# ABCD is a rectangle formed by the points

Question:

$A B C D$ is a rectangle formed by the points $A(-1,-1), B(-1,4), C(5,4)$ and $D(5,-1)$. If $P, Q, R$ and $S$ be the mid points of $A B, B C, C D$ and $D A$ respectively, show that $P Q R S$ is a rhombus.

Solution:

Here, the points PQR and S are the mid points of ABBCCD and DA respectively. Then

Coordinates of $P=\left(\frac{-1-1}{2}, \frac{-1+4}{2}\right)=\left(-1, \frac{3}{2}\right)$

Coordinates of $Q=\left(\frac{-1+5}{2}, \frac{4+4}{2}\right)=(2,4)$

Coordinates of $R=\left(\frac{5+5}{2}, \frac{4-1}{2}\right)=\left(5, \frac{3}{2}\right)$

Coordinates of $S=\left(\frac{-1+5}{2}, \frac{-1-1}{2}\right)=(2,-1)$

Now

$P Q=\sqrt{(2+1)^{2}+\left(4-\frac{3}{2}\right)^{2}}=\sqrt{9+\frac{25}{4}}=\frac{\sqrt{61}}{2}$

$Q R=\sqrt{(5-2)^{2}+\left(\frac{3}{2}-4\right)^{2}}=\sqrt{9+\frac{25}{4}}=\frac{\sqrt{61}}{2}$

$R S=\sqrt{(5-2)^{2}+\left(\frac{3}{2}+1\right)^{2}}=\sqrt{9+\frac{25}{4}}=\frac{\sqrt{61}}{2}$

$S P=\sqrt{(2+1)^{2}+\left(-1-\frac{3}{2}\right)^{2}}=\sqrt{9+\frac{25}{4}}=\frac{\sqrt{61}}{2}$

$P R=\sqrt{(5+1)^{2}+\left(\frac{3}{2}-\frac{3}{2}\right)^{2}}=\sqrt{36}=6$

$Q S=\sqrt{(2-2)^{2}+(-1-4)^{2}}=\sqrt{25}=5$

Thus, $P Q=Q R=R S=S P$ and $P R \neq Q S$ therefore $P Q R S$ is a rhombus.