Question
Show that points $\mathrm{A}(-3,2), \mathrm{B}(1,-2)$ and $\mathrm{C}(9,-10)$ are collinear.

Answer

If the sum of any two distances out of $d(A, B), d(B, C)$ and $d(A, C)$ is equal to the third, then the three points $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are collinear.
$\therefore$ we will find $\mathrm{d}(\mathrm{A}, \mathrm{B}), \mathrm{d}(\mathrm{B}, \mathrm{C})$ and $\mathrm{d}(\mathrm{A}, \mathrm{C})$.
Co-ordinates of A Co-ordinates of B Distance formula
$ (-3,2)$
$(1,-2)$
$\mathrm{d}(\mathrm{A}, \mathrm{B})=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
$\left(x_1, y_1\right)$
$\left(x_2, y_2\right)$
$\therefore d(A, B)=\sqrt{[1-(-3)]^2+[(-2)-2]^2}$
$=\sqrt{(1+3)^2+(-4)^2}$
$=\sqrt{16+16}$
$=\sqrt{32}=4 \sqrt{2}$
$d(B, C)=\sqrt{(9-1)^2+(-10+2)^2}$
$=\sqrt{64+64}=8 \sqrt{2}$
$\text { and } d(A, C)=\sqrt{(9+3)^2+(-10-2)^2}$
$=\sqrt{144+144} \quad=12 \sqrt{2}$
$\therefore \text { from(I), (II) and (III) } 4 \sqrt{2}+8 \sqrt{2}=12 \sqrt{2}$
$\therefore \mathrm{d}(\mathrm{A}, \mathrm{B})+\mathrm{d}(\mathrm{B}, \mathrm{C})=\mathrm{d}(\mathrm{A}, \mathrm{C})$
$\therefore \text { Points A, B, C are collinear. }$
from distance formula

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