Chapter 13.1, Problem 27WE

To determine

To show that the triangle is isosceles triangle

Explanation of Solution

Given information:

The triangle has vertices

  $A\left(-3,4\right),M\left(3,1\right),\text{ and }Y\left(0,2\right)$

Proof:

The triangle is isosceles triangle can be show as

In geometry, an isosceles triangle is a triangle that has two sides of equal length.

So, use distance formula to show that $AM=AY$

  $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Let, $A\left(x_1,y_1\right)\text{ be }A\left(-3,4\right),M\left(x_2,y_2\right)\text{ be }M\left(3,1\right)\text{ and }Y\left(x_3,y_3\right)\text{ be }Y\left(0,-2\right)$

Therefore,

  $AM=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Put the values of $x_1=-3,y_1=4,x_2=3\text{ and }{y}_2=1$ on the above equation

  $\begin{array}{l}AM=\sqrt{{\left(3-\left(-3\right)\right)}^2+{\left(1-4\right)}^2}\\ AM=\sqrt{{\left(3+3\right)}^2+{\left(-3\right)}^2}\\ AM=\sqrt{{\left(6\right)}^2+{\left(-3\right)}^2}\\ AM=\sqrt{36+9}\\ AM=\sqrt{45}\end{array}$

And,

  $AY=\sqrt{{\left(x_3-x_1\right)}^2+{\left(y_3-y_1\right)}^2}$

Put the values of $x_1=-3,y_1=4,x_3=0\text{ and }{y}_3=-2$ on the above equation

  $\begin{array}{l}AY=\sqrt{{\left(0-\left(-3\right)\right)}^2+{\left(-2-4\right)}^2}\\ AY=\sqrt{{\left(0+3\right)}^2+{\left(-2-4\right)}^2}\\ AY=\sqrt{{\left(3\right)}^2+{\left(-6\right)}^2}\\ AY=\sqrt{9+36}\\ AY=\sqrt{45}\end{array}$

Hence,

  $AM=AY$ and triangle AMY is isosceles triangle.