Question

Dados os pontos A(1,1), B(3,5), C(x,7), determine x para que o triangulo ABC seja isosceles (AC=BC).

Answer 1

${(x_{B}-x_{A}) }^{2}={(3-1)}^{2}={2}^{2}=4$

${(y_{B}-y_{A}) }^{2}={(5-1)}^{2}={4}^{2}=16$

$d_{A, B}=\mathsf{\sqrt{{(x_{B}-x_{A}) }^{2}+{(y_{B}-y_{A}) }^{2}}}\\\mathsf{d_{A, B}=\sqrt{4+16}=\sqrt{20}}$

${(x_{C}-x_{A}) }^{2}={(x-1)}^{2}$

${(y_{C}-y_{A}) }^{2}={(7-1)}^{2}={6}^{2}=36$

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

$\mathsf{d_{A,C}=\sqrt{{(x-1)}^{2}+36}}$

Para ser isósceles $d_{A, B}=d_{B,C}$

$\mathsf{\sqrt{20}=\sqrt{{(x-1)}^{2}+36}}$

Elevando ambos membros ao quadrado

$\mathsf{20={(x-1)}^{2}+36}\\\mathsf{{(x-1)}^{2}=20-36}$

$\mathsf{{(x-1)}^{2}=-16}\\\mathsf{x-1=\pm\sqrt{-16}}\\\mathsf{x-1=\pm4i}$

$\mathsf{x_{1}=1+4i}\\\mathsf{x_{2}=1-4i}$