Find the value of p for which the points (-2, 5), (-5, -10) and (p, -13) are collinear.

$-\frac{28}{5}$

We have point A , B & C points (-2, 5), (-5, -10) and (p, -13) RESPECTIVELY ,

A , B & C are collinear

So,  slope of AB = slope of BC

\(\frac{Y2 - Y1 }{X2 - X1}\) = \(\frac{Y3 - Y2}{X3 - X2}\)

\(\frac{- 10 - 5 }{- 5 - (-2)}\) = \(\frac{-13 - (-10)}{P -(-5)}\)

\(\frac{- 15 }{-3}\) = \(\frac{-3}{P + 5}\)

\(\frac{5 }{3}\) = \(\frac{-1}{P + 5}\)

5P + 25 = - 3

5P = 28

P = \(\frac{28 }{5}\)

The correct answer is Option (3) → $-\frac{28}{5}$