Problem 27 - 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. $$ (-2,2,4),(-2,2,6),(-2,4,8) $$

Step-by-Step Solution

Answer

The lengths of the sides of the triangle are 2, \(2\sqrt{2}\) and \(2\sqrt{2}\). The triangle is a right triangle but not an isosceles triangle.

Step 1: Calculate the distances between vertices

First, calculate the distances (lengths of sides) between each pair of points using the distance formula. Let's denote the three vertices as A(-2,2,4), B(-2,2,6), C(-2,4,8). The lengths will be: \[ AB= \sqrt{(-2-(-2))^2+(2-2)^2+(6-4)^2}=2 \] \[ BC= \sqrt{(-2-(-2))^2+(4-2)^2+(8-6)^2}=2\sqrt{2} \] \[ AC= \sqrt{(-2-(-2))^2+(4-2)^2+(8-4)^2}=2\sqrt{2} \]

Step 2: Determine the type of triangle

Now that each of the lengths is determined, use these lengths to figure out the triangle's type. Here, AB^2 + AC^2 = BC^2, so it’s a right triangle. Also, as AB ≠ AC ≠ BC, it’s not an isosceles triangle.