question_answer

(0, -1) and (0, 3) are two opposite vertices of a square. The other two vertices are   [Karnataka CET 2005]

A) (0, 1), (0, -3)

B) (3, -1) (0, 0)

C) (2, 1), (-2, 1)

D) (2, 2), (1, 1)

Correct Answer: C

Solution :

Length of diagonal = 4 Now, \[A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\]   \[A{{C}^{2}}=2A{{B}^{2}}\Rightarrow 8=A{{B}^{2}}\]   \[AB=BC=2\sqrt{2}\] Now, let \[B\,(x,y)\]; \[\therefore A{{B}^{2}}=B{{C}^{2}}\] Þ\[{{(x-0)}^{2}}+{{(y+1)}^{2}}={{(x-0)}^{2}}+{{(y-3)}^{2}}\] \[{{x}^{2}}+{{y}^{2}}+2y+1={{x}^{2}}+{{y}^{2}}-6y+9\] Þ \[y=1;\,\,\,\therefore {{x}^{2}}+{{(2)}^{2}}=8;\Rightarrow {{x}^{2}}=4\Rightarrow x=\pm \,2\] \[\therefore \] other vertices are (2, 1),(-2, 1).