Question

If $A(x,-1),B(2,1),C(0,3)$ and $D(-2,1)$ are the vertices of a square, then find the value of $x$.

Answer 1

As $ABCD$ is a square $\Rightarrow AC=BD$

$\Rightarrow \sqrt{x^2+(-4{)}^2}=\sqrt{4^2}$

$\Rightarrow x^2=4^2-4^2=0$

$\Rightarrow x=0$