Question

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. $(\frac{a}{\sqrt{2}},0)$
• B. $(0,\frac{a}{\sqrt{2}})$
• C. $(a\sqrt{2},0)$
• D. $(-\frac{a}{\sqrt{2}},0)$

Answer 1

The correct option is C $(a\sqrt{2},0)$

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.