Question

The value of p for which the points (–1, 3), (2, p) and (5, –1) are collinear is

(a) –1 

(b) 2 

(c) \(\frac13\) 

(d) 1

Answer 1

(d) 1

For collinearity of points A (–1, 3), B (2, p), C (5, –1) area of ΔABC should be zero, i.e.,

\(\frac12\) | - 1(p +1) + 2(-1 -3) + 5(3-p) | = 0 

⇒ – p – 1 – 8 + 15 – 5 p = 0 

⇒ – 6 p + 6 = 0 ⇒ –6 p = – 6 ⇒ p = 1.