Chapter 1, Problem 49RE

(a)

To determine

To show that 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 using the converse of the Pythagorean Theorem.

Explanation of Solution

Given information:

The points $A=\left(-2,0\right),B=\left(-4,4\right),\text{ and }C=\left(8,5\right)$ .

Concept Used:

Pythagorean Theorem states that in a right triangle, the sum of the squares of the legs is equal to the square of the hypotenuse.

Calculation:

In order to show that the given points are the vertices of a right triangle, first find the length of the sides connecting the points using the distance formula. So, the distance between the points A and B is given by

  $\begin{array}{c}AB=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(-4-\left(-2\right)\right)}^2+{\left(4-0\right)}^2}\\ =\sqrt{{\left(-4+2\right)}^2+{\left(4\right)}^2}\\ =\sqrt{{\left(-2\right)}^2+16}\\ =\sqrt{4+16}\\ =\sqrt{20}\end{array}$

Similarly, the distance between the points B and C is given by

  $\begin{array}{c}BC=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(8-\left(-4\right)\right)}^2+{\left(5-4\right)}^2}\\ =\sqrt{{\left(8+4\right)}^2+{\left(1\right)}^2}\\ =\sqrt{{\left(12\right)}^2+1}\\ =\sqrt{144+1}\\ =\sqrt{145}\end{array}$

And, the distance between the points A and C is given by

  $\begin{array}{c}AC=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(8-\left(-2\right)\right)}^2+{\left(5-0\right)}^2}\\ =\sqrt{{\left(8+2\right)}^2+{\left(5\right)}^2}\\ =\sqrt{{\left(10\right)}^2+25}\\ =\sqrt{100+25}\\ =\sqrt{125}\end{array}$

Now using the converse of the Pythagorean Theorem, it gives

  $\begin{array}{c}{\left(AB\right)}^2+{\left(AC\right)}^2={\left(BC\right)}^2\\ {\left(\sqrt{20}\right)}^2+{\left(\sqrt{125}\right)}^2={\left(\sqrt{145}\right)}^2\\ 20+125=145\\ 145=145\text{True statement}\end{array}$

It implies that the given points are the vertices of a right triangle.

(b)

To determine

To show that 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 using the slopes of the lines joining the vertices.

Explanation of Solution

Given information:

The points $A=\left(-2,0\right),B=\left(-4,4\right),\text{ and }C=\left(8,5\right)$ .

Concept Used:

The slope of a line connecting the points $\left(x_1,y_1\right),\text{ and }\left(x_2,y_2\right)$ is given by the formula

  $\text{Slope }\left(m\right)=\frac{y_2-y_1}{x_2-x_1}$ .

Calculation:

In order to show that the given points are the vertices of a right triangle, find the slopes of the line connecting the given points. Then use the concept that the slope of two perpendicular lines is negative reciprocal of each other.

The slope of the line joining the points A and B is given by

  $\begin{array}{c}\text{Slope }\left(AB\right)=\frac{y_2-y_1}{x_2-x_1}\\ =\frac{4-0}{-4-\left(-2\right)}\\ =\frac{4}{-4+2}\\ =\frac{4}{-2}\\ =-2\end{array}$

Similarly, the slope of the line joining the points B and C is given by

  $\begin{array}{c}\text{Slope }\left(BC\right)=\frac{y_2-y_1}{x_2-x_1}\\ =\frac{5-4}{8-\left(-4\right)}\\ =\frac{9}{8+4}\\ =\frac{9}{12}\\ =\frac{3}{4}\end{array}$

And, the slope of the line joining the points A and C is given by

  $\begin{array}{c}\text{Slope }\left(AC\right)=\frac{y_2-y_1}{x_2-x_1}\\ =\frac{5-0}{8-\left(-2\right)}\\ =\frac{5}{8+2}\\ =\frac{5}{10}\\ =\frac{1}{2}\end{array}$

Here observe that the slopes of the line AB and AC are negative reciprocal of each other, so these lines are perpendicular to each other.

Thus, the given points are the vertices of a right triangle.