Find the value of P for which the points A(-1, 3), B(2, P) and C(5, -1) are collinear :

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Solution

The correct option is B 1
Since the given points are collinear, they do not form a
triangle, which means area of the triangle is Zero.

Area of a triangle with vertices $(x_{1}, y_{1})$ ; $(x_{2}, y_{2})$ and $(x_{3}, y_{3})$ is $∣ ∣ ∣ \frac{x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})}{2} ∣ ∣ ∣$
Hence, substituting the points $(x_{1}, y_{1}) = (- 1, 3)$ ; $(x_{2}, y_{2}) = (2, P)$ and $(x_{3}, y_{3}) = (5, - 1)$
in the area formula, we get
$∣ ∣ ∣ \frac{(- 1) (P + 1) + 2 (- 1 - 3) + 5 (3 - P)}{2} ∣ ∣ ∣ = 0$
$=> - P - 1 - 8 + 15 - 5 P = 0$
$=> 6 P = 6$
$=> P = 1$

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