Question

6. Sabendo que a reta y1 passa pelos pontos (1,5) e (-2,-3) . Escreva a equação da reta y que é perpendicular a reta y1 e passa pelo ponto (2, 3))

Answer 1

Se duas retas são perpendiculares, implica que o produto dos coeficientes angulares delas resulta em -1.

Façamos :

$\text m_{\text y_1} = \text{coeficiente angular da reta y}_1\\\\ \text m_{\text y} = \text{coeficiente angular da reta y } \\\\ \underline{\text{se a reta y}_1 \ \text{{\'e perpendicular {\`a} reta y, ent{\~a}o}}}: \\\\ \text{m}_{\text y_1}.\text{m}_\text y = -1$

Coeficiente angular da reta  y1 passando pelos pontos (1,5) e (-2,-3)

$\displaystyle \text m_{\text y_1}=\frac{\text y_2-\text y_1}{\text x_2-\text x_1} \to \text m_{\text y_1}=\frac{-3-5}{-2-1}\to \boxed{\text{m}_{\text y_1}=\frac{8}{3}}$

Coeficiente angular da reta y que é perpendicular à reta y1 :

$\displaystyle \text m_\text y.\text m_{\text y_1}=-1 \to \text{m}_\text y.\frac{8}{3}=-1 \\\\ \therefore \\\\ \boxed{\text m_\text y=\frac{-3}{8}}$

Equação da reta y que passa pelo ponto (2,3) :

$\displaystyle \text{reta y :} \ \text y-\text y_\text o=\text m_\text y(\text x-\text x_\text o) \\\\ \text{reta y : } \text y - 3 = \frac{-3}{8}(\text x-\text 2) \\\\\ \text{reta y : }8\text y -24=-3\text x+6 \\\\\\ \huge\boxed{\text{reta y : } 3\text x+8\text y -30=0\ }\checkmark$