question_answer

Point of intersection of the diagonals of square is at origin and coordinate axis are drawn along the diagonals. If the side is of length a, then one which is not the vertex of square is

A) \[(a\sqrt{2},0)\]

B) \[\left( 0,\frac{a}{\sqrt{2}} \right)\]

C) \[\left( \frac{a}{\sqrt{2}},0 \right)\]

D) \[\left( -\frac{a}{\sqrt{2}},0 \right)\]

Correct Answer: A

Solution :

  Obviously, from right angled triangle BOA \[OA=OB=\frac{a}{\sqrt{2}}\] Hence the vertex \[(a\sqrt{2},\,\,0)\] is not the vertex of square.