Find the value of p for which the points (-2, 5), (-5, -10) and (p, -13) are collinear.
Answer & explanation
Correct answer: option 3
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) → $-\frac{28}{5}$