question_answer

(0, -1) and (0, 3) are two opposite vertices of a square. The other two vertices are:

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).