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

The Correct Option is C

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:

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