Twice the difference of a number and 7 is 6

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davi113001

18.10.2022

Mathematics

$A=(-4,3),B=(-1,1),C=(1,3)\\
|AB|=\sqrt{(-4+1)^2+(3-1)^2}=\sqrt{9+4}=\sqrt{13}\\
|BC|=\sqrt{(-1-1)^2+(1-3)^2}=\sqrt{4+4}=\sqrt{8}\\
|AC|=\sqrt{(-4-1)^2+(3-3)^2}=\sqrt{25+0}=5\\
L=\sqrt{13}^2+\sqrt{8}^2=13+8=21\neq5^2\neq\ R$

It's impossible to form right triangle using thes points.

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