Write the equation of the line that passes through the point (-6, 5) with slope 2/3.

Step-by-step solution

1. 1

   Start from point-slope form. Given a point $(x_1,y_1)$ and a slope $m$, the line is

   $y-y_1=m\left(x-x_1\right)$

   This form needs no algebra to write down — the work is entirely in simplifying it afterwards.

2. 2

   Substitute the given values, minding the double negative. With $x_1=-6$, $y_1=5$ and $m=\tfrac23$:

   $y-5=\frac{2}{3}\left(x-(-6)\right)=\frac{2}{3}(x+6)$

   Writing $(x-6)$ instead of $(x+6)$ here is the most common mistake and would shift the line by eight units vertically.

3. 3

   Distribute the slope. The fraction clears exactly because $6$ is a multiple of $3$:

   $y-5=\frac{2}{3}x+\frac{2}{3}\cdot 6=\frac{2}{3}x+4$

4. 4

   Solve for $y$ to reach slope-intercept form.

   $y=\frac{2}{3}x+4+5=\frac{2}{3}x+9$

   so the slope is $\tfrac23$ and the $y$-intercept is $(0,9)$.

5. 5

   Verify the point lies on the line. Substituting $x=-6$:

   $y=\frac{2}{3}(-6)+9=-4+9=5\;\checkmark$

   A second check on the slope: moving from $x=-6$ to $x=-3$ (a run of $3$) raises $y$ from $5$ to $7$ (a rise of $2$), giving $\tfrac{2}{3}\;\checkmark$. In standard form the same line is $2x-3y=-27$.