Q12 of 157 Page 753

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?

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

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.

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.

Show that the following points are collinear:

A(2, – 2), B(–3, 8) and C(–1, 4)

Show that the following points are collinear:

A(–5, 1), B(5, 5) and C(10, 7)