Show that the point P (−4, 2) lies on the line segment

We have to show that point P (−4, 2) lies on line segment AB with points A (−4, 6) and

B (−4, −6)

If P (−4, 2) lies on the line segment joining A (−4, 6) and B (−4, −6), then the three points

must be collinear.

Let the three points be not collinear and form a triangle PAB

We know that area of a triangle with coordinates of vertices $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right),\left(x_{3}, y_{3}\right)$

$=\frac{x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)}{2}$

Therefore area of triangle PAB $=\frac{-4(6+6)-4(-6-2)-4(2-6)}{2}$

$=\frac{-48+36+16}{2}$

$=\frac{0}{2}$

$=0$

Since area of the triangle is 0, no triangle exists.

Therefore points P (−4, 2), A (−4, 6) and B (−4, −6) are collinear.