Q.Show that the points $A(4, 3)$, $B(-4, 3)$ and $C(0, -3)$ are the vertices of an isosceles triangle.

A triangle is isosceles if at least two of its three sides are equal in length, so compute all three side lengths with the distance formula and compare them.

$AB=\sqrt{(-4-4)^2+(3-3)^2}=\sqrt{64}=8$.

$BC=\sqrt{(0-(-4))^2+(-3-3)^2}=\sqrt{16+36}=\sqrt{52}=2\sqrt{13}$.

$CA=\sqrt{(4-0)^2+(3-(-3))^2}=\sqrt{16+36}=\sqrt{52}=2\sqrt{13}$.

Since $BC=CA$, two sides are equal.

Using the distance formula $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ on each pair of vertices:

Side $AB$ (from $A(4,3)$ to $B(-4,3)$):

Side $BC$ (from $B(-4,3)$ to $C(0,-3)$):

Side $CA$ (from $C(0,-3)$ to $A(4,3)$):

Since $BC = CA = 2\sqrt{13}$ (while $AB=8$ is different), two of the three sides are equal in length. A triangle with two equal sides is, by definition, isosceles.

Students sometimes stop after computing only two of the three sides and assume the triangle is scalene, or make a sign error when one coordinate is negative (e.g. computing $-3-3=0$ instead of $-6$ for the $y$-difference in side $BC$).