Question

If the points A(2, 3), B(5, k) and C(6, 7) are collinear, then

(a) k = 4
(b) k = 6
(c) $k=\frac{-3}{2}$
(d) $k=\frac{11}{4}$

Answer 1

(b) k = 6
The given points are A(2, 3), B(5, k) and C(6, 7).
Here, $\left(x_1=2,y_1=3\right),\left(x_2=5,y_2=k\right)\text{and}\left(x_3=6,y_3=7\right).$
Points A,B and C are collinear. Then,
$$\begin{gathered} x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)=0 \\ \Rightarrow 2\left(k-7\right)+5\left(7-3\right)+6\left(3-k\right)=0 \\ \Rightarrow 2k-14+20+18-6k=0 \\ \Rightarrow -4k=-24 \\ \Rightarrow k=6 \end{gathered}$$