Question

$(0, - 1)$ 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)

Answer 1

The Correct Option is C

Let the points be $B \left( x _{1}, y _{1}\right)$ and $D \left( x _{2}, y _{2}\right)$ 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 $E$.
$\Rightarrow \frac{ x _{1}+ x _{2}}{2}=0$ and $\frac{ y _{1}+ y _{2}}{2}=1$
$\Rightarrow x _{1}+ x _{2}=0$ ...(i)
and $y_{1}+y_{2}=$ ... (ii)
Slope of $BD \times$ slope of $AC =-1$
$\frac{\left( y _{1}- y _{2}\right)}{\left( x _{1}- x _{2}\right)} \times \frac{(3+1)}{(0-0)}=-1$
$\Rightarrow y _{1}- y _{2}=0$ ... (iii)
Solving Eqs. (ii) and (iii), we get
$y _{1}=1, y _{2}=1$
Now, slope of $A B \times$ slope of $B C=-1$
$\Rightarrow \frac{\left( y _{1}+1\right)}{\left( x _{1}-0\right)} \times \frac{\left( y _{1}-3\right)}{\left( x _{1}-0\right)}=-1$
$\Rightarrow\left( y _{1}+1\right)\left( y _{1}-3\right)=- x _{1}^{2}$
$\Rightarrow 2(-2)=- x _{1}^{2}\left[\because y _{1}=1\right]$
$\Rightarrow x _{1}=\pm 2$
$\therefore$ The required points are $(2,1)$ and $(-2,1)$.