The vertices of a triangle are (0, 6), (-5, 3) and (3, 2). The triangle is:
Answer: D
We find the lengths of the three sides using the distance formula.
Side 1 (between (0,6) and (-5,3)): \(\sqrt{(-5-0)^2 + (3-6)^2} = \sqrt{(-5)^2 + (-3)^2} = \sqrt{25+9} = \sqrt{34}\).
Side 2 (between (-5,3) and (3,2)): \(\sqrt{(3-(-5))^2 + (2-3)^2} = \sqrt{8^2 + (-1)^2} = \sqrt{64+1} = \sqrt{65}\).
Side 3 (between (3,2) and (0,6)): \(\sqrt{(0-3)^2 + (6-2)^2} = \sqrt{(-3)^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5\).
Since all three sides have different lengths (\(\sqrt{34}, \sqrt{65}, 5\)), the triangle is scalene.