Chapter 1.1, Problem 33AYU

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

55. $A=\left(4,-\text{3}\right);B=\left(0,-3\right);C=\left(4,\text{ 2}\right)$

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 33AYU

$\text{d}\left(\text{A, B}\right)\text{ = }\sqrt{\text{16}}\text{; d}\left(\text{B, C}\right)\text{ = }\sqrt{\text{41}}\text{; d}\left(\text{C, A}\right)\text{ =}\sqrt{\text{25}}$

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

$\text{Area = 10 sq}\text{. units}$

Explanation of Solution

Given:

$\text{A }=\left(4,-3\right),\text{ B }=\left(0,-3\right),\text{ C }=\left(4,2\right)$

Formula used:

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

Calculation:

$\text{d}\left(\text{A, B}\right)=\sqrt{{(0-4)}^2+{(-3-\left(-3\right))}^2}$

$= \sqrt{{(-4)}^2+{(-3+3)}^2}$

$= \sqrt{{(-4)}^2+{(0)}^2}$

$= \sqrt{16+0}$

$= \sqrt{16}$

$\text{d}\left(\text{B, C}\right)=\sqrt{{(4-0)}^2+{(2-\left(-3\right))}^2}$

$= \sqrt{{(4)}^2+{(2+3)}^2}$

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

$= \sqrt{16+25}$

$= \sqrt{41}$

$\text{d}\left(\text{C, A}\right)=\sqrt{{(4-4)}^2+{(-3-2)}^2}$

$= \sqrt{{(0)}^2+{(-5)}^2}$

$= \sqrt{0+25}$

$= \sqrt{25}$

To prove 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.

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

$\Rightarrow (\sqrt{16})²+(\sqrt{25})²=(\sqrt{41})²$

$\Rightarrow 16+25=41$

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

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

Its area is

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

$= \frac{1}{2}(\sqrt{16})(\sqrt{25})$

$= \frac{1}{2}\left(4\right)\left(5\right)$

$= \frac{20}{2}$

$= \text{10}$ square units