Question

Show that the points $A(5,6),B(1,5),C(2,1)$ and $D(6,2)$ are the vertices of a square.

Answer 1

Given, $A(5,6)$ $B(1,5)$ $C(2,1)$ and $D(6,2)$

So by distance formula we have,

Distance between two points = $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

$\therefore AB=\sqrt{(1-5{)}^2+(5-6{)}^2}$$=\sqrt{(-4{)}^2+1}=\sqrt{17}$

$BC=\sqrt{(2-1{)}^2+(1-5{)}^2}=\sqrt{17}$

$CD=\sqrt{(6-2{)}^2+(2-1{)}^2}=\sqrt{17}$

$DA=\sqrt{(5-6{)}^2+(6-2{)}^2}=\sqrt{17}$

$\therefore AB=BC=CD=DA$...........(1)

Now,

Slope of side AB, $\left(m_1\right)=\frac{6-5}{5-1}$$=\frac{1}{4}$

Slope of side BC $\left(m_2\right)=\frac{5-1}{1-2}=-4$

$\therefore m_1{m}_2=\frac{1}{4}\times (-4)=-1$

Hence, $AB\perp BC$..........(2)

from (1) and (2)

$\therefore ABCD$ is a square (proved)