Key Concepts and Formulas

• Parametric Form of a Line: The parametric equations of a line passing through point $(x_1, y_1)$ and making an angle $\theta$ with the positive x-axis are given by $x = x_1 + r\cos\theta$ and $y = y_1 + r\sin\theta$, where $r$ is the directed distance from $(x_1, y_1)$ to any point $(x, y)$ on the line.
• Slope and Angle: The slope $m$ of a line is related to the angle $\theta$ it makes with the positive x-axis by $m = \tan\theta$.
• Rationalizing the Denominator: To simplify an expression with a radical in the denominator, multiply both the numerator and denominator by the conjugate of the denominator.

Step-by-Step Solution

Step 1: Understand the Problem and Identify Given Information

We are given a point $A(2, 3)$, a line $L_1: 2x - 3y + 28 = 0$, and a direction line $L_2: \sqrt{3}x - y + 1 = 0$. We need to find the distance from point $A$ to line $L_1$, measured parallel to line $L_2$.

Step 2: Determine the Slope and Angle of the Direction Line

The direction line $L_2$ is given by $\sqrt{3}x - y + 1 = 0$. We can rewrite this in slope-intercept form as $y = \sqrt{3}x + 1$. Therefore, the slope of $L_2$ is $m = \sqrt{3}$. Since $m = \tan\theta$, we have $\tan\theta = \sqrt{3}$. This implies that $\theta = \frac{\pi}{3}$ (or $60^\circ$).

Step 3: Calculate $\cos\theta$ and $\sin\theta$

Since $\theta = \frac{\pi}{3}$, we have: $\cos\theta = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$ $\sin\theta = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$

Step 4: Write the Parametric Equations of the Line Through A

Using the parametric form of a line passing through $(x_1, y_1) = (2, 3)$ with $\cos\theta = \frac{1}{2}$ and $\sin\theta = \frac{\sqrt{3}}{2}$, we have: $x = 2 + r\left(\frac{1}{2}\right) = 2 + \frac{r}{2}$ $y = 3 + r\left(\frac{\sqrt{3}}{2}\right) = 3 + \frac{\sqrt{3}r}{2}$

Step 5: Find the Intersection Point with the Target Line

The intersection point must lie on the line $L_1: 2x - 3y + 28 = 0$. Substitute the parametric coordinates of the point $(x, y)$ into the equation of $L_1$: $2\left(2 + \frac{r}{2}\right) - 3\left(3 + \frac{\sqrt{3}r}{2}\right) + 28 = 0$

Step 6: Solve for r

Simplify and solve the equation for $r$: $4 + r - 9 - \frac{3\sqrt{3}r}{2} + 28 = 0$ $23 + r - \frac{3\sqrt{3}r}{2} = 0$ $r\left(1 - \frac{3\sqrt{3}}{2}\right) = -23$ $r\left(\frac{2 - 3\sqrt{3}}{2}\right) = -23$ $r = \frac{-46}{2 - 3\sqrt{3}}$

Step 7: Rationalize the Denominator

To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator, which is $2 + 3\sqrt{3}$: $r = \frac{-46}{2 - 3\sqrt{3}} \cdot \frac{2 + 3\sqrt{3}}{2 + 3\sqrt{3}}$ $r = \frac{-46(2 + 3\sqrt{3})}{4 - (3\sqrt{3})^2}$ $r = \frac{-46(2 + 3\sqrt{3})}{4 - 27}$ $r = \frac{-46(2 + 3\sqrt{3})}{-23}$ $r = 2(2 + 3\sqrt{3})$ $r = 4 + 6\sqrt{3}$

Common Mistakes & Tips

• Remember that the problem asks for the distance parallel to a line, not the perpendicular distance.
• Pay close attention to the signs when simplifying the equation and rationalizing the denominator.
• Ensure you use the correct trigonometric values for the angle $\theta$.

Summary

To find the distance of the point (2,3) from the line 2x - 3y + 28 = 0, measured parallel to the line $\sqrt{3}x - y + 1 = 0$, we first found the angle the parallel line makes with the x-axis. Then we wrote the parametric equations of the line through (2,3) with that angle. Finally, we substituted these equations into the equation of the target line and solved for the directed distance $r$, which after simplification and rationalization, gave us $r = 4 + 6\sqrt{3}$.

Final Answer

The final answer is $\boxed{4 + 6\sqrt{3}}$, which corresponds to option (C).