Answer to Question #138116 in Geometry for J

Question #138116
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
1
Expert's answer
2020-10-13T19:34:05-0400


In the taxicab metric the length of edges:

AB=AxBx+AyBy=AB=|A_{x}-B_{x}|+|A_{y}-B_{y}|=

=2(1)+21=4=|2-(-1)|+|2-1|=4 or AB=2Δy+3(2Δx)=20.5+60.5=4;AB=2\Delta y+3 (2\Delta x)=2\cdot0.5+6\cdot0.5=4;

BC=BxCx+ByCy=BC=|B_{x}-C_{x}|+|B_{y}-C_{y}|=

=11+1(1)=4;=|-1-1|+|1-(-1)|=4;

AC=AxCx+AyCy=AC=|A_{x}-C_{x}|+|A_{y}-C_{y}|=

=21+2(1)=4=|2-1|+|2-(-1)|=4 .

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


In the euclidean metric the length of edges:

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

=(2(1))2+(21)2=32+12==\sqrt{(2-(-1))^2+(2-1)^2}=\sqrt{3^2+1^2}=

=103.16;=\sqrt{10}\approx3.16;

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

=(11)2+(1(1))2=(2)2+22==\sqrt{(-1-1)^2+(1-(-1))^2}=\sqrt{(-2)^2+2^2}=

=82.83;=\sqrt{8}\approx2.83;

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

=(21)2+(2(1))2=12+32==\sqrt{(2-1)^2+(2-(-1))^2}=\sqrt{1^2+3^2}=

=103.16=\sqrt{10}\approx3.16 .

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


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