Question

Let $A (a, b), B(3, 4)$ and $(-6, -8)$ respectively denote the centroid, circumcentre, and orthocentre of a triangle. Then, the distance of the point $P(2a + 3, 7b + 5)$ from the line $2x + 3y - 4 = 0$ measured parallel to the line $x - 2y - 1 = 0$ is

• A. $\frac{15 \sqrt{5}}{7}$
• B. $\frac{17 \sqrt{5}}{6}$
• C. $\frac{17 \sqrt{5}}{7}$
• D. $\frac{\sqrt{5}}{17}$

Answer 1

Answer: (C)

$\frac{17 \sqrt{5}}{7}$

Answer 2

Given:

$A(a, b), \quad B(3, 4), \quad C(-6, -8)$

Since $A$ is the centroid, we have:

$a = 0, \quad b = 0 \implies P(3, 5)$

To find the distance of point $P$ from the line $2x + 3y - 4 = 0$ measured parallel to the line $x - 2y - 1 = 0$, we first find the direction cosine.

Let the line $x - 2y - 1 = 0$ represent:

$x = 3 + r \cos \theta, \quad y = 5 + r \sin \theta$

where $\theta$ is the angle such that:

$\tan \theta = \frac{1}{2}$

For the line parallel:

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

Thus:

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