Engineering
Mathematics
Introduction to Determinants
Question
If the points (– 1, 3), (2, p) and (5, – 1) are collinear, the value of p is
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Solution
Let A(– 1, 3), B(2, p) and C(5, – 1) be the given points.
Three points are collinear if x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) = 0
Therefore, ⇒ x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) = 0
⇒ – 1(p – ) (– 1) + 2 (– 1 – 3) + 5(3 – p) = 0
⇒ – 1(p + 1) + 2(– 4) + 5(3 – p) = 0
⇒ – p – 1 – 8 + 15 – 5p = 0
⇒ – 6p + 6 = 0
⇒ – 6p = – 6
⇒ p = 1
Hence p = 1
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