Question

For which value of '$k$' the points $(7,-2),(5,1),(3,k)$ are collinear?

• A. $-4$
• B. $4$
• C. $8$
• D. $-8$

Answer 1

The correct option is B $4$

Since points are collinear,

so, $x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)=0$

Take ($7,-2$) as $(x_1,y_1)$, ($5,1$) as $(x_2,y_2)$ and ($3,k$) as $(x_3,y_3)$, we have

$7(1-k)+5(k+2)+3(-2-1)=0$

$\Rightarrow$ $7-7k+5k+10-9=0$

$\Rightarrow$ $2k=8$

$\Rightarrow$ $k=4$