Question 10 – Find Relation Between x and y for Equidistant Point
Find a relation between x and y such that the point \((x, y)\) is equidistant from the points \((3, 6)\) and \((-3, 4)\).
🖼️ Visual Representation
Figure: The line 3x + y = 5 is the perpendicular bisector of AB, containing all points equidistant from A and B
📋 Given
- Point A: \((3, 6)\)
- Point B: \((-3, 4)\)
- Point P: \((x, y)\) — any general point
- Condition: PA = PB (P is equidistant from A and B)
🎯 To Find
A relation (equation) between x and y that represents all points equidistant from A and B.
💡 This relation will be the equation of a straight line called the perpendicular bisector of line segment AB.
📐 Formula
Condition: If PA = PB, then \(PA^2 = PB^2\)
📝 Solution
Since P(x, y) is equidistant from A(3, 6) and B(-3, 4):
\[ PA = PB \]Squaring both sides to eliminate square roots:
\[ PA^2 = PB^2 \]Using the distance formula with P(x, y) and A(3, 6):
\[ PA^2 = (x – 3)^2 + (y – 6)^2 \]Using the distance formula with P(x, y) and B(-3, 4):
\[ PB^2 = (x – (-3))^2 + (y – 4)^2 \] \[ PB^2 = (x + 3)^2 + (y – 4)^2 \]Left side:
\[ x^2 – 6x + 9 + y^2 – 12y + 36 \] \[ = x^2 + y^2 – 6x – 12y + 45 \]Right side:
\[ x^2 + 6x + 9 + y^2 – 8y + 16 \] \[ = x^2 + y^2 + 6x – 8y + 25 \]Cancel \(x^2\) and \(y^2\) from both sides:
\[ -6x – 12y + 45 = 6x – 8y + 25 \]Bring all terms to one side:
\[ -6x – 12y + 45 – 6x + 8y – 25 = 0 \] \[ -12x – 4y + 20 = 0 \]Or in standard form:
\[ 3x + y = 5 \]or equivalently: \( 3x + y – 5 = 0 \)
✓ Verification
Let’s verify with a point on the line \(3x + y = 5\):
Example: Take point P(1, 2) which satisfies \(3(1) + 2 = 5\) ✓
| Distance | Calculation | Result |
|---|---|---|
| PA | \(\sqrt{(1-3)^2 + (2-6)^2} = \sqrt{4 + 16}\) | \(\sqrt{20} = 2\sqrt{5}\) |
| PB | \(\sqrt{(1-(-3))^2 + (2-4)^2} = \sqrt{16 + 4}\) | \(\sqrt{20} = 2\sqrt{5}\) |
Since PA = PB, our relation is verified! ✓
💡 Geometric Interpretation
Understanding the Result:
- The equation \(3x + y = 5\) represents a straight line
- This line is the perpendicular bisector of line segment AB
- Every point on this line is equidistant from A(3, 6) and B(-3, 4)
- The line passes through the midpoint M of AB
Finding the Midpoint M:
\[ M = \left(\frac{3 + (-3)}{2}, \frac{6 + 4}{2}\right) = (0, 5) \]Check: Does M lie on the line \(3x + y = 5\)?
\[ 3(0) + 5 = 5 \]Yes! ✓ The midpoint lies on our line.
Slope Verification:
- Slope of AB = \(\frac{4-6}{-3-3} = \frac{-2}{-6} = \frac{1}{3}\)
- Slope of perpendicular bisector = \(-3\) (from \(3x + y = 5\), or \(y = -3x + 5\))
- Product of slopes = \(\frac{1}{3} \times (-3) = -1\) ✓ (perpendicular condition)
⚠️ Common Mistakes
✅ Correct: Always square both sides first: \(PA^2 = PB^2\).
✅ Correct: \((x + 3)^2 = x^2 + 6x + 9\).
✅ Correct: \(-6x – 6x = -12x\).
✅ Correct: Divide by common factors to get \(3x + y = 5\).
💡 Key Points
- The locus of points equidistant from two fixed points is a straight line
- This line is called the perpendicular bisector
- The perpendicular bisector passes through the midpoint of the line segment
- It is perpendicular to the line joining the two points
- This concept is used in finding circumcenters of triangles
- When PA = PB, then \(PA^2 = PB^2\) (easier to work with)
- Always expand algebraic expressions carefully
- \(x^2\) and \(y^2\) terms cancel out, leaving a linear equation
- The result is always a first-degree equation (straight line)