Find the values of k so that the area of the triangle with vertices (1, -1), (-4, 2k), and (-k, -5) is 24 sq. units.

Given area of triangle whose vertices are (1, -1), (-4, 2k), and (-k, -5) is 24 sq units
We know that the area of triangle = 1/2[x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)]
∴ 24 = 1/2[1(2k - (-5)) - 4(-5 - (-1)) + (-k)(-1 - 2k)]
⇒ 48 = [2k + 5 + (-4)(-4) + k(1 + 2k)]
⇒ 48 = [2k + 5 + 16 + k + 2k²]
⇒ 2k² + 3k + 5 + 16 - 48 = 0
⇒ 2k² + 3k - 27 = 0
⇒ 2k² + 9k - 6k - 27 = 0
⇒ k(2k + 9) - 3(2k + 9) = 0
⇒ (k - 3)(2k + 9) = 0
⇒ k - 3 = 0 or 2k + 9 = 0
⇒ k = 3 or k = -9/2