What is an equation of the line that passes through the points (-6, -2) and (6,8)? NEED ASAP!!

Answer:[tex]y-8=\dfrac{2}{3}(x-6)\ -\ \text{point-slope form}\\\\y=\dfrac{2}{3}x+4\ -\ \text{slope-intercept form}\\\\2x-3y=-12\ -\ \text{standard form}[/tex]Step-by-step explanation:The point-slope form of an equation of a line:[tex]y-y_1=m(x-x_1)[/tex]m - slopeThe formula of a slope:[tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]=============================================We have two points (-6, -2) and (6, 8).Calculate the slope:[tex]m=\dfrac{8-(-2)}{6-(-6)}=\dfrac{8}{12}=\dfrac{8:4}{12:4}=\dfrac{2}{3}[/tex]Put it and the coordinates of the point (6, 8) to the equation of a line:[tex]y-8=\dfrac{2}{3}(x-6)[/tex]Convert to the slope-intercept form y = mx + b:[tex]y-8=\dfrac{2}{3}(x-6)[/tex]     use the distributive property a(b + c) = ab + ac[tex]y-8=\dfrac{2}{3}x-\left(\dfrac{2}{3\!\!\!\!\diagup_1}\right)(6\!\!\!\!\diagup^2)[/tex][tex]y-8=\dfrac{2}{3}x-4[/tex]            add 8 to both sides[tex]y=\dfrac{2}{3}x+4[/tex]Convert to the standard form Ax + By = C:[tex]y=\dfrac{2}{3}x+4[/tex]            multiply both sides by 3[tex]3y=3\!\!\!\!\diagup^1\cdot\dfrac{2}{3\!\!\!\!\diagup_1}x+(3)(4)[/tex][tex]3y=2x+12[/tex]        subtract 2x from both sides[tex]-2x+3y=12[/tex]          change the signs[tex]2x-3y=-12[/tex]