Is This Trapezium Isosceles
\[(x_1,y_1), \: (x_2,y_2)\]
, we can find the length \[d\]
using \[d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\]
.\[|AB|= \sqrt{(2-1)^2+(3-1)^2}=\sqrt{5}\]
.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.