For what value of k(k 0) is the area of the triangle with vertices (–2,5) and (k, –4) and (2k + 1,10) equal to 53 square units?

Question

For what value of k(k > 0) is the area of the triangle with vertices (–2,5) and (k, –4) and (2k + 1,10) equal to 53 square units?

Philoid · Accepted Answer

Given the area of triangle, Δ = 53

⇒ Δ = 1/2{x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)}

53 = 1/2{–2(–4–10) + k(10–5) + 2k + 1(5 + 4)}

53 = 1/2{28 + 5k + 9(2k + 1)}

106 = (28 + 5k + 18k + 9)

37 + 3k = 106

23k = 69

k = 3