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 $(2,1),(-2,1)$

Let the points be $B(x_1,y_1)$ and $D(x_2,y_2)$ mid point of
$BD=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$
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 $\frac{y_1+y_2}{2}=1$
$\Rightarrow x_1+x_2=0.....\left(i\right)$
and $y_1+y_2=2.....\left(ii\right)$
Slope of $BD\times$ slope of $AC=-1$
$\frac{(y_1-y_2)}{(x_1-x_2)}\times \frac{(3+1)}{(0-0)}=-1$
$\Rightarrow y_1-y_2=0.....\left(iii\right)$
Solving Eqs. (ii) and (iii), we get
$y_1=1,y_2=1$
Now, slope of $AB\times$ 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 $(-2,1)$.