Q. 52 - Precalculus Enhanced with Graphing Utilities

Question

Plot each point and form the triangle ABC. Verify that the triangle is a right triangle. Find its area. 

$A=\left(-6,3\right); B=\left(3,-5\right) ; C=\left(-1,5\right)$

Step-by-Step Solution

Answer

Part (a). The triangle are 

Part (b). The length of sides are $d\left(A,B\right)=\sqrt{145}, d\left(B,C\right)=\sqrt{116}, d\left(A,C\right)=\sqrt{29}$

Part (c). The triangle is right triangle: ${\left(\sqrt{145}\right)}^2={\left(\sqrt{29}\right)}^2+{\left(\sqrt{116}\right)}^2$

Part (d). $\mathrm{Area}=29 \mathrm{square} \mathrm{units}$

Step 1. Given information.

Plot each point and form the triangle ABC. Verify that the triangle is a right triangle. Find its area.

$A=\left(-6,3\right); B=\left(3,-5\right) ; C=\left(-1,5\right)$

Part (a) Step 1. Plot each point and form the triangle ABC.

Below figure shows the points A, B, C and the triangle ABC.

Part (b) Step 1. Find the length of each side of the triangle.

To find the length of each side of the triangle, we use the distance formula. 

$$\begin{gathered} d(A,B)=\sqrt{{\left(-6-3\right)}^2+{\left(3-(-5)\right)}^2}=\sqrt{{\left(-9\right)}^2+8^2}=\sqrt{81+64}=\sqrt{145} \\ d(B,C)=\sqrt{{\left(3-(-1)\right)}^2+{\left(-5-5\right)}^2}=\sqrt{4^2+{(-10)}^2}=\sqrt{16+100}=\sqrt{116} \\ d(A,C)=\sqrt{{\left(-6-(-1)\right)}^2+{\left(3-5\right)}^2}=\sqrt{{(-5)}^2+{(-2)}^2}=\sqrt{25+4}=\sqrt{29} \end{gathered}$$

Part (c) Step 1. Verify that the triangle is a right triangle.

To show that the triangle is a right triangle, we need to show that the sum of the squares of the lengths of two of the sides equals the square of the length of the third side. 

From the above graph, we can see that AB is longest side, hence the hypotenuse.

${\left[d\left(A,B\right)\right]}^2={\left[d\left(B,C\right)\right]}^2+{\left[d\left(A,C\right)\right]}^2$

Using the results from part (b),

$$\begin{gathered} {\left(\sqrt{145}\right)}^2={\left(\sqrt{116}\right)}^2+{\left(\sqrt{29}\right)}^2 \\ 145=116+29 \\ 145=145 \end{gathered}$$

It follows from the converse of the Pythagorean Theorem that triangle ABC is a right triangle.

Part (d) Step 1. Find the area of the triangle.

Because the right angle is at vertex C, the sides AC and BC form the base and height of the triangle. Its area is 

$\mathrm{Area}=\frac{1}{2}\left(\mathrm{Base}\right)\left(\mathrm{height}\right)=\frac{1}{2}\sqrt{29}·\sqrt{116}=\frac{\sqrt{29}\sqrt{4·29}}{2}=\frac{\sqrt{4}{\left(\sqrt{29}\right)}^2}{2}=29 \mathrm{square} \mathrm{units}$