Chapter 13.1, Problem 42WE

To determine

To discover and prove something about the triangle.

Explanation of Solution

Given information:

The triangle has vertices

  $K\left(-3,4\right),M\left(3,1\right)\text{ and }J\left(-6,-2\right)$

Proof:

The triangle can be discover as

Use distance formula to find the value of $KM,KJ\text{ and }MJ$

The distance formula is

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

Therefore,

  $\begin{array}{l}KM=\sqrt{{\left(3-\left(-3\right)\right)}^2+{\left(1-4\right)}^2}\left[\text{since, }K\left(-3,4\right)\text{ and }M\left(3,1\right)\right]\\ KM=\sqrt{{\left(3+3\right)}^2+{\left(1-4\right)}^2}\\ KM=\sqrt{{\left(6\right)}^2+{\left(-3\right)}^2}\\ KM=\sqrt{36+9}\\ KM=\sqrt{45}\\ KM=3\sqrt{5}\end{array}$

  $\begin{array}{l}KJ=\sqrt{{\left(-6-\left(-3\right)\right)}^2+{\left(-2-4\right)}^2}\left[\text{since, }K\left(-3,4\right)\text{ and }J\left(-6,-2\right)\right]\\ KJ=\sqrt{{\left(-6+3\right)}^2+{\left(-2-4\right)}^2}\\ KJ=\sqrt{{\left(-3\right)}^2+{\left(-6\right)}^2}\\ KJ=\sqrt{9+36}\\ KJ=\sqrt{45}\\ KJ=3\sqrt{5}\end{array}$

  $\begin{array}{l}MJ=\sqrt{{\left(-6-3\right)}^2+{\left(-2-1\right)}^2}\left[\text{since, }M\left(3,1\right)\text{ and }J\left(-6,-2\right)\right]\\ MJ=\sqrt{{\left(-9\right)}^2+{\left(-3\right)}^2}\\ MJ=\sqrt{81+9}\\ MJ=\sqrt{90}\\ MJ=3\sqrt{10}\end{array}$

Since, pairs of two sides are equal but the length of one sidesare different

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

So, KMJ is an isosceles triangle

Hence,

The triangleKMJ is an isosceles triangle.