Describing Triangles

Let's start by drawing the triangle in a coordinate plane.

Classify by Sides

To classify a triangle by its sides means to classify it as either scalene, isosceles, or equilateral. To do that we have to calculate the length of all sides using the distance formula. Let's begin by finding the distance between $A(1,9)$ and $B(4,8).$ This will give us $AB.$ The length of $\overline{AB}$ is $\sqrt{10}.$ We can find the rest of the sides using the same method.

Side	Points	$\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$	Length
$\overline{AB}$	$A(1,9)\&B(4,8)$	$\sqrt{{\left(4-1\right)}^2+{\left(8-9\right)}^2}$	$\sqrt{10}$
$\overline{AC}$	$A(1,9)\&C(2,5)$	$\sqrt{{\left(2-1\right)}^2+{\left(5-9\right)}^2}$	$\sqrt{17}$
$\overline{BC}$	$B(4,8)\&C(2,5)$	$\sqrt{{\left(2-4\right)}^2+{\left(5-8\right)}^2}$	$\sqrt{13}$

As we can see, each side of the triangle has a different length, so $△ABC$ is a scalene triangle.

Right Triangle?

In our diagram, we see that $\angle A$ and $\angle C$ are acute angles. Therefore, if $△ABC$ is a right triangle, the right angle must be $\angle B.$ To determine if this is the case, we will first calculate the slope of $\overline{AB}$ and $\overline{BC}$ by using the Slope Formula.

Side	Points	$\frac{y_2-y_1}{x_2-x_1}$	Slope	Simplified Slope
$\overline{AB}$	$A(1,9)\&B(4,8)$	$\frac{8-9}{4-1}$	$\frac{\text{-}1}{3}$	$\text{-}\frac{1}{3}$
$\overline{BC}$	$B(4,8)\&C(2,5)$	$\frac{5-8}{2-4}$	$\frac{\text{-}3}{\text{-}2}$	$\frac{3}{2}$

Since $\text{-}\frac{1}{3}$ and $\frac{3}{2}$ are not opposite reciprocals, $\overline{AB}$ is not perpendicular to $\overline{BC}.$ Therefore, $△ABC$ is not a right triangle.