Chapter 1, Problem 32RE

Show that the points $A=\left(-2,0\right)$, $B=\left(-4,4\right)$, and $C=\left(8,5\right)$ are the vertices of a right triangle in two ways:

(a) By using the converse of the Pythagorean Theorem

(b) By using the slopes of the lines joining the vertices

To determine

To show: The points $A = \left(-2, 0\right), B = \left(-4, 4\right)\text{and}C = \left(8, 5\right)$ are the vertices of a right triangle in two ways: (a) By using the converse of the Pythagorean Theorem (b) By using the slopes of the lines joining the vertices.

Answer to Problem 32RE

Explanation of Solution

If these three points are to make a right triangle, then that triangle must have a hypotenuse.

There are only three possibilities for the hypotenuse: the line $AB$, the line $BC$, or the line $AC$.

$A{B}^2={\left(-4-\left(-2\right)\right)}^2+{\left(4-0\right)}^2=4+16=20$

$A{C}^2={\left(8-\left(-2\right)\right)}^2+{\left(5-0\right)}^2=100+25=125$

$B{C}^2={\left(8-\left(-4\right)\right)}^2+{\left(5-4\right)}^2=144+1=145$

Here $B{C}^2=A{B}^2+A{C}^2$

Therefore the points are the vertices of a right triangle by Pythagorean Theorem.

(b) By using the slopes of the lines joining the vertices

Slope between $A$ and $B$

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{4-0}{-4+2}=\frac{4}{-2}=-2$

Slope between $B$ and $C$

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{5-4}{8-\left(-4\right)}=\frac{1}{12}$

Slope between $C$ and $A$

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{5-0}{8-\left(-2\right)}=\frac{5}{10}=\frac{1}{2}$

Multiplication of slope between $AB$ and $CA$ is $-2\times \frac{1}{2}=-1$

Therefore the points are the vertices of a right triangle by using the slopes of the lines joining the vertices.