Problem 28 - College Algebra and Calculus: An Applied Approach

Question

Find the lengths of the sides of the triangle with the given vertices, and determine whether the triangle is a right triangle, an isosceles triangle, or neither of these. $$ (5,0,0),(0,2,0),(0,0,-3) $$

Step-by-Step Solution

Answer

The lengths of the sides of the triangle are \(\sqrt{29}\), \(\sqrt{34}\), and \(\sqrt{13}\) units, respectively. The triangle is neither a right triangle nor an isosceles triangle.

Step 1: Compute the lengths of the sides using the distance formula

Calculate the lengths of the sides, which are the distances between each pair of vertices: \(AB = \sqrt{(5-0)^2 + (0-2)^2 + (0-0)^2} = \sqrt{25 + 4} = \sqrt{29}\),\(AC = \sqrt{(5-0)^2 + (0-0)^2 + (0+3)^2} = \sqrt{25 + 9} = \sqrt{34}\),and \(BC = \sqrt{(0-0)^2 + (2-0)^2 + (0+3)^2} = \sqrt{4 + 9} = \sqrt{13}\).

Step 2: Check if the triangle is a right triangle

Apply the Pythagorean Theorem (a² + b² = c²) with the longest side as 'c'. After checking, \(AB^2 + BC^2 = 29 + 13 = 42\) is not equal to \(AC^2 = 34\), so the triangle is not a right triangle.

Step 3: Check if the triangle is an isosceles triangle

An isosceles triangle has two sides of the same length. Comparing the previously calculated side lengths, none of the sides are of equal length, so the triangle is not an isosceles triangle.