Question

Suppose a point \(P\) moves so that \[ BP^2-AP^2=121, \] where \(A\) and \(B\) are \((2,5)\) and \((5,11)\) respectively. Then the locus of \(P\) is a straight line, whose slope is:

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

Hint:

For locus problems involving difference of squares of distances from two fixed points, use the distance formula and simplify. The resulting equation is usually a straight line.

Answer 1

The Correct Option is C

Step 1: Set up coordinates for P.
Let $P = (x, y)$ be the moving point. The fixed points are $A = (2, 5)$ and $B = (5, 11)$.

Step 2: Write the given condition.
We are told \[ BP^2 - AP^2 = 121. \]

Step 3: Expand $BP^2$.
Using the distance formula, \[ BP^2 = (x-5)^2 + (y-11)^2. \] Expanding, \[ BP^2 = x^2 - 10x + 25 + y^2 - 22y + 121. \]

Step 4: Expand $AP^2$.
Similarly, \[ AP^2 = (x-2)^2 + (y-5)^2 = x^2 - 4x + 4 + y^2 - 10y + 25. \]

Step 5: Subtract and simplify.
\[ BP^2 - AP^2 = (x^2 - 10x + 25 + y^2 - 22y + 121) - (x^2 - 4x + 4 + y^2 - 10y + 25). \] \[ = -10x + 25 - 22y + 121 + 4x - 4 + 10y - 25. \] \[ = -6x - 12y + 117. \]

Step 6: Apply the condition and find the locus.
Setting this equal to 121: \[ -6x - 12y + 117 = 121. \] \[ -6x - 12y = 4. \] \[ 6x + 12y + 4 = 0. \] \[ 3x + 6y + 2 = 0. \] Rewrite as $y = -\frac{3}{6}x - \frac{2}{6} = -\frac{1}{2}x - \frac{1}{3}$. So the slope is $-\dfrac{1}{2}$.
\[ \boxed{-\dfrac{1}{2}} \]