In the xy-plane, the perpendicular bisector of segment AB passes through points \(\mathrm{P(1, 6)}\) and \(\mathrm{Q(5, 2)}\). If A has...

1. TRANSLATE the problem information

• Given information:
  • Perpendicular bisector of segment AB passes through \(\mathrm{P(1, 6)}\) and \(\mathrm{Q(5, 2)}\)
  • Point A has coordinates \(\mathrm{(5, 6)}\)
  • Need to find coordinates of point B

2. INFER the key geometric relationship

• Since P and Q lie on the perpendicular bisector of AB, both points must be equidistant from A and B
• This means: distance from P to A = distance from P to B, and distance from Q to A = distance from Q to B
• We can use this to set up equations to solve for B's coordinates

3. Calculate distances from P and Q to point A

• Distance from \(\mathrm{P(1, 6)}\) to \(\mathrm{A(5, 6)}\):
  \(\mathrm{\sqrt{(1-5)^2 + (6-6)^2}}\)
  \(\mathrm{= \sqrt{16 + 0}}\)
  \(\mathrm{= 4}\)
• Distance from \(\mathrm{Q(5, 2)}\) to \(\mathrm{A(5, 6)}\):
  \(\mathrm{\sqrt{(5-5)^2 + (2-6)^2}}\)
  \(\mathrm{= \sqrt{0 + 16}}\)
  \(\mathrm{= 4}\)

4. Set up distance equations for point B

• Let \(\mathrm{B = (x, y)}\)
• Distance from P to B must equal 4: \(\mathrm{(1-x)^2 + (6-y)^2 = 16}\)
• Distance from Q to B must equal 4: \(\mathrm{(5-x)^2 + (2-y)^2 = 16}\)

5. SIMPLIFY by expanding both equations

• First equation:
  \(\mathrm{1 - 2x + x^2 + 36 - 12y + y^2 = 16}\)
  Simplifying: \(\mathrm{x^2 - 2x + y^2 - 12y = -21}\)
• Second equation:
  \(\mathrm{25 - 10x + x^2 + 4 - 4y + y^2 = 16}\)
  Simplifying: \(\mathrm{x^2 - 10x + y^2 - 4y = -13}\)

6. SIMPLIFY further by eliminating variables

• Subtract the first equation from the second:
  \(\mathrm{-8x + 8y = 8}\)
  Therefore: \(\mathrm{y = x + 1}\)

7. SIMPLIFY by substitution to find x

• Substitute \(\mathrm{y = x + 1}\) into the first equation:
  \(\mathrm{x^2 - 2x + (x+1)^2 - 12(x+1) = -21}\)
• Expanding:
  \(\mathrm{x^2 - 2x + x^2 + 2x + 1 - 12x - 12 = -21}\)
• Combining:
  \(\mathrm{2x^2 - 12x - 11 = -21}\)
• Simplifying:
  \(\mathrm{2x^2 - 12x + 10 = 0}\), or \(\mathrm{x^2 - 6x + 5 = 0}\)
• Factoring:
  \(\mathrm{(x - 5)(x - 1) = 0}\)
• Solutions: \(\mathrm{x = 5}\) or \(\mathrm{x = 1}\)

8. APPLY CONSTRAINTS to select the valid solution

• If \(\mathrm{x = 5}\), then \(\mathrm{y = 6}\), giving \(\mathrm{B = (5, 6) = A}\) (invalid - same point)
• If \(\mathrm{x = 1}\), then \(\mathrm{y = 2}\), giving \(\mathrm{B = (1, 2)}\) (valid)

Answer: \(\mathrm{B = (1, 2)}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize the equidistant property of perpendicular bisectors, instead trying to find the equation of the perpendicular bisector line and work from there. This approach is much more complex and often leads to algebraic mistakes or getting overwhelmed by the calculations. This causes them to get stuck and randomly select an answer.

Second Most Common Error:

Inadequate SIMPLIFY execution: Students set up the distance equations correctly but make algebraic errors when expanding or combining terms, particularly when substituting \(\mathrm{y = x + 1}\) back into the quadratic. Common mistakes include sign errors or incorrect expansion of \(\mathrm{(x+1)^2}\). This may lead them to select Choice A (-1, 2) or get confused values that don't match any answer choice.

The Bottom Line:

This problem tests whether students can connect the geometric concept of perpendicular bisectors to coordinate geometry through distance relationships. The key insight is recognizing that the equidistant property gives you the system of equations you need to solve.