Geometry: Common Core (15th Edition)

We use the distance formula to determine what type of triangle is pictured. The vertices of the triangle are $A(1, 3)$, $B(3, 1)$, and $C(-2, -2)$. The distance formula is given by the following formula: $d = \sqrt {(x_2 - x_1)^2 + (y_2 - y_1)^2}$ Let's determine the lengths of the different sides of the triangle. We'll look at $AB$ first: $AB = \sqrt {(3 - 1)^2 + (1 - 3)^2}$ Simplify within the parentheses: $AB = \sqrt {(2)^2 + (-2)^2}$ Evaluate the exponents: $AB = \sqrt {4 + 4}$ Add what is underneath the radical: $AB = \sqrt {8}$ Rewrite $8$ as the product of a perfect square and another number: $AB = \sqrt {4 • 2}$ Take the square root of $4$ to simplify the radical: $AB = 2 \sqrt {2}$ Let's look at the next side, $BC$: $BC = \sqrt {(-2 - 3)^2 + (-2 - 1)^2}$ Simplify within the parentheses: $BC = \sqrt {(-5)^2 + (-3)^2}$ Evaluate the exponents: $BC = \sqrt {25 + 9}$ Add what is underneath the radical: $BC = \sqrt {34}$ Let's look at $CA$: $CA = \sqrt {(-2 - 1)^2 + (-2 - 3)^2}$ Simplify within the parentheses: $CA = \sqrt {(-3)^2 + (-5)^2}$ Evaluate the exponents: $CA = \sqrt {9 + 25}$ Add what is underneath the radical: $CA = \sqrt {34}$ $BC = CA$; therefore, this triangle is isosceles.