Chapter 1.1, Problem 32AYU

In Problems 51-56, plot each point and form the triangle $ABC$. Verify that the triangle is a right triangle. Find its area.

54. $A=\left(-6,\text{ 3}\right);B=\left(3,-5\right);C=\left(-1,\text{ 5}\right)$

To determine

To find: To plot the given points and form the triangle, length of the sides of the triangle, to verify that the triangle is a right triangle and find the area of the triangle.

Answer to Problem 32AYU

$\text{d}\left(\text{A, B}\right)\text{ = }\sqrt{\text{145}}\text{; d}\left(\text{B, C}\right)\text{ = }\sqrt{\text{116}}\text{; d}\left(\text{C, A}\right)\text{ = }\sqrt{\text{29}}$

$\text{ABC is a right angle triangle}$

$\text{Area = 29 sq}\text{. units}$

Explanation of Solution

Given:

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

Formula used:

$\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

Calculation:

$\text{d}\left(\text{A, B}\right)=\sqrt{{(3-\left(-6\right))}^2+{(-5-3)}^2}$

$= \sqrt{{(3+6)}^2+{(-8)}^2}$

$= \sqrt{{(9)}^2+{(-8)}^2}$

$= \sqrt{81+64}$

$= \sqrt{145}$

$\text{d}\left(\text{B, C}\right)\text{=}\sqrt{{(-1-3)}^2+{(5-\left(-5\right))}^2}$

$=\sqrt{{(-4)}^2+{(5+5)}^2}$

$= \sqrt{{(-4)}^2+{(10)}^2}$

$= \sqrt{16+100}$

$= \sqrt{116}$

$\text{d}\left(\text{C, A}\right)= \sqrt{{(-1-\left(-6\right))}^2+{(5-3)}^2}$

$= \sqrt{{(-1+6)}^2+{(2)}^2}$

$= \sqrt{{(5)}^2+{(2)}^2}$

$= \sqrt{25+4}$

$= \sqrt{29}$

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.

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

$\Rightarrow (\sqrt{29})²+(\sqrt{116})²=(\sqrt{145})²$

$\Rightarrow 29+116=145$

It follows from the converse of the Pythagorean Theorem that triangle $\text{ABC}$ is right triangle and right angle is at vertex $\text{C}$.

Because the right angle is at vertex $\text{C}$, the sides $\text{AC}$ and $\text{BC}$ form the base and height of the triangle, respectively.

Its area is

Area $= \frac{1}{2}\left(\text{Base}\right)\left(\text{Height}\right)$

$= \frac{1}{2}(\sqrt{29})(\sqrt{116})$

$= \frac{1}{2}(\sqrt{29})(\sqrt{4*29})$

$= \frac{1}{2}(\sqrt{29})(2\sqrt{29})$

$= \text{29}$ square units