【Rpta.】a) La ecuación de la recta es y =2x + 4 b) La ecuación de la recta es y= (-3/2)x - 5/2
[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]
a) A(0,4) y B(2,8)
Del problema tenemos que:
[tex]\begin{array}{cccccccccccccccc}\boldsymbol{\bigcirc \kern-8pt \triangleright}\:\:\:\:\mathsf{\boldsymbol{\mathsf{A=}(}\:\overbrace{\boldsymbol{0}}^{x_1}\:\boldsymbol{,}\:\underbrace{\boldsymbol{4}}_{y_1}\:\boldsymbol{)}}&&&&&&&\boldsymbol{\bigcirc \kern-8pt \triangleright}\:\:\:\:\mathsf{\boldsymbol{\mathsf{B=}(}\:\overbrace{\boldsymbol{2}}^{x_2}\:\boldsymbol{,}\:\underbrace{\boldsymbol{8}}_{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-(4) = \left(\dfrac{8-(4)}{2-(0)}\right)\big(x - (0)\big)}\\\\\mathsf{y-4 = \left(\dfrac{8- 4}{2-0}\right)(x - 0)}\\\\\mathsf{y-4 = \left(\dfrac{4}{2}\right)(x )}\\\\\mathsf{y-4 = 2x}\\\\\boxed{\boxed{\boldsymbol{\sf{y=2x+4}}}} \end{array}[/tex]
b) A(-3,2) y B(1, -4)
Del problema tenemos que:
[tex]\begin{array}{cccccccccccccccc}\boldsymbol{\bigcirc \kern-8pt \triangleright}\:\:\:\:\mathsf{\boldsymbol{\mathsf{A=}(}\:\overbrace{\boldsymbol{-3}}^{x_1}\:\boldsymbol{,}\:\underbrace{\boldsymbol{2}}_{y_1}\:\boldsymbol{)}} &&&&&&&\boldsymbol{\bigcirc \kern-8pt \triangleright}\:\:\:\:\mathsf{\boldsymbol{\mathsf{B=}(}\:\overbrace{\boldsymbol{1}}^{x_2}\:\boldsymbol{,}\:\underbrace{\boldsymbol{-4}}_{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-(2) = \left(\dfrac{-4-(2)}{1-(-3)}\right)\big(x - (-3)\big)}\\\\\mathsf{y-2 = \left(\dfrac{-4- 2}{1+3}\right)(x + 3)}\\\\\mathsf{y-2 = \left(\dfrac{-\not \!6}{\not \!4}\right)(x + 3)}\\\\\mathsf{(2)y-2 = (-3)(x + 3)}\\\\\mathsf{2y-4 = -3x-9}\\\\\boxed{\boxed{\boldsymbol{\mathsf{y=-\dfrac{3}{2}x-\dfrac{5}{2}}}}}\end{array}[/tex]
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