Question

Find the value of 'k', for which the points (8, 1), (k, -4), (2, -5) are collinear.

• A. 2
• B. 4
• C. 3
• D. 1

Answer 1

The correct option is C

3

The three points will be collinear if the $\Delta$ formed by these points has an area equal to zero.
Let A(8, 1), B(k, -4) and C(2, -5) be the vertices of a triangle.
$\therefore$ The given points will be collinear, if ar $\left(\Delta ABC\right)=0$
$\because$ Area of triangle having coordinates $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is $\frac{1}{2}\times |\left(x_1\right)(y_2-y_3)+\left(x_2\right)(y_3-y_1)+\left(x_3\right)(y_1-y_2)|$
$\therefore \frac{1}{2}\left[8\right(-4+5)+k(-5-1)+2(1+4\left)\right]=0\Rightarrow 8-6k+10=0\Rightarrow 6k=18\Rightarrow k=3$