Find the equation of the line through the points (10, -5) and (1, -8).

Step-by-step solution

1. 1

   Compute the slope, minding the double negatives. With $(x_{1},y_{1})=(10,-5)$ and $(x_{2},y_{2})=(1,-8)$:

   $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-8-(-5)}{1-10}=\frac{-3}{-9}=\frac{1}{3}$

   Both differences are negative, so the slope is positive. Subtracting in the same order top and bottom is what guarantees the correct sign.

2. 2

   Write point-slope form using either point. Taking $(10,-5)$:

   $y-(-5)=\frac{1}{3}(x-10)\ \Longrightarrow\ y+5=\frac{1}{3}(x-10)$

   The formula subtracts the coordinate, so $y-(-5)$ becomes $y+5$.

3. 3

   Distribute the slope.

   $y+5=\frac{1}{3}x-\frac{10}{3}$

4. 4

   Isolate $y$, converting 5 to thirds.

   $y=\frac{1}{3}x-\frac{10}{3}-5=\frac{1}{3}x-\frac{10}{3}-\frac{15}{3}=\frac{1}{3}x-\frac{25}{3}$

   Keep $-\tfrac{25}{3}$ exact; writing $-8.33$ loses information and will not verify cleanly.

5. 5

   Check both original points. At $x=10$: $\tfrac{10}{3}-\tfrac{25}{3}=-\tfrac{15}{3}=-5$ ✓. At $x=1$: $\tfrac{1}{3}-\tfrac{25}{3}=-\tfrac{24}{3}=-8$ ✓. Testing both points, not just the one used to build the equation, is what actually validates the slope.