Question

ABCD 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 midpoints 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

The points are A(-1, -1), B(-1, 4), C(5, 4) and D(5, -1)

Co-ordinates of P $=(\frac{-1-1}{2},\frac{-1+4}{2})=(-1,\frac{3}{2})$

Co-ordinates of Q $=(\frac{-1+5}{2},\frac{4+4}{2})=(2,4)$

Co-ordinates of R $=(\frac{5+5}{2},\frac{4-1}{2})=(5,\frac{3}{2})$

Co-ordinates of S $=(\frac{-1+5}{2},\frac{-1-1}{2})=(2,-1)$

Now,

$PQ=\sqrt{(2+1{)}^2+(4-\frac{3}{2}{)}^2}=\sqrt{9+\frac{25}{4}}=\frac{\sqrt{61}}{2}$

$QR=\sqrt{(5-2{)}^2+(\frac{3}{2}-4{)}^2}=\sqrt{9+\frac{25}{4}}=\frac{\sqrt{61}}{2}$

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

$SP=\sqrt{(2+1{)}^2+(-1-\frac{3}{2}{)}^2}=\sqrt{9+\frac{25}{4}}=\frac{\sqrt{61}}{2}$

$PR=\sqrt{(5+1{)}^2+(\frac{3}{2}-\frac{3}{2}{)}^2}=\sqrt{36}=6$

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

Thus $PQ=QR=RS=SP$ and $PR\ne QS$.

Therefore PQRS is a rhombus