Chapter 9.5, Problem 17PSB

To determine

To show:The Triangle is isosceles.

Explanation of Solution

Given information:Here, we have given that vertices of triangle are $(-2,1)$ , $(5,5)$ and $(-1,-7)$ .

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

Proof: Let vertices be A $(-2,1)$ , B $(5,5)$ and C $(-1,-7)$ .

A triangle is isosceles when lengths of any of its two is equal.

So, AB = $\sqrt{{[5-(-2)]}^2+{(5-1)}^2}$

  $\begin{array}{l}\Rightarrow \sqrt{{(7)}^2+{(4)}^2}\\ \Rightarrow \sqrt{49+16}\\ \Rightarrow \sqrt{65}\end{array}$

Now, BC = $\sqrt{{(-1-5)}^2+{(-7-5)}^2}$

  $\begin{array}{l}\Rightarrow \sqrt{{(-6)}^2+{(-12)}^2}\\ \Rightarrow \sqrt{36+144}\\ \Rightarrow \sqrt{180}\end{array}$

Similarly, AC = $\sqrt{{[-1-(-2)]}^2+{(-7-1)}^2}$

  $\begin{array}{l}\Rightarrow \sqrt{{(1)}^2+{(-8)}^2}\\ \Rightarrow \sqrt{1+64}\\ \Rightarrow \sqrt{65}\end{array}$

Now, as AB = AC

  $\Rightarrow \Delta ABC$ is isosceles.