Question

ABCD is a rectangle formed by the points $A\left(-1,-1\right),B\left(-1,4\right),C\left(5,4\right)\mathrm{and}D\left(5,-1\right)$. If P, Q, R and S be the mid points of
AB, BC, CD and DA respectively, show that PQRS is a rhombus.

Answer 1

Here, the points P, Q, R and S are the mid points of AB, BC, CD and DA respectively. Then
$$\begin{gathered} \mathrm{Coordinates}\mathrm{of}P=\left(\frac{-1-1}{2},\frac{-1+4}{2}\right)=\left(-1,\frac{3}{2}\right) \\ \mathrm{Coordinates}\mathrm{of}Q=\left(\frac{-1+5}{2},\frac{4+4}{2}\right)=\left(2,4\right) \\ \mathrm{Coordinates}\mathrm{o}fR=\left(\frac{5+5}{2},\frac{4-1}{2}\right)=\left(5,\frac{3}{2}\right) \\ \mathrm{Coordinates}\mathrm{of}S=\left(\frac{-1+5}{2},\frac{-1-1}{2}\right)=\left(2,-1\right) \end{gathered}$$
Now
$$\begin{gathered} PQ=\sqrt{{\left(2+1\right)}^2+{\left(4-\frac{3}{2}\right)}^2}=\sqrt{9+\frac{25}{4}}=\frac{\sqrt{61}}{2} \\ QR=\sqrt{{\left(5-2\right)}^2+{\left(\frac{3}{2}-4\right)}^2}=\sqrt{9+\frac{25}{4}}=\frac{\sqrt{61}}{2} \\ RS=\sqrt{{\left(5-2\right)}^2+{\left(\frac{3}{2}+1\right)}^2}=\sqrt{9+\frac{25}{4}}=\frac{\sqrt{61}}{2} \\ SP=\sqrt{{\left(2+1\right)}^2+{\left(-1-\frac{3}{2}\right)}^2}=\sqrt{9+\frac{25}{4}}=\frac{\sqrt{61}}{2} \\ PR=\sqrt{{\left(5+1\right)}^2+{\left(\frac{3}{2}-\frac{3}{2}\right)}^2}=\sqrt{36}=6 \\ QS=\sqrt{{\left(2-2\right)}^2+{\left(-1-4\right)}^2}=\sqrt{25}=5 \end{gathered}$$
Thus, PQ = QR = RS = SP and $PR\ne QS$ therefore PQRS is a rhombus.