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).You need to login to perform this action.
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