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) $k=\frac{11}{4}$
Solution:
(b) k = 6
The given points are A(2, 3), B(5, k) and C(6, 7).
Here, $\left(x_{1}=2, y_{1}=3\right),\left(x_{2}=5, y_{2}=k\right)$ and $\left(x_{3}=6, y_{3}=7\right)$
Points A,B and C are collinear. Then,
$x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)=0$
$\Rightarrow 2(k-7)+5(7-3)+6(3-k)=0$
$\Rightarrow 2 k-14+20+18-6 k=0$
$\Rightarrow-4 k=-24$
$\Rightarrow k=6$