A) (0,1),(0,-3)
B) (3,-1),(0,0)
C) (2,1), (-2,1)
D) (2, 2), (1,1)
Correct Answer: C
Solution :
Let the points be \[B\,({{x}_{1}},\,{{y}_{1}})\] and \[D\,({{x}_{2}},\,{{y}_{2}})\] Mid point of \[BD=\left( \frac{{{x}_{1}}+{{x}_{2}}}{2},\frac{{{y}_{1}}+{{y}_{2}}}{2} \right)\] and mid point of AC = (0,1) We know, mid point of both the diagonal lie on the same point E \[\frac{{{x}_{1}}+{{x}_{2}}}{2}=0\] and \[\frac{{{y}_{1}}+{{y}_{2}}}{2}=1\] \[\Rightarrow \] \[{{x}_{1}}+{{x}_{2}}=0\] ... (i) and \[{{y}_{1}}+{{y}_{2}}=2\] ... (ii) slope of \[BD\times \] slope of \[AC=-1\] \[\frac{({{y}_{1}}+{{y}_{2}})}{({{x}_{1}}-{{x}_{2}})}\times \frac{(3+1)}{(0-0)}=-1\] \[\Rightarrow \] \[{{y}_{1}}-{{y}_{2}}=0\] ?. (iii) solving equations (ii) and (iii), we get \[{{y}_{1}}=1,\,\,{{y}_{2}}=1\] Now, slope of AB x slope of \[BC=-1\] \[\Rightarrow \] \[\frac{({{y}_{1}}+1)}{({{x}_{1}}-0)}\times \frac{({{y}_{1}}-3)}{({{x}_{1}}-0)}=-1\] \[\Rightarrow \] \[({{y}_{1}}+1)\,({{y}_{1}}-3)=-x_{1}^{2}\] \[\Rightarrow \] \[2\,(-2)=-x_{1}^{2}\] \[[\because \,{{y}_{1}}=1]\] \[\Rightarrow \] \[{{x}_{1}}=\pm 2\] \[\therefore \] The required points are (2,1) and (-2,1).You need to login to perform this action.
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