Question

On the locus of the point \( P(x, y) \) equidistant from \( (3, 0) \) and \( (0, 4) \), if \( A \) and \( B \) are two points that satisfy \( 4x = 3y \) and \( x = y \) respectively, then the distance between A and B is:

• A. \( \frac{5}{2} \)
• B. \( 5 \)
• C. \( \frac{25}{4} \)
• D. \( 25 \)

Hint:

Locus of Equidistant Points}
Use distance formula to set up equations
Substituting lines into the locus helps locate specific points
Be precise with algebra to simplify fractional coordinates

Answer 1

The Correct Option is A

The locus of point \( P(x, y) \) equidistant from \( (3, 0) \) and \( (0, 4) \) is the perpendicular bisector. Use distance formula: \[ \sqrt{(x - 3)^2 + y^2} = \sqrt{x^2 + (y - 4)^2} \Rightarrow (x - 3)^2 + y^2 = x^2 + (y - 4)^2 \Rightarrow x^2 - 6x + 9 + y^2 = x^2 + y^2 - 8y + 16 \Rightarrow -6x + 9 = -8y + 16 \Rightarrow 6x + 8y = 7 \] Now find intersection with: - Line 1: \( 4x = 3y \Rightarrow y = \frac{4}{3}x \) - Line 2: \( x = y \) Substitute into locus: \[ 6x + 8 \cdot \frac{4}{3}x = 7 \Rightarrow 6x + \frac{32}{3}x = 7 \Rightarrow \frac{50x}{3} = 7 \Rightarrow x = \frac{21}{50} \Rightarrow y = \frac{28}{50} \] Point A: \( \left( \frac{21}{50}, \frac{28}{50} \right) \) Point B (on \( x = y \)): Substitute into same equation: \[ 6x + 8x = 7 \Rightarrow 14x = 7 \Rightarrow x = \frac{1}{2}, y = \frac{1}{2} \] Now compute distance: \[ AB = \sqrt{\left( \frac{21}{50} - \frac{1}{2} \right)^2 + \left( \frac{28}{50} - \frac{1}{2} \right)^2} = \sqrt{\left( \frac{-4}{50} \right)^2 + \left( \frac{3}{50} \right)^2} = \sqrt{\frac{25}{100}} = \frac{5}{10} = \frac{5}{2} \]