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

Now we can classify the triangle by calculating the side lengths and then we can determine if it is a right triangle.

Classify by Sides

To classify a triangle by its sides we need to determine if it is scalene, isosceles, or equilateral. To do that, we have to calculate the length of the sides using the Distance Formula. Let's start by finding the distance between

A (0, 0)

and

B (3, 3) .

This will give us the value of

A B .

d_{A B} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

SubstitutePoints

Substitute $(0, 0)$ & $(3, 3)$

d_{A B} = (3 - 0)^{2} + (3 - 0)^{2}

▼

Evaluate right-hand side

SubTerms

Subtract terms

d_{A B} = 3^{2} + 3^{2}

CalcPow

Calculate power

d_{A B} = 9 + 9

AddTerms

Add terms

d_{A B} = 18

SplitIntoFactors

Split into factors

d_{A B} = 9 \cdot 2

RootProd

$a \cdot b = a \cdot b$

d_{A B} = 9 \cdot 2

CalcRoot

Calculate root

d_{A B} = 32

We can find the lengths of the other sides in the same way.

Side	Points	$(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$	Length
$\overline{A B}$	$(0, 0) & (3, 3)$	$(3 - 0)^{2} + (3 - 0)^{2}$	$32$
$\overline{A C}$	$(0, 0) & (- 3, 3)$	$(- 3 - 0)^{2} + (3 - 0)^{2}$	$32$
$\overline{B C}$	$(3, 3) & (- 3, 3)$	$(- 3 - 3)^{2} + (3 - 3)^{2}$	$6$

As we can see, $\overline{A B}$ and $\overline{A C}$ have the same length. Since two of the sides are congruent, the triangle is isosceles.

Right Triangle?

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

Side	Points	$\frac{y _{2} - y _{1}}{x _{2} - x _{1}}$	Slope	Simplified Slope
$\overline{C A}$	$(- 3, 3) & (0, 0)$	$\frac{0 - 3}{0 - ( - 3 )}$	$\frac{- 3}{3}$	$- 1$
$\overline{A B}$	$(0, 0) & (3, 3)$	$\frac{3 - 0}{3 - 0}$	$\frac{3}{3}$	$1$

Since $1$ and $- 1$ are opposite reciprocals, we know that $\overline{A C}$ is perpendicular to $\overline{A B} .$ Therefore, $△ A B C$ is in fact a right triangle.

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