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

Answer: (D)

$4 + 6 \sqrt{3}$

Answer 2

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:$4 + 6\sqrt{3}$