Question

If the points P(–3, 9), Q(a, b) and R(4, – 5) are collinear and a + b = 1, find the values of a and b.

Answer 1

The given points are P(–3, 9), Q(a, b) and R(4, –5).

Since the given points are collinear, the area of triangle PQR is 0.

∴1/2[x₁(y₂−y₃)+x₂(y₃−y₁)+x₃(y₁−y₂)]=0

Here, x₁=−3, y₁=9, x₂=a, y₂=b and x₃=4, y₃=−5

∴1/2[−3(b+5)+a(−5−9)+4(9−b)]=0

⇒−3b−15−14a+36−4b=0

⇒−7b−14a+21=0

⇒2a+b=3 ...(1)

Given:
a+b=1 ...(2)

Subtracting equation (2) from (1), we get:

a = 2

Putting a = 2 in (2), we get:

b = 1 − 2 = − 1

Thus, the values of a and b are 2 and −1, respectively.