A) An equilateral triangle
B) An isosceles triangle
C) A right angled triangle
D) None of these
Correct Answer: B
Solution :
Let \[A\,\left( \frac{a}{\sqrt{3}},\,a \right),\,\,B\,\left( \frac{2a}{\sqrt{3}},\,\,2a \right)\] and \[C\,\left( \frac{a}{\sqrt{3}},\,\,3a \right)\] Then \[A{{B}^{2}}={{\left( \frac{a}{\sqrt{3}}-\frac{2a}{\sqrt{3}} \right)}^{2}}+{{(a-2a)}^{2}}=\frac{{{a}^{2}}}{3}+{{a}^{2}}=\frac{4{{a}^{2}}}{3}\] Similarly \[B{{C}^{2}}=\frac{4{{a}^{2}}}{3}\] and \[A{{C}^{2}}=4{{a}^{2}}\] Hence it is an isosceles triangle.
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