# The coordinates for a rhombus are given as (2a, 0) (0, 2b), (-2a, 0), and (0.-2b). How do you write a plan to prove that the midpoints of the sides of a rhombus determine a rectangle using coordinate geometry?

Jan 12, 2017

#### Explanation:

Let the points of rhombus be $A \left(2 a , 0\right) , B \left(0 , 2 b\right) , C \left(- 2 a , 0\right)$ and $D \left(0. - 2 b\right)$.

Let midpoints of $A B$ be $P$ and its coordinates are $\left(\frac{2 a + 0}{2} , \frac{0 + 2 b}{2}\right)$ i.e. $\left(a , b\right)$. Similarly midpoint of $B C$ is $Q \left(- a , b\right)$; midpoint of $C D$ is $R \left(- a , - b\right)$ and midpoint of $D A$ is $S \left(a , - b\right)$.

It is apparent that while $P$ lies in Q1 (first quadrant), $Q$ lies in Q2, $R$ lies in Q3 and $S$ lies in Q4.

Further, $P$ and $Q$ are reflection of each other in $y$-axis, $Q$ and $R$ are reflection of each other in $x$-axis, $R$ and $S$ are reflection of each other in $y$-axis and $S$ and $P$ are reflection of each other in $x$-axis.

Hence $P Q R S$ or midpoints of the sides of a rhombus $A B C D$ form a rectangle.