Question

For 'k' =, the following points (7,-2), (5, 1), (3, k) are collinear.

• A. 4
• B. 3
• C. 5

Answer 1

The correct option is A 4

Let $A(7,-2),B(5,1)\&C(3,k)$ be the vertices of a triangle.

Condition for collinear points, $ar\left(△ABC\right)=0$
Area of triangle having coordinates $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is
$\frac{1}{2}\times |x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$


$\therefore \frac{1}{2}|7(1-k)+5(k+2)+3(-2-1)|=07-7k+5k+10+(-6)-3=017-9+5k-7k=08-2k=0\Rightarrow 2k=8\Rightarrow k=\frac{8}{2}\therefore k=4$