Question

Find the value of p for which the points (-5‚1) (1‚p) (4‚-2)are collinear

Answer 1

Answer:

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Answer 2

The value of p is - 1

Step-by-step explanation :

Given -

$\bf (-5,1), (1,p), (4,-2)\;are\;collinear.$

To Find -

The value of p.

Solution :

We know that :

$\bigstar\;\boxed{\bf x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)}$

So,

$\implies \sf (-5,1) = (x1,y1)\\ \\ \\ \implies \sf (1,p) = (x2,y2)\\ \\ \\ \implies \sf (4,-2) = (x3,y3)\\ \\ \\ \bf Now,\\ \\ \\ \implies \sf -5(p + 2) + 1(-2 - 1) + 4(1 - p) = 0\\ \\ \\ \implies \sf -5p - 10 - 3 + 4 - 4p = 0\\ \\ \\ \implies \sf 9p = -9\\ \\ \\ \implies \sf p = -1$

Hence,

The value of p is - 1.