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

Answer: (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 $AB\times$ slope of $BC=-1$
$\Rightarrow \frac{\left(y_1+1\right)}{\left(x_1-0\right)}\times \frac{\left(y_1-3\right)}{\left(x_1-0\right)}=-1$
$\Rightarrow \left(y_1+1\right)\left(y_1-3\right)=-x_1^2$
$\Rightarrow 2(-2)=-x_1^2\left[\because y_1=1\right]$
$\Rightarrow x_1=\pm 2$
$\therefore$ The required points are $(2,1)$ and $(-2,1)$.