Question

Write various of the equation of a line that passes through (-6, 3) and has a slope of - 1/3

Answer 1

Answer:

$\large\boxed{y-3=-\dfrac{1}{3}(x+6)-\text{point-slope form}}\\\boxed{y=-\dfrac{1}{3}x+1-\text{slope-intercept form}}\\\boxed{x+3y=3-\text{standard form}}$

Step-by-step explanation:

The point-slope form of an equation of a line:

$y-y_1=m(x-x_1)$

m - slope

We have

$m=-\dfrac{1}{3},\ (-6,\ 3)\to x_1=-6,\ y_1=3$

Substitute:

$y-3=-\dfrac{1}{3}(x-(-6))\\\\y-3=-\dfrac{1}{3}(x+6)$

Convert to the slope-intercept form

$y=mx+b$

$y-3=-\dfrac{1}{3}(x+6)$           use the distributive property

$y-3=-\dfrac{1}{3}x-2$           add 3 to both sides

$y=-\dfrac{1}{3}x+1$

Convert to the standard form

$Ax+By=C$

$y=-\dfrac{1}{3}x+1$           multiply both sides by 3

$3y=-x+3$             add x to both sides

$x+3y=3$