Question

Using the distance formula, prove that the points A (-2, 3), B(1, 2) and C(7,0) are collinear.

Answer 1

We have

AB = $\sqrt{(1+2{)}^2+(2-3{)}^2}=\sqrt{3^2+(-1{)}^2}=\sqrt{10}\text{units};$

BC = $\sqrt{(7-1{)}^2+(0-1{)}^2}=\sqrt{6^2+(-2{)}^2}=\sqrt{40}=2\sqrt{10}\text{units};$

AC = $\sqrt{(7+2{)}^2+(0-3{)}^2}=\sqrt{9^2+(-3{)}^2}=\sqrt{90}=3\sqrt{10}\text{units};$

$\therefore AB+BC=(\sqrt{10}+2\sqrt{10})\text{units}=3\sqrt{10}\text{units}=AC.$

Thus, AB + BC = AC, showing that the points A, B, C are collinear.