**Question:**

*ABCD* is a rectangle formed by joining the points *A* (−1, −1), *B*(−1 4) *C* (5 4) and *D* (5, −1). *P*, *Q*, *R* and *S* are the mid-points of sides *AB*, *BC*, *CD* and *DA* respectively. Is the quadrilateral *PQRS* a square? a rectangle? or a rhombus? Justify your answer.

**Solution:**

We have a rectangle ABCD formed by joining the points A (−1,−1); B (−1, 4); C (5, 4) and D (5,−1). The mid-points of the sides AB, BC, CD and DA are P, Q, R, S respectively.

We have to find that whether PQRS is a square, rectangle or rhombus.

In general to find the mid-point $\mathrm{P}(x, y)$ of two points $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ we use section formula as,

$\mathrm{P}(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$

Therefore mid-point P of side AB can be written as,

$P(x, y)=\left(\frac{-1-1}{2}, \frac{4-1}{2}\right)$

Now equate the individual terms to get,

$x=-1$

$y=\frac{3}{2}$

So co-ordinates of $P$ is $\left(-1, \frac{3}{2}\right)$

Similarly mid-point Q of side BC can be written as,

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

Now equate the individual terms to get,

$x=2$

$y=4$

So co-ordinates of Q is (2, 4)

Similarly mid-point R of side CD can be written as,

$R(x, y)=\left(\frac{5+5}{2}, \frac{4-1}{2}\right)$

Now equate the individual terms to get,

$x=5$

$y=\frac{3}{2}$

So co-ordinates of $R$ is $\left(5, \frac{3}{2}\right)$

Similarly mid-point S of side DA can be written as,

$\mathrm{S}(x, y)=\left(\frac{5-1}{2}, \frac{-1-1}{2}\right)$

Now equate the individual terms to get,

$x=2$

$y=-1$

So co-ordinates of S is (2,−1)

So we should find the lengths of sides of quadrilateral PQRS.

$\mathrm{PQ}=\sqrt{(2+1)^{2}+\left(4-\frac{3}{2}\right)^{2}}$

$=\sqrt{9+\frac{25}{4}}$

$=\frac{\sqrt{61}}{2}$

$\mathrm{QR}=\sqrt{(2-5)^{2}+\left(4-\frac{3}{2}\right)^{2}}$

$=\sqrt{9+\frac{25}{4}}$

$=\frac{\sqrt{61}}{2}$

$\mathrm{RS}=\sqrt{(5-2)^{2}+\left(\frac{3}{2}+1\right)^{2}}$

$=\sqrt{9+\frac{25}{4}}$

$=\frac{\sqrt{61}}{2}$

$\mathrm{SP}=\sqrt{(2+1)^{2}+\left(-1-\frac{3}{2}\right)^{2}}$

$=\sqrt{9+\frac{25}{4}}$

$=\frac{\sqrt{61}}{2}$

All the sides of quadrilateral are equal.

So now we will check the lengths of the diagonals.

$\mathrm{PR}=\sqrt{(5+1)^{2}+\left(\frac{3}{2}-\frac{3}{2}\right)^{2}}$

$=6$

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

$=5$

All the sides are equal but the diagonals are unequal. Hence ABCD is a rhombus.

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