Question

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

• A. 6
• B. 1
• C. 4
• D. 8

Answer 1

The correct option is B

1

For 3 points to be collinear the area of the triangle formed by these three points should be zero.

And, area of a triangle formed by $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is given by
$\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$

According to the question:
The given points A(-1, 3), B(2, p), C(5, -1) are collinear.

$\therefore$ Area ΔABC formed by these points should be zero.

⇒ The area of ΔABC = 0

$\Rightarrow \frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]=0$

⇒ -1(p + 1) + 2 (-1 - 3) + 5(3 - p) = 0

⇒ -p - 1 - 8 + 15 - 5p = 0

⇒ -6p + 15 - 9 = 0 ⇒ -6p = -6 ⇒ p = 1

Hence, the value of p is 1.