Question

Suppose a point \( P \) moves such that: \[ BP^2 - AP^2 = 121 \] where \( A = (2, 5) \), \( B = (5, 11) \). Then the locus of \( P \) is a straight line. What is the slope of this line?

• A. \( \frac{1}{2} \)
• B. \( -2 \)
• C. \( -\frac{1}{2} \)
• D. \( 2 \)

Hint:

Whenever a condition relates squared distances, simplify it algebraically to identify the geometric nature of the locus.

Answer 1

The Correct Option is C

Let \( P = (x, y) \). Then: \[ BP^2 = (x - 5)^2 + (y - 11)^2,\quad AP^2 = (x - 2)^2 + (y - 5)^2 \] Compute: \[ BP^2 - AP^2 = [(x - 5)^2 + (y - 11)^2] - [(x - 2)^2 + (y - 5)^2] = 121 \] Simplify each: \[ (x^2 - 10x + 25 + y^2 - 22y + 121) - (x^2 - 4x + 4 + y^2 - 10y + 25) = 121 \] Cancel common terms: \[ (-6x - 12y + 117) = 121 \Rightarrow -6x - 12y = 4 \Rightarrow 3x + 6y = -2 \Rightarrow x + 2y = -\frac{2}{3} \] This is the equation of a straight line. Slope is: \[ \text{Slope} = -\frac{1}{2} \]