Algebra 2 points

Question

Find all the possible values of k so that (-1,2), (-10,5), and (-4,k) are the vertices of a right triangle

(I want someone to show me the work and also how to do this problem because I have a test tomorrow on this)

Answer 1

Let A = (-1,2), B = (-10,5), and C = (-4,k)

Slope AB = (5-2) / (-10-(-1)) = -1/3

Slope AC = (k+1) / (-4-2) = (k+1) / (-6)

Slope BC = (k-5) / (-4-(-10)) = (k-5) / 6

If the angle with vertex B is a right angle, then (k-5) / 6 = 3. So, k = 23.

If the angle with vertex A is a right angle, then (k+1) / (-6) = 3. So, k = -19.

If the angle with vertex C is a right angle, then (k+1) / (-6) = -6 / (k-5).

So, (k+1)(k-5) = 36

k² - 4k - 41 = 0

k = [4 ± √180] / 2

The possible values of k are 23, -19, (4+√180) / 2, and (4-√180)/2.