A) Vertices of an equilateral triangle
B) Vertices of a right angled triangle
C) Vertices of an isosceles triangle
D) Collinear
Correct Answer: D
Solution :
\[{{l}_{1}}=\sqrt{{{(2a)}^{2}}+{{(2b)}^{2}}}=2\sqrt{{{a}^{2}}+{{b}^{2}}}\] \[{{l}_{2}}=\sqrt{{{({{a}^{2}}-a)}^{2}}+{{b}^{2}}{{(a-1)}^{2}}}=(a-1)\,\sqrt{{{a}^{2}}+{{b}^{2}}}\] \[{{l}_{3}}=\sqrt{{{({{a}^{2}}+a)}^{2}}+{{b}^{2}}{{(a+1)}^{2}}}=(a+1)\,\sqrt{{{a}^{2}}+{{b}^{2}}}\] Now \[{{l}_{1}}+{{l}_{2}}={{l}_{3}}.\] Hence points are collinear.
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