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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.
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[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.
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