# If the points <sp

Given points P(-3, 9), Q(a, b) and R(4, -5) are collinear

And a + b = 1 … (1)

We have to find the values of a and b.

Since the given points are collinear, the area of the triangle formed by them must be 0.

We know that the area of Triangle =

= 0

[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = 0

[-3(b – (-5)) + a(-5 – 9) + 4(9 – b)] = 0

[-3(b + 5) -14a + 4(9 – b)] = 0

[-3b – 15 - 14a + 36 – 4b] = 0

-14a - 7b + 21 = 0

-14a – 7b = -21

Dividing by -7, we get

2a + b = 3

b = 3 – 2a … (2)

Substituting (2) in (1),

a + (3 – 2a) = 1

-a + 3 = 1

-a = -2

a = 2

Substituting value of a in (1),

2 + b = 1

b = -1

The values of a and b are 2 and -1 respectively.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses