Question

dados os pontos abaixo, encontre a equação da reta que passa pelos pontos P (-1,3) e Q (1/2; -5)

Answer 1

$\Large\boxed{\begin{array}{l}\underline{\rm Coeficiente\,angular\,da\,reta}\\\underline{\rm que\,passa\,pelos\,pontos\,A(x_A.y_A)\,e\,B(x_B,y_B)}\\\sf m=\dfrac{y_B-y_A}{x_B-x_A}\\\underline{\rm Equac_{\!\!,}\tilde ao\,da\,reta\,que\,possui}\\\underline{\rm\,coeficiente\,angular\,m\,e\,passa}\\\underline{\rm pelos\,ponto~P(x_0,y_0)}\\\sf y=y_0+m(x-x_0)\end{array}}$

$\Large\boxed{\begin{array}{l}\sf P(-1,3)~~Q\bigg(\dfrac{1}{2},-5\bigg)\\\\\sf m=\dfrac{y_Q-y_P}{x_Q-x_P}\\\\\sf m=\dfrac{-5-3}{\dfrac{1}{2}-[-1]}=\dfrac{-8}{\dfrac{1}{2}+1}\\\\\sf m=\dfrac{-8}{\dfrac{3}{2}}=-8\cdot\dfrac{2}{3}=-\dfrac{16}{3}\end{array}}$

$\Large\boxed{\begin{array}{l}\sf Adotanto\,o\,ponto\,P\,temos:\\\sf y=3-\dfrac{16}{3}(x-[-1])\\\\\sf y= 3-\dfrac{16}{3}(x+1)\\\\\sf y=3-\dfrac{16}{3}x-\dfrac{16}{3}\\\\\sf y=\dfrac{9-16x-16}{3}\\\\\sf y=-\dfrac{16}{3}x-\dfrac{7}{3}\end{array}}$