#### Point P (– 4, 2) lies on the line segment joining the points A (– 4, 6) and B (– 4, – 6).

Solution.
The given points A (–4, 6) and B (–4, –6)
(x1, y1) = (-4, 6)                           (x2, y2) = (-4, -6)        (x3,y3)=(-4,2)
Point p(-4,2) lie on line AB if the area of triangle ABP =0
$\frac{1}{2}\left [ x_{1}\left ( y_{2}-y_{3} \right )+x_{2}\left ( y_{3}-y_{2}+ \right )+x_{3}\left ({y_{1}-{y_{2}}} \right ) \right ]= 0$
$\frac{1}{2}\left [ -4\left ( -6-2 \right )-4\left ( 2-6 \right ) -4\left ( 6+6 \right )\right ]= 0$
$\frac{1}{2}\left [ -4\left ( -8 \right )-4\left ( -4 \right ) -4\left ( 12 \right )\right ]= 0$
$\left [ +32+16-48 \right ]= 0$
so we can say that p(-4,2) must lie on line joining, AB