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.

We have the following triangle,

Let the vertices of the triangle be A(1,-3), B(4,p) and C(-9,7) be denoted by A(x₁,y₁), B(x₂,y₂) and C(x₃,y₃).

Area of the triangle where the vertices are given is,

Area = 1/2 [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

And Area is given as 15 sq. units.

So,

1/2 [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)] = �15

⇒ [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)] = �30

Substituting corresponding values in the equation, we get

[1(p – 7) + 4(7 – (-3)) + (-9)(-3 – p) = �30

⇒ [p – 7 + 40 + 27 + 9p = �30

⇒ 60 + 10p = �30

⇒ 10p = �30 – 60

⇒ p = ( �30 – 60)/10

So we have two solution, lets explore them.

I case: p = (30 – 60)/10

⇒ p = -30/10

⇒ p = -3

II case: p = (-30 – 60)/10

⇒ p = -90/10

⇒ p = -9

Hence, p has two values, they are -3 and -9.