Question

write slope intercept form of equation for the line passing through the given points. and equation (-2,8) and (1,-3)

Answer 1

$\bf (\stackrel{x_1}{-2}~,~\stackrel{y_1}{8})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{-3}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-3}-\stackrel{y1}{8}}}{\underset{run} {\underset{x_2}{1}-\underset{x_1}{(-2)}}}\implies \cfrac{-11}{1+2}\implies -\cfrac{11}{3}$

$\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{8}=\stackrel{m}{-\cfrac{11}{3}}[x-\stackrel{x_1}{(-2)}]\implies y-8=-\cfrac{11}{3}(x+2) \\\\\\ y-8=-\cfrac{11}{3}x-\cfrac{22}{3}\implies y=-\cfrac{11}{3}x-\cfrac{22}{3}+8\implies y=-\cfrac{11}{3}x+\cfrac{2}{3}$

Answer 2

Answer:

$y = -\dfrac{11}{3}x + \dfrac{2}{3}$

Step-by-step explanation:

The slope-intercept equation for a straight line is

y = mx + b

where m is the slope of the line and b is the y-intercept.

The line passes through the points (-2, 8) and (1, -3)

(a) Calculate the slope of the line

$\begin{array}{rcl}m & = & \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\ & = & \dfrac{-3 - 8}{1 - (-2)}\\\\& = & \dfrac{-11}{1 + 2}\\\\& = & \mathbf{-\dfrac{11}{3}}\\\\\end{array}$

(b) Find the y-intercept

Insert the coordinates of one of the points into the equation

$\begin{array}{rcl}y & = & mx + b\\-3 & = & -\dfrac{11}{3}\times1 + b\\ \\b & = & -3 + \dfrac{11}{3}\\\\& = & -\dfrac{9}{3}+ \dfrac{11 }{3}\\\\& = &\mathbf{\dfrac{2}{3}}\\\end{array}$

(c) Write the equation for the line

$\mathbf{y} = \mathbf{-\dfrac{11}{3}x + \dfrac{2}{3}}$

The diagram below shows your graph passing through the two points with a slope of -11/3 and a y-intercept of ⅔.