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