\mspace{6mu}\left( \frac{a}{\sqrt{3}},3a \right)are the vertices of.

A

An equilateral triangle

B

An isosceles triangle

C

A right angled triangle

D

None of these

Answer

An isosceles triangle

Explanation

Solution

Let A(a3,a),B(2a3,2a)A \left( \frac { a } { \sqrt { 3 } } , a \right) , B \left( \frac { 2 a } { \sqrt { 3 } } , 2 a \right) and C(a3,3a)C \left( \frac { a } { \sqrt { 3 } } , 3 a \right)

Then AB2=(a32a3)2+(a2a)2=a23+a2=4a23A B ^ { 2 } = \left( \frac { a } { \sqrt { 3 } } - \frac { 2 a } { \sqrt { 3 } } \right) ^ { 2 } + ( a - 2 a ) ^ { 2 } = \frac { a ^ { 2 } } { 3 } + a ^ { 2 } = \frac { 4 a ^ { 2 } } { 3 }

Similarly BC2=4a23B C ^ { 2 } = \frac { 4 a ^ { 2 } } { 3 } and AC2=4a2A C ^ { 2 } = 4 a ^ { 2 }

Hence it is an isosceles triangle.