Question

The two opposite vertices of a square are $(-1,2)$ and $(3,2)$. Find the coordinates of the other two vertices.

Answer 1

Step 1:Find the value of $\mathrm{x}$

Let $\mathrm{ABCD}$ is a square where two opposite vertices are$\mathrm{A}(-1,2)$ and $\mathrm{C}(3,2)$.

Let $\mathrm{B}(\mathrm{x},\mathrm{y})$ and $\mathrm{D}({\mathrm{x}}_1,{\mathrm{y}}_1)$ be the other two vertices.
In Square $\mathrm{ABCD}$
$\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}$
Hence AB=BC
$\sqrt{{\left(\mathrm{x}+1\right)}^2+{\left(\mathrm{y}-2\right)}^2}=\sqrt{{\left(3-\mathrm{x}\right)}^2+{\left(2-\mathrm{y}\right)}^2}$ (by distance formula)
Squaring both sides
$(\mathrm{x}+1{)}^2+(\mathrm{y}-2{)}^2=(3-\mathrm{x}{)}^2+(2-\mathrm{y}{)}^2$
${\mathrm{x}}^2+1+2\mathrm{x}+{\mathrm{y}}^2+4-4\mathrm{y}=9+{\mathrm{x}}^2-6\mathrm{x}+4+{\mathrm{y}}^2-4\mathrm{y}$
$2\mathrm{x}+5=13-6\mathrm{x}$
$2\mathrm{x}+6\mathrm{x}=13-5$
$8\mathrm{x}=8$
$\mathrm{x}=1$

Step 2:Find the value of $\mathrm{y}$

In $∆\mathrm{ABC}$, $\mathrm{\angle }B={90}^{\circ }$
(All angles of the square are ${90}^{\circ }$)
Then according to the Pythagoras theorem
${\mathrm{AB}}^2+{\mathrm{BC}}^2={\mathrm{AC}}^2$
$2{\mathrm{AB}}^2={\mathrm{AC}}^2(\mathrm{since}\mathrm{AB}=\mathrm{BC})$
$2{\left(\sqrt{(\mathrm{x}+1{)}^2+(\mathrm{y}-2{)}^2}\right)}^2={\left(\sqrt{{\left(3-(-1{)}^2\right)}^2+(2-2{)}^2}\right)}^2$
$2\left((\mathrm{x}+1{)}^2+(\mathrm{y}-2{)}^2\right)=(3+1{)}^2+(2-2{)}^2$
$2\left({\mathrm{x}}^2+2\mathrm{x}+1+{\mathrm{y}}^2+4-4\mathrm{y}\right)=4^2$
put $\mathrm{x}=1$
$2\left(1^2+2+1+{\mathrm{y}}^2+4-4\mathrm{y}\right)=16$
$2\left({\mathrm{y}}^2-4\mathrm{y}+8\right)=16$
$2{\mathrm{y}}^2-8\mathrm{y}+16=16$
$2{\mathrm{y}}^2-8\mathrm{y}=0$
$2\mathrm{y}(\mathrm{y}-4)=0$
$\mathrm{y}=0,4$

Therefore, the coordinates of the two vertices of square $\mathrm{ABCD}$ are $(1,0)$ and $(1,4)$.