Chapter 9.5, Problem 16PSB

To determine

To show: Parallelogram with given vertices is not a rhombus.

Explanation of Solution

Given information:Here, we have given that vertices of parallelogram are $(-1,-3)$ , $(2,1)$ , $(3,-2)$ and $(-2,0)$ .

Formula used: Distance formula: d = $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

Proof: Let vertices be A $(-1,-3)$ , B $(2,1)$ , C $(3,-2)$ and D $(-2,0)$ .

A quadrilateral is rhombus when lengths all sides is equal.

So, AB = $\sqrt{{[2-(-1)]}^2+{[1-(-3)]}^2}$

  $\begin{array}{l}\Rightarrow \sqrt{{(3)}^2+{(4)}^2}\\ \Rightarrow \sqrt{9+16}\\ \Rightarrow \sqrt{25}\\ \Rightarrow 5\end{array}$

Now, BC = $\sqrt{{(3-2)}^2+{(-2-1)}^2}$

  $\begin{array}{l}\Rightarrow \sqrt{{(1)}^2+{(-3)}^2}\\ \Rightarrow \sqrt{1+9}\\ \Rightarrow \sqrt{10}\end{array}$

As, AB is not equal to BC.

  $\Rightarrow$ ABCD is not a rhombus.