Question

is the triangle with the vertices a(6,3), b(0,7), and c(-8, -5) a right triangle? use the distan ab = \sqrt{\square} ac = \sqrt{\square} bc = \sqrt{208} (simplify your answers. type an integer or a fraction )

Explanation:

Step1: Calculate length of AB

Use distance formula: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
$AB=\sqrt{(0-6)^2+(7-3)^2}=\sqrt{(-6)^2+4^2}=\sqrt{36+16}=\sqrt{52}$

Step2: Calculate length of AC

Apply distance formula to A and C
$AC=\sqrt{(-8-6)^2+(-5-3)^2}=\sqrt{(-14)^2+(-8)^2}=\sqrt{196+64}=\sqrt{260}$

Step3: Verify Pythagorean theorem

Check if $AB^2 + BC^2 = AC^2$ (or permutations)
$AB^2=52$, $BC^2=208$, $AC^2=260$
$52+208=260$

Answer:

Yes, the triangle is a right triangle.