Question

If the perpendicular distances from the points \( (2,3) \), \( (4,a) \), and \( (\alpha, \beta) \) onto the line \( 3x + 4y - 3 = 0 \) are equal and \( 4\alpha - 3\beta + 1 = 0 \), then the sum of all possible values of \( \alpha \), \( \alpha \) and \( \beta \) is:

• A. \( \frac{-79}{10} \)
• B. \( \frac{83}{15} \)
• C. \( \frac{-73}{5} \)
• D. \( \frac{28}{15} \)

Hint:

For problems involving perpendicular distances, use the standard formula: \[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \] Carefully solve equations for unknowns by equating distances.

Answer 1

The Correct Option is B

The perpendicular distance formula for a point \( (x_0, y_0) \) from a line \( Ax + By + C = 0 \) is: \[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \] Applying to given points, equating distances and solving for \( \alpha \) and \( \beta \), we find their sum: \[ \sum \alpha = \frac{83}{15} \]