Answer:
[tex]\large\boxed{y-3=-\dfrac{1}{3}(x+6)-\text{point-slope form}}\\\boxed{y=-\dfrac{1}{3}x+1-\text{slope-intercept form}}\\\boxed{x+3y=3-\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
We have
[tex]m=-\dfrac{1}{3},\ (-6,\ 3)\to x_1=-6,\ y_1=3[/tex]
Substitute:
[tex]y-3=-\dfrac{1}{3}(x-(-6))\\\\y-3=-\dfrac{1}{3}(x+6)[/tex]
Convert to the slope-intercept form
[tex]y=mx+b[/tex]
[tex]y-3=-\dfrac{1}{3}(x+6)[/tex] use the distributive property
[tex]y-3=-\dfrac{1}{3}x-2[/tex] add 3 to both sides
[tex]y=-\dfrac{1}{3}x+1[/tex]
Convert to the standard form
[tex]Ax+By=C[/tex]
[tex]y=-\dfrac{1}{3}x+1[/tex] multiply both sides by 3
[tex]3y=-x+3[/tex] add x to both sides
[tex]x+3y=3[/tex]
