Q11 of 157 Page 753

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

A(6, 1), B(8, 2) and C(9, 4) are the vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area of ΔADE.

If the vertices of LABC be A(1, –3), B(4, p) and C(–9, 7) and its area is 15 square units, find the values of p.

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?

Show that the following points are collinear:

A(2, – 2), B(–3, 8) and C(–1, 4)