Let A (a, b), B (3, 4) and (- 6, - 8) respectively denote the...

Given: $A$\left(a, b\right), B$\left(3, 4\right), C\left(- 6, - 8\right)$ denotes centroid, circumcentre and orthocentre respectively.

We know that, the line joining orthocentre and circumcentre is divided by centroid in the ratio $2: 1$, where distance from orthocentre being the larger part.

$⇒ a = \dfrac{2 \times 3 + 1\left(- 6\right)}{2 + 1}, b = \dfrac{2 \times 4 + 1\left(- 8\right)}{2 + 1}$

$⇒ a = 0, b = 0$

$⇒ P ≡ \left(2 \times 0 + 3, 7 \times 0 + 5\right)$

$⇒ P ≡ \left(3, 5\right)$

Distance from $P$ measured along $x - 2y - 1 = 0$

$⇒ x = 3 + rcos\theta, y = 5 + rsin\theta$

Where, $tan\theta = \dfrac{1}{2}$

So, the distance of point from $2x + 3y - 4 = 0$ will be,

$2\left(3 + rcos\theta\right) + 3\left(5 + rsin\theta\right) - 4 = 0$

$⇒ r\left(2cos\theta + 3sin\theta\right) = - 17$

$⇒ r = \left|- \dfrac{17}{\left(2 \cdot \dfrac{2}{\sqrt{5}} + 3 \cdot \dfrac{1}{\sqrt{5}}\right)}\right|$

$⇒ r = \left|\dfrac{- 17\sqrt{5}}{7}\right|$

$⇒ r = \dfrac{17\sqrt{5}}{7}$