If the vertices of a triangle are (1,-3), (4, p) and (-9, 7) and its area is 15 sq. units, find the value(s) of p.

Question

Philoid · Accepted Answer

Let A(1, −3), B(4, p) and C(−9, 7) be the vertices of the ∆ABC.

Area of the triangle having vertices (x₁,y₁), (x₂,y₂) and (x₃,y₃)

= |x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)|

Given that area of ∆ABC = 15

∴ 15 = |1(p – 7) +4 (7 – (-3)) - 9(-3 - p)|

∴ 30 = |p – 7 + 40 + 27 + 9p|

∴ 30 = | 10p + 60|

∴ 10p + 60 = ± 30

Taking positive sign,

10p + 60 = 30

p = -3

Taking negative sign,

10p + 60 = - 30

p = -9

Hence, the value of p are -3 and -9