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 length between $A(1,2)$ and $B(6,2).$ The length of $\overline{AB}$ is $5.$ 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
$\overline{AB}$	$A(1,2)\&B(6,2)$	$\sqrt{{\left(6-1\right)}^2+{\left(2-2\right)}^2}$	$5$
$\overline{AC}$	$A(1,2)\&C(1,7)$	$\sqrt{{\left(1-1\right)}^2+{\left(7-2\right)}^2}$	$5$
$\overline{BC}$	$B(6,2)\&C(1,7)$	$\sqrt{{\left(1-6\right)}^2+{\left(7-2\right)}^2}$	$5\sqrt{2}$

As we can see, $\overline{AB}$ and $\overline{AC}$ have the same length, so $△ABC$ is an isosceles triangle.

Classify by Angles

In our diagram, we see that $\angle B$ and $\angle C$ are acute angles. We now need to determine what type of angle $\angle A$ is. To determine this, we will first calculate the slope of $\overline{AB}$ and $\overline{AC}$ by using the slope formula.

Side	Points	$\frac{y_2-y_1}{x_2-x_1}$	Slope	Simplified Slope
$\overline{AB}$	$A(1,2)\&B(6,2)$	$\frac{2-2}{6-1}$	$\frac{0}{5}$	$0$
$\overline{AC}$	$A(1,2)\&C(1,7)$	$\frac{7-2}{1-1}$	$\frac{5}{0}$	Undefined

Looking at the table, we can deduce that $\overline{AB}$ is a horizontal segment and $\overline{AC}$ is a vertical segment. This means that the intersection of $\overline{AB}$ and $\overline{AC}$ forms a right angle. Thus, we know that $△ABC$ is a right triangle.