Question

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

Answer 1

Let $ABCD$ is a square where two opposite vertices are $A(-1,2)andC(3,2)$.

Let B$(x,y)$ and D$(x_1.y_1)$ ids the other two vertices.

In Square $ABCD$

$AB=BC=CD=DA$

Hence $AB=BC$

$\Rightarrow \sqrt{(x+1{)}^2+(y-2{)}^2}=\sqrt{(3-x{)}^2+(2-y{)}^2}$ [by distance formula]

Squaring both sides

$\Rightarrow (x+1{)}^2+(y-2{)}^2=(3-x{)}^2+(2-y{)}^2$

$\Rightarrow x^2+2x+1+y^2+4-4y=9+x^2-6x+4+y^2-4y$

$\Rightarrow 2x+5=13-6x$

$\Rightarrow 2x+6x=13-5$

$\Rightarrow 8x=8$

$\Rightarrow x=1$

In $△ABC,\angle B={90}^{\circ }$ [all angles of square are ${90}^{\circ }$]

Then according to the Pythagorean theorem

$A{B}^2+B{C}^2=A{C}^2$

As $AB=BC$

$\therefore 2A{B}^2=A{C}^2$

$\Rightarrow 2{\left(\sqrt{x+1{)}^2+(y-2{)}^2}\right)}^2={\left(\sqrt{(3-(-1){)}^2+(2-2{)}^2}\right)}^2$

$\Rightarrow 2\left(\right(x+1{)}^2+(y-2{)}^2)=(3+1{)}^2+(2-2{)}^2$

$\Rightarrow 2(x^2+2x+1+y^2+4-4y)=(4{)}^2$

Put the value of $x=1$

$\Rightarrow 2(1^1+2\times 1+1+y^2+4-4y)=16$

$\Rightarrow 2(y^2-4y+8)=16$

$\Rightarrow 2y^2-8y+16=16$

$\Rightarrow 2y^2-8y=0$

$\Rightarrow 2y(y-4)=0$

$\Rightarrow (y-4)=0$

Hence $y=0$ or $4$.

As diagonals of a square are equal in length and bisect each other at ${90}^{\circ }$

Let $P$ is the midpoint of $AC$

$\therefore$CO-ordinates of $P=(\frac{3-1}{2},\frac{2+2}{2})=(1,2)$

P is also the midpoint of $BD$

then co-ordinates of mid-point of $BD$=co-ordinates of $P$

$\Rightarrow (x_1,y_1)=1,2$

$\therefore x_1=1,y_1=2$

Then other two vertices of square $ABCD$ are $(1,0)$ or $(1,4)$ and $(1,2)$.