Question

The distance of the point \( (2, 3) \) from the line \( 2x - 3y + 28 = 0 \), measured parallel to the line \( \sqrt{3}x - y + 1 = 0 \), is equal to

• A. \( 4 \sqrt{2} \)
• B. \( 6 \sqrt{3} \)
• C. \( 3 + 4 \sqrt{2} \)
• D. \( 4 + 6 \sqrt{3} \)

Answer 1

The Correct Option is D

Let the point \( A = (2, 3) \) and the line \( 2x - 3y + 28 = 0 \). We want to find the distance from \( A \) to this line, measured parallel to the line \( \sqrt{3}x - y + 1 = 0 \).

Step 1. Write point \( P \) in terms of parametric coordinates along the direction of \( \sqrt{3}x - y + 1 = 0 \):
  The direction ratios of this line are \( \cos\theta = \sqrt{3} \) and \( \sin\theta = 1 \), so the point \( P \) can be written as:  
 \(P \left( 2 + \frac{r\sqrt{3}}{2}, 3 + \frac{r}{2} \right)\)

Step 2. Condition for \( P \) to lie on the line \( 2x - 3y + 28 = 0 \): Substitute \( P \) into the equation \( 2x - 3y + 28 = 0 \):  
 \(2 \left( 2 + \frac{r\sqrt{3}}{2} \right) - 3 \left( 3 + \frac{r}{2} \right) + 28 = 0\)

Step 3. Simplifying, we get:
\(4 + r\sqrt{3} - 9 - \frac{3r}{2} + 28 = 0\)
\(r = 4 + 6\sqrt{3}\)

Thus, the required distance is .

The Correct Answer is:\( 4 + 6\sqrt{3} \)