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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*.

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#### Solution

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[x_{1}(y_{2}−y_{3})+x_{2}(y_{3}−y_{1})+x_{3}(y_{1}−y_{2})]=0

Here, x_{1}=−3, y_{1}=9, x_{2}=a, y_{2}=b and x_{3}=4, y_{3}=−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.

Concept: Area of a Triangle

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