Question: The point $A$ has coordinates $(2, 0)$ and the point $B$ has coordinates $(3, 1)$. The line $l_1$ passes through $A$ and is perpendicular to $AB$. Find an equation for the line $l_1$. Give your answer in the form $y = mx + c$.

1. **State the problem:** We need to find the equation of the line $l_1$ that passes through point $A(2,0)$ and is perpendicular to the line segment $AB$ where $B(3,1)$. 2. **Find the slope of line $AB$:** The slope formula is $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ Substitute $A(2,0)$ and $B(3,1)$: $$m_{AB} = \frac{1 - 0}{3 - 2} = \frac{1}{1} = 1$$ 3. **Find the slope of line $l_1$:** Since $l_1$ is perpendicular to $AB$, its slope $m_{l_1}$ is the negative reciprocal of $m_{AB}$: $$m_{l_1} = -\frac{1}{m_{AB}} = -\frac{1}{1} = -1$$ 4. **Use point-slope form to find the equation of $l_1$:** The point-slope form is: $$y - y_1 = m(x - x_1)$$ Using point $A(2,0)$ and slope $m_{l_1} = -1$: $$y - 0 = -1(x - 2)$$ Simplify: $$y = -1(x - 2) = -x + 2$$ 5. **Final equation in slope-intercept form:** $$\boxed{y = -x + 2}$$ This is the equation of the line $l_1$ passing through $A$ and perpendicular to $AB$.