Question

Considere os pontos A(1,5) e B(-3,-1).
Determine a equação do lugar geométrico dos pontos P sabendo-se que o
coeficiente angular da reta que passa por A e Pé 4 unidade(s) maior que o
coeficiente angular da reta que passa por B e p.​

Answer 1

$\displaystyle \text{Sejam os pontos :}\\\\ \text A(1,5) \ , \text B(-3,-1) \ , \text P(\text x, \text y) \\\\ \text{Lugar geom{\'e}trico dos pontos P, tal que :}\\\\\ \text m_{\text{AP}} = \text{m}_{\text{BP}}+4 \\\\ \text{onde}: \\\\ \text m_{\text{AP}} = \text{coeficiente angular da reta que passa por A e P} \\\\ \text m_{\text{BP}} = \text{coeficiente angular da reta que passa por B e P}$

Coeficiente angular da reta :

$\displaystyle \text m = \frac{\text y_2-\text y_1}{\text x_2-\text x_1} \\\\\\\ 1) \ \text m_{\text{AP}} = \frac{\text y-5}{\text x-1} \\\\\\ 2) \ \text{m}_{\text{BP}} = \frac{\text y-(-1)}{\text x-(-3)} \to \text{m}_{\text{BP}}=\frac{\text y+1}{\text x+3} \\\\\\\ \underline{\text{Temos}} : \\\\\ \text m_{\text{AP}} =\text{m}_{\text{BP}}+4 \\\\\\ \frac{\text y-5}{\text x-1}=\frac{\text y+1}{\text x+3}+4 \\\\\\ \frac{\text y-5}{\text x-1}=\frac{\text y+1+4(\text x+3)}{\text x+3}$

$\displaystyle \frac{\text y-5}{\text x-1}=\frac{\text y+1+4\text x+12}{\text x+3} \\\\\\ \frac{\text y-5}{\text x-1}=\frac{\text y+4\text x+13}{\text x+3} \\\\\\ (\text y-5)(\text x+3)=(\text x-1)(\text y+4\text x+13) \\\\ \text{xy}+3\text y-5\text x-15=\text {xy}+4\text x^2+13\text x-\text y-4\text x-13 \\\\ 4\text y = 4\text x^2+14\text x-28\\\\\\ \huge\boxed{\ \text y = \text x^2+\frac{7\text x}{2}-7\ }$

O lugar geométrica dos pontos P é uma Parábola