Question:
Suppose \(P\) and \(Q\) are the midpoints of the sides \(AB\) and \(BC\) of a triangle where \(A(1,3)\), \(B(3,7)\) and \(C(7,15)\) are vertices. Then the locus of \(R\) satisfying
\[
AC^2+QR^2=PR^2
\]
is
Suppose \(P\) and \(Q\) are the midpoints of the sides \(AB\) and \(BC\) of a triangle where \(A(1,3)\), \(B(3,7)\) and \(C(7,15)\) are vertices. Then the locus of \(R\) satisfying
\[
AC^2+QR^2=PR^2
\]
is
Show Hint
For locus problems involving distances, assume the variable point as \(R=(x,y)\), write all squared distances using the distance formula, and simplify.
Updated On: Jun 22, 2026
- \(6x+12y=297\)
- \(6x+12y+297=0\)
- \(12x+6y=297\)
- \(12x+6y+297=0\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Find the midpoints \(P\) and \(Q\).
Since \(P\) is the midpoint of \(AB\), \[ P=\left(\frac{1+3}{2},\frac{3+7}{2}\right)=(2,5) \] Since \(Q\) is the midpoint of \(BC\), \[ Q=\left(\frac{3+7}{2},\frac{7+15}{2}\right)=(5,11) \]
Step 2: Find \(AC^2\).
\[ AC^2=(7-1)^2+(15-3)^2 \] \[ =6^2+12^2 \] \[ =36+144=180 \]
Step 3: Let \(R=(x,y)\).
Then, \[ PR^2=(x-2)^2+(y-5)^2 \] and \[ QR^2=(x-5)^2+(y-11)^2 \] Given, \[ AC^2+QR^2=PR^2 \] So, \[ 180+(x-5)^2+(y-11)^2=(x-2)^2+(y-5)^2 \]
Step 4: Simplify the equation.
\[ PR^2-QR^2=180 \] \[ [(x-2)^2+(y-5)^2]-[(x-5)^2+(y-11)^2]=180 \] \[ 6x+12y-117=180 \] \[ 6x+12y=297 \]
Step 5: Final conclusion.
Hence, the locus of \(R\) is \[ \boxed{6x+12y=297} \] which corresponds to option (1).
Since \(P\) is the midpoint of \(AB\), \[ P=\left(\frac{1+3}{2},\frac{3+7}{2}\right)=(2,5) \] Since \(Q\) is the midpoint of \(BC\), \[ Q=\left(\frac{3+7}{2},\frac{7+15}{2}\right)=(5,11) \]
Step 2: Find \(AC^2\).
\[ AC^2=(7-1)^2+(15-3)^2 \] \[ =6^2+12^2 \] \[ =36+144=180 \]
Step 3: Let \(R=(x,y)\).
Then, \[ PR^2=(x-2)^2+(y-5)^2 \] and \[ QR^2=(x-5)^2+(y-11)^2 \] Given, \[ AC^2+QR^2=PR^2 \] So, \[ 180+(x-5)^2+(y-11)^2=(x-2)^2+(y-5)^2 \]
Step 4: Simplify the equation.
\[ PR^2-QR^2=180 \] \[ [(x-2)^2+(y-5)^2]-[(x-5)^2+(y-11)^2]=180 \] \[ 6x+12y-117=180 \] \[ 6x+12y=297 \]
Step 5: Final conclusion.
Hence, the locus of \(R\) is \[ \boxed{6x+12y=297} \] which corresponds to option (1).
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