Answer:
[tex]_{\text{point-slope form}}[/tex]
[tex]\huge\boxed{y-5=\dfrac{1}{3}(x+12)}[/tex]
[tex]_{\text{slope-intercept form}}[/tex]
[tex]\huge\boxed{y=\dfrac{1}{3}x+9}[/tex]
[tex]_{\text{standard form}}[/tex]
[tex]\huge\boxed{x-3y=-27}[/tex]
Step-by-step explanation:
The point-slope form of an equation of a line:
[tex]y-y_1=m(x-x_1)[/tex]
where
[tex]m[/tex] - slope
[tex](x_1;\ y_1)[/tex] - a point on a line
We have
[tex]m=\dfrac{1}{3};\ (-12;\ 5)\to x_1=-12;\ y_1=5[/tex]
Substitute:
[tex]y-5=\dfrac{1}{3}(x-(-12))\\\\y-5=\dfrac{1}{3}(x+12)\\==================[/tex]
[tex]y-5=\dfrac{1}{3}(x+12)\qquad|\text{use the distributive property}\ a(b+c)=ab+ac\\\\y-5=\dfrac{1}{3}x+4\qquad|\text{add 5 to both sides}\\\\y=\dfrac{1}{3}x+9\\=======================[/tex]
[tex]y=\dfrac{1}{3}x+9\qquad|\text{multiply both sides by 3}\\\\3y=x+27\qquad|\text{subtract}\ x\ \text{from both sides}\\\\-x+3y=27\qquad|\text{change the signs}\\\\x-3y=-27[/tex]
