# Find the value of k so that the area of the triangle with verticesn5 rticles1 A(k + 1,1), B(4, –3) and C(7, –k) is 6 square units.

Δ = 6

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

6 = 1/2(x1(y2−y3) + x2(y3−y1) + x3(y1−y2))

6 = 1/2(k + 1(–3 + k) + 4(–k–1) + 7(1 + 3))

6 = 1/2(k2–2k–3–4k–4 + 28)

k2 – 6k + 9 = 0

k = 3

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