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Chapter 32: Problem 1039

Show that the points \(\mathrm{A}(-2,4), \mathrm{B}(-3,-8)\), and \(\mathrm{C}(2,2)\) are vertices of a right triangle.

Short Answer

Expert verified
The points A(-2, 4), B(-3, -8), and C(2, 2) form a right triangle because the distances between each pair of points satisfy the Pythagorean theorem. Specifically, we found that \(AC^2 + BC^2 = AB^2 (\sqrt{20}^2 + \sqrt{125}^2 = \sqrt{145}^2)\).

Step by step solution

01

Calculate the distance between each pair of points

Use the distance formula to calculate the distance between points A and B, A and C, and B and C: - Distance between A and B (AB): \(AB = \sqrt{(-2 - (-3))^2 + (4 - (-8))^2}\) - Distance between A and C (AC): \(AC = \sqrt{(-2 - 2)^2 + (4 - 2)^2}\) - Distance between B and C (BC): \(BC = \sqrt{(-3 - 2)^2 + (-8 - 2)^2}\)
02

Simplify the distances

Simplify the distances calculated in step 1: - Distance between A and B (AB): \(AB = \sqrt{(1)^2 + (12)^2} = \sqrt{145}\) - Distance between A and C (AC): \(AC = \sqrt{(-4)^2 + (2)^2} = \sqrt{20}\) - Distance between B and C (BC): \(BC = \sqrt{(-5)^2 + (-10)^2} = \sqrt{125}\)
03

Check for Pythagorean theorem

Check if any combination of the three distances satisfies the Pythagorean theorem: - Check if \(AB^2 + AC^2 = BC^2\) \(145 + 20 = 125\) \(165 \neq 125\) - Check if \(AB^2 + BC^2 = AC^2\) \(145 + 125 = 20\) \(270 \neq 20\) - Check if \(AC^2 + BC^2 = AB^2\) \(20 + 125 = 145\) \(145 = 145\) Since the last combination (AC虏 + BC虏 = AB虏) satisfies the Pythagorean theorem, we can conclude that the points A(-2, 4), B(-3, -8), and C(2, 2) indeed form a right triangle.

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