Question

$(1,6),(3.-2)$ and $(-2,K)$ are collinear points. What is $K$?

• A. $-6$
• B. $2$
• C. $8$
• D. $10$
• E. $18$

Answer 1

Answer: (E)

$18$

If three points are collinear then the area of triangle is zero.

Therefore, if the points $(1,6)$, $(3,-2)$ and $(-2,K)$ are collinear, then their area must be zero that is:

$A=\frac{1}{2}\left|x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right|\Rightarrow 0=\frac{1}{2}\left|1\right(-2-K)+3(K-6)-2(6+2\left)\right|\Rightarrow 0=|-2-K+3K-18-16|\Rightarrow 0=|2K-36|\Rightarrow 2K-36=0\Rightarrow 2K=36\Rightarrow K=18$

Hence, $K=18$