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 - slope
The formula of a slope:
[tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]
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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]
