Question

Show that the points $A(-3,3)$, $B(0,0)$, $C(3,-3)$ are collinear.

Answer 1

Let the points $(-3,3),(0,0)$, and $(3,-3)$ be representing the vertices A, B, and C of the given triangle respectively.
Let $A=(-3,3)$, $B=(0,0)$ and $C=(3,-3)$
AB= = $\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}$
= $\sqrt{(-3-0{)}^2+(3-0{)}^2}$ = $\sqrt{18}$
BC = $\sqrt{(0-3{)}^2+(0-(-3){)}^2}$ = $\sqrt{18}$
AC =$\sqrt{(-3-3{)}^2+(3-(-3){)}^2}$ = $\sqrt{36}$
AB + BC is equal to AC Therefore, the points $(-3,3),(0,0)$, and $(3,-3)$ are collinear.