If a trapezium is isosceles, when reflected in the line of symmetry, the slant heights must be reflected onto one another. This means they must have the same length, a given the endpoints of the lines.

\[(x_1,y_1), \: (x_2,y_2)\]

, we can find the length \[d\]

using \[d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\]

.This distance from A to B is

\[|AB|= \sqrt{(2-1)^2+(3-1)^2}=\sqrt{5}\]

.This distance from A to B is

\[|CD|= \sqrt{(10-8)^2+(4-5)^2}=\sqrt{5}\]

.The distances are the same, and so the trapezium is isosceles.