Chapter 1.1, Problem 29AYU

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

51. $A=\left(-2,5\right);B=\left(1,3\right);C=\left(-1,0\right)$

To determine

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

Answer to Problem 29AYU

$\text{d}\left(\text{A, B}\right)\text{ = }\sqrt{\text{13;}}\text{ d}\left(\text{B, C}\right)\text{ = }\sqrt{\text{13}}\text{; d}\left(\text{C, A}\right)\text{ = }\sqrt{\text{26}}\text{;}$

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

$\text{Area = }\frac{\text{13}}{\text{2}}\text{ sq}\text{. units}$

Explanation of Solution

Given:

$\text{A }=\left(-2,5\right),\text{ B }=\left(1,3\right),\text{ C }=\left(-1,0\right)$

Formula used:

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

Calculation:

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

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

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

$= \sqrt{9+4}$

$=\sqrt{13}$

$\text{d}\left(\text{B, C}\right)=\sqrt{{(-1-1)}^2+{(0-3)}^2}$

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

$= \sqrt{4+9}$

$= \sqrt{13}$

$\text{d}\left(\text{C, A}\right)=\sqrt{\left(-1+2\right){)}^2+{(0-5)}^2}$

$= \sqrt{{(1)}^2+{(-5)}^2}$

$= \sqrt{1+25}$

$= \sqrt{26}$

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.

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

$\Rightarrow (\sqrt{13})²+(\sqrt{13})²=(\sqrt{26})²$

$\Rightarrow 13+13=26$

It follows from the converse of the Pythagorean Theorem that triangle $\text{ABC}$ is right triangle and the 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 $= \frac{1}{2}\left(\text{Base}\right)(\text{Height})$

$= \frac{1}{2}(\sqrt{13})(\sqrt{13})$

$= \frac{13}{2}\text{square units}$