Question

you do: check for understanding
consider a triangle with vertices at the following points: x(-7,2), y(-6,4), and z(-3,1).
use the lengths of the triangle to classify it as acute, obtuse, or right.

Explanation:

Step1: Calculate side lengths

First, use the distance formula $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ for each side:

• $XY$: $\sqrt{(-6-(-7))^2+(4-2)^2}=\sqrt{1^2+2^2}=\sqrt{1+4}=\sqrt{5}$
• $YZ$: $\sqrt{(-3-(-6))^2+(1-4)^2}=\sqrt{3^2+(-3)^2}=\sqrt{9+9}=\sqrt{18}$
• $XZ$: $\sqrt{(-3-(-7))^2+(1-2)^2}=\sqrt{4^2+(-1)^2}=\sqrt{16+1}=\sqrt{17}$

Step2: Square all side lengths

To avoid square roots, use squared lengths:

• $(XY)^2 = 5$
• $(YZ)^2 = 18$
• $(XZ)^2 = 17$

Step3: Compare to Pythagorean theorem

Identify the largest squared length ($18$), then check:
$5 + 17 = 22$, and $22 > 18$
When $a^2 + b^2 > c^2$ (where $c$ is the longest side), the triangle is acute.

Answer:

The triangle is acute.