Question

find the value of k so that points collinear (7,-2) (5,1) (3,k).

Answer 1

$\begin{array}{c}A\left(7,-2\right),B\left(5,1\right),C\left(3,k\right)\\ SlopeAB=SlopeofBC\\ \Rightarrow \frac{1-\left(-2\right)}{5-7}=\frac{k-1}{3-5}\\ \Rightarrow \frac{3}{-2}=\frac{k-1}{-2}\\ \Rightarrow 3=k-1\\ \therefore k=4\end{array}$