Show that $(10, -18), (3, 6)$ and $(-5, 2)$ are the vertices of an isosceles triangle.

(N/A) Let the vertices of the triangle be $A(10, -18), B(3, 6)$ and $C(-5, 2)$.
To determine if the triangle is isosceles,we calculate the lengths of the sides using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$1$. Length of side $AB$:
$AB = \sqrt{(3 - 10)^2 + (6 - (-18))^2} = \sqrt{(-7)^2 + (24)^2} = \sqrt{49 + 576} = \sqrt{625} = 25$ units.
$2$. Length of side $BC$:
$BC = \sqrt{(-5 - 3)^2 + (2 - 6)^2} = \sqrt{(-8)^2 + (-4)^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5}$ units.
$3$. Length of side $AC$:
$AC = \sqrt{(-5 - 10)^2 + (2 - (-18))^2} = \sqrt{(-15)^2 + (20)^2} = \sqrt{225 + 400} = \sqrt{625} = 25$ units.
Since the lengths of sides $AB$ and $AC$ are equal ($AB = AC = 25$ units),the triangle is an isosceles triangle.