1. **State the problem:** Find the standard and general form of the line passing through the points $(-10,1)$ and $(-7,2)$. 2. **Formula for slope:** The slope $m$ of a line through points $(x_1,y_1)$ and $(x_2,y_2)$ is given by $$m=\frac{y_2 - y_1}{x_2 - x_1}.$$ 3. **Calculate the slope:** Substitute $x_1=-10$, $y_1=1$, $x_2=-7$, $y_2=2$: $$m=\frac{2 - 1}{-7 - (-10)}=\frac{1}{-7 + 10}=\frac{1}{3}.$$ 4. **Use point-slope form:** The equation of the line is $$y - y_1 = m(x - x_1).$$ Using point $(-10,1)$ and $m=\frac{1}{3}$: $$y - 1 = \frac{1}{3}(x + 10).$$ 5. **Simplify to standard form:** Multiply both sides by 3 to clear the fraction: $$3(y - 1) = x + 10.$$ 6. **Expand:** $$3y - 3 = x + 10.$$ 7. **Rearrange to standard form $Ax + By = C$:** $$-x + 3y = 13.$$ Multiply both sides by $-1$ to make $A$ positive: $$x - 3y = -13.$$ 8. **General form:** The general form is $Ax + By + C = 0$, so rewrite as $$x - 3y + 13 = 0.$$ **Final answers:** - Standard form: $x - 3y = -13$. - General form: $x - 3y + 13 = 0$.