【Rpta.】La ecuación de la recta es 4y - 3x - 21 = 0
[tex]\green{{\hspace{50 pt}\above 1.2pt}\boldsymbol{\mathsf{Procedimiento}}{\hspace{50pt}\above 1.2pt}}[/tex]
Para determinar la ecuación de la recta que pasa por dos puntos usaremos lo siguiente:
[tex]\blue{\begin{array}{c}\boxed{\:\:\boldsymbol{\mathsf{\vphantom{\rule{1pt}{20pt}_{\rule{1pt}{17pt}}}y-y_1 = \left(\dfrac{y_2-y_1}{x_2-x_1}\right)(x - x_1)}}\:\:}\\\\\mathsf{Siendo\ (x_1,y_1)\ y \ (x_2,y_2)\ los\ puntos}\end{array}}[/tex]
Entonces del problema tenemos que:
[tex]\begin{array}{cccccccccccccc}\boldsymbol{\bigcirc \kern-8pt \triangleright}\:\:\:\:\mathsf{\boldsymbol{\mathsf{A=}(}\:\overbrace{\boldsymbol{-3}}^{x_1}\:\boldsymbol{,}\:\underbrace{\boldsymbol{3}}_{y_1}\:\boldsymbol{)}}&&&&&&&&&&\boldsymbol{\bigcirc \kern-8pt \triangleright}\:\:\:\:\mathsf{\boldsymbol{\mathsf{B=}(}\:\overbrace{\boldsymbol{5}}^{x_2}\:\boldsymbol{,}\:\underbrace{\boldsymbol{9}}_{y_2}\:\boldsymbol{)}}\end{array}[/tex]
Reemplazamos
[tex]\begin{array}{c}\mathsf{y-y_1 = \left(\dfrac{y_2-y_1}{x_2-x_1}\right)(x - x_1)}\\\\\mathsf{y-(3) = \left(\dfrac{9-(3)}{5-(-3)}\right)\big(x - (-3)\big)}\\\\\mathsf{y-3 = \left(\dfrac{9- 3}{5+3}\right)(x + 3)}\\\\\mathsf{y-3 = \left(\dfrac{\not\!6}{\not\!8}\right)(x + 3)}\\\\\mathsf{y-3 = \left(\dfrac{3}{4}\right)(x + 3)}\\\\\mathsf{(4)(y-3) = (3)(x+3)}\\\\\mathsf{4y-12 = 3x + 9}\\\\\boxed{\boxed{\boldsymbol{\mathsf{4y - 3x -21= 0}}}}\end{array}[/tex]
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