Question

Show that the triangle with vertices A(2,2), B(-1,1) and C (1,-1) is equilateral under the taxicab metric, but not under the euclidean metric

Answer 1

In the taxicab metric the length of edges:

$AB=|A_{x}-B_{x}|+|A_{y}-B_{y}|=$

$=|2-(-1)|+|2-1|=4$ or $AB=2\Delta y+3 (2\Delta x)=2\cdot0.5+6\cdot0.5=4;$

$BC=|B_{x}-C_{x}|+|B_{y}-C_{y}|=$

$=|-1-1|+|1-(-1)|=4;$

$AC=|A_{x}-C_{x}|+|A_{y}-C_{y}|=$

$=|2-1|+|2-(-1)|=4$ .

$AB=BC=AC=4$ therefore the triangle is equilateral under the taxicab metric.

In the euclidean metric the length of edges:

$AB=\sqrt{(A_{x}-B_{x})^2+(A_{y}-B_{y})^2}=$

$=\sqrt{(2-(-1))^2+(2-1)^2}=\sqrt{3^2+1^2}=$

$=\sqrt{10}\approx3.16;$

$BC=\sqrt{(B_{x}-C_{x})^2+(B_{y}-C_{y})^2}=$

$=\sqrt{(-1-1)^2+(1-(-1))^2}=\sqrt{(-2)^2+2^2}=$

$=\sqrt{8}\approx2.83;$

$AC=\sqrt{(A_{x}-C_{x})^2+(A_{y}-C_{y})^2}=$

$=\sqrt{(2-1)^2+(2-(-1))^2}=\sqrt{1^2+3^2}=$

$=\sqrt{10}\approx3.16$ .

$AB=AC\neq BC$ therefore the triangle is not equilateral under the euclidean metric.