Question

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

Answer 1

Answer:

Given: A=(1, -1),  B=(-4, 2k),  C=(-k, -5)

Area of ΔABC = 24 sq . units

Substituting the values in the above formula,

24 = l 1(2k+5) + (-4)(-5+1) + (-k)(-1-2k) / 2 l

⇒ 24 = l 2k + 5 + 20 - 4 + k + 2k2 / 2 l

⇒48 = 2k2 + 3k + 21

⇒ 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

Step-by-step explanation: