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:
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:
Show 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.
Updated On: Jun 24, 2026
- \(\dfrac{1}{2}\)
- \(-2\)
- \(-\dfrac{1}{2}\)
- \(2\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Assume coordinates of moving point \(P\).
Let \[ P=(x,y) \] Given points are \[ A=(2,5) \] and \[ B=(5,11) \]
Step 2: Write \(AP^2\) and \(BP^2\).
Using distance formula, \[ AP^2=(x-2)^2+(y-5)^2 \] Similarly, \[ BP^2=(x-5)^2+(y-11)^2 \]
Step 3: Use the given condition.
Given, \[ BP^2-AP^2=121 \] Substitute the values: \[ (x-5)^2+(y-11)^2-\left[(x-2)^2+(y-5)^2\right]=121 \]
Step 4: Expand and simplify.
Now, \[ (x^2-10x+25)+(y^2-22y+121) - \left[(x^2-4x+4)+(y^2-10y+25)\right] =121 \] \[ x^2-10x+25+y^2-22y+121-x^2+4x-4-y^2+10y-25=121 \] Cancel common terms: \[ -6x-12y+117=121 \] \[ -6x-12y=4 \] \[ 6x+12y+4=0 \] Divide by \(2\): \[ 3x+6y+2=0 \]
Step 5: Find the slope of the straight line.
The equation of the locus is \[ 3x+6y+2=0 \] For a line \[ Ax+By+C=0, \] the slope is \[ -\frac{A}{B} \] Here, \[ A=3,\quad B=6 \] Therefore, \[ \text{slope}=-\frac{3}{6} \] \[ \text{slope}=-\frac{1}{2} \]
Step 6: Final conclusion.
Hence, the slope of the locus is \[ \boxed{-\frac{1}{2}} \]
Let \[ P=(x,y) \] Given points are \[ A=(2,5) \] and \[ B=(5,11) \]
Step 2: Write \(AP^2\) and \(BP^2\).
Using distance formula, \[ AP^2=(x-2)^2+(y-5)^2 \] Similarly, \[ BP^2=(x-5)^2+(y-11)^2 \]
Step 3: Use the given condition.
Given, \[ BP^2-AP^2=121 \] Substitute the values: \[ (x-5)^2+(y-11)^2-\left[(x-2)^2+(y-5)^2\right]=121 \]
Step 4: Expand and simplify.
Now, \[ (x^2-10x+25)+(y^2-22y+121) - \left[(x^2-4x+4)+(y^2-10y+25)\right] =121 \] \[ x^2-10x+25+y^2-22y+121-x^2+4x-4-y^2+10y-25=121 \] Cancel common terms: \[ -6x-12y+117=121 \] \[ -6x-12y=4 \] \[ 6x+12y+4=0 \] Divide by \(2\): \[ 3x+6y+2=0 \]
Step 5: Find the slope of the straight line.
The equation of the locus is \[ 3x+6y+2=0 \] For a line \[ Ax+By+C=0, \] the slope is \[ -\frac{A}{B} \] Here, \[ A=3,\quad B=6 \] Therefore, \[ \text{slope}=-\frac{3}{6} \] \[ \text{slope}=-\frac{1}{2} \]
Step 6: Final conclusion.
Hence, the slope of the locus is \[ \boxed{-\frac{1}{2}} \]
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