Question

Determine o perímetro do triângulo ABC cujas coordenadas são: A(3,3) B(-5-6) C(4-2).​

Answer 1

$\mathtt{{(x_{B}-x_{A})}^{2}={(-5-3)}^{2}={(-8)}^{2}=64}$

$\mathtt{{(y_{B}-y_{A})}^{2}={(-6-3)}^{2}={(-9)}^{2}=81}$

$\mathtt{d_{A, B}=\sqrt{{(x_{B}-x_{A})}^{2}+{(y_{B}-y_{A})}^{2}}}$

$\mathtt{d_{A, B}=\sqrt{64+81}=\sqrt{145}}$

$\mathtt{{(x_{C}-x_{B})}^{2}={(4-(-5))}^{2}={(4+5)}^{2}=81}$

$\mathtt{{(y_{C}-y_{B})}^{2}={(-2-(-6))}^{2}={(-2+6)}^{2}=16}$

$\mathtt{d_{B, C}=\sqrt{{(x_{C}-x_{B})}^{2}+{(y_{C}-y_{B})}^{2}}}$

$\mathtt{d_{B, C}=\sqrt{81+16}=\sqrt{97}}$

$\mathtt{{(x_{C}-x_{A})}^{2}={(4-3)}^{2}={(1)}^{2}=1}$

$\mathtt{{(y_{C}-x_{A})}^{2}={(-2-3)}^{2}={(-5)}^{2}=25}$

$\mathtt{d_{A, C}=\sqrt{{(x_{C}-x_{A})}^{2}+{(y_{C}-y_{A})}^{2}}}$

$\mathtt{d_{A, C}=\sqrt{1+25}=\sqrt{26}}$

$\mathtt{p=\sqrt{145}+\sqrt{97}+\sqrt{26}}$