Chapter 1.1, Problem 30AYU

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

52. $A=\left(-2,5\right);B=\left(12,3\right);C=\left(10,-11\right)$

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 30AYU

$\text{d}\left(\text{A, B}\right)\text{ = }\sqrt{\text{200}}\text{; d}\left(\text{B, C}\right)\text{ = }\sqrt{\text{200}}\text{; d}\left(\text{C, A}\right)\text{ = }\sqrt{\text{400}}$

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

$\text{Area = 100 sq}\text{. units}$

Explanation of Solution

Given:

$\text{A }=\left(-2,5\right),\text{ B }=\left(12,3\right),\text{ C }=(10,\text{ <mo>−</mo><mn>11</mn><mo>)</mo>}$

Formula used:

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

Calculation:

$\text{d}\left(\text{A, B}\right)=\sqrt{{(12-\left(-2\right))}^2+{(3-5)}^2}$

$= \sqrt{{(12+2)}^2+{(-2)}^2}$

$= \sqrt{{(14)}^2+{(-2)}^2}$

$= \sqrt{196+4}$

$= \sqrt{200}$

$\text{d}\left(\text{B, C}\right)=\sqrt{{(10-12)}^2+{(-11-3)}^2}$

$= \sqrt{{(2)}^2+{(-14)}^2}$

$= \sqrt{4+196}$

$= \sqrt{200}$

$\text{d}\left(\text{C, A}\right)=\sqrt{(10-{(-2)}^2+{(-11-5)}^2}$

$= \sqrt{{(10+2)}^2+{(-16)}^2}$

$= \sqrt{{(12)}^2+{(-16)}^2}$

$= \sqrt{144+256}$

$= \sqrt{400}$

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.

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

$\Rightarrow (\sqrt{200})²+(\sqrt{200})²=(\sqrt{400})²$

$\Rightarrow 200+200=400$

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

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

Its area is

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

$= \frac{1}{2}(\sqrt{200})(\sqrt{200})$

$= \frac{200}{2}$

$= \text{100 square units}$