The problem asks for the equation of a line in the $xy$-plane that passes through the point $(3, 1)$ and makes an angle of $60^\circ$ with the $x$-axis.

Solution:

The slope $m$ of a line is given by: $m = \tan(\theta),$ where $\theta$ is the angle the line makes with the positive $x$-axis.

Step 1: Calculate the slope

Given $\theta = 60^\circ$: $m = \tan(60^\circ) = \sqrt{3}.$

Step 2: Use the point-slope form

The equation of a line passing through a point $(x_1, y_1)$ with slope $m$ is: $y - y_1 = m(x - x_1).$

Here, the point is $(x_1, y_1) = (3, 1)$ and the slope $m = \sqrt{3}$. Substituting these values: $y - 1 = \sqrt{3}(x - 3).$

Step 3: Simplify the equation

Distribute $\sqrt{3}$: $y - 1 = \sqrt{3}x - 3\sqrt{3}.$ Add 1 to both sides: $y = \sqrt{3}x - 3\sqrt{3} + 1.$

Thus, the equation of the line is: $\boxed{y = \sqrt{3}x - 3\sqrt{3} + 1.}$

Would you like further clarification or assistance with similar problems?

Follow-up Questions:

1. How can you generalize this problem for other angles with the $x$-axis?
2. What happens to the line equation if the angle is $90^\circ$?
3. How would the equation change if the line passes through a different point?
4. Can this method be extended to 3D space? How?
5. What is the relationship between slope and the perpendicular line?

Tip:

Always visualize the geometry of the line and confirm that your slope sign matches the line's direction. For example, angles between $90^\circ$ and $180^\circ$ will result in a negative slope.