The distance of the point ( 2, 3 ) from the line 2x - 3y + 28 = 0 ,...

Let, ${L}_{1}: \sqrt{3}x - y + 1 = 0$ and {L}_{2}: 2x - 3y + 28 = 0.

Slope, ${m}_{1} = \sqrt{3} = tan\theta$

$⇒ \theta = {60}^{°}$

$⇒ sin\theta = \dfrac{\sqrt{3}}{2}, cos\theta = \dfrac{1}{2}$

Now, equation of line passing through $\left(2, 3\right)$ and having slope $\sqrt{3}$ is,

$\dfrac{x - 2}{\dfrac{1}{2}} = \dfrac{y - 3}{\dfrac{\sqrt{3}}{2}} = r$

$⇒ 2x - 3y + 28 = 0$

$⇒ 4 + r - 9 - \dfrac{3\sqrt{3}}{2}r + 28 = 0$

$⇒ \left(\dfrac{2 - 3\sqrt{3}}{2}\right)r + 23 = 0$

$⇒ r = \dfrac{46}{3\sqrt{3} - 2} \times \dfrac{3\sqrt{3} + 2}{3\sqrt{3} + 2}$

$⇒ r = \dfrac{46}{23} \times \dfrac{3\sqrt{3} + 2}{1} = 4 + 6\sqrt{3}$