Question 7 – Find Point on X-axis Equidistant from Two Points
🖼️ Diagram
Figure 1: Point P(-7,0) on x-axis is equidistant from A(2,-5) and B(-2,9)
📋 Given Information
| Item | Details |
|---|---|
| Point A | \((2, -5)\) |
| Point B | \((-2, 9)\) |
| Required Point | On x-axis (form: \((x, 0)\)) |
| Condition | Equidistant from A and B |
🎯 To Find
The coordinates of the point P on the x-axis such that \(PA = PB\), where A and B are the given points.
📐 Formula Used
Key Concept: Any point on the x-axis has coordinates of the form \((x, 0)\), where the y-coordinate is always 0.
📝 Step-by-Step Solution
Since P lies on the x-axis, its y-coordinate is 0. So, P has coordinates \((x, 0)\).
Since P is equidistant from A and B, we have:
\[ PA = PB \]Squaring both sides (to eliminate the square root):
\[ PA^2 = PB^2 \]Using points \(P(x, 0)\) and \(A(2, -5)\):
\[ PA^2 = (x – 2)^2 + (0 – (-5))^2 \] \[ PA^2 = (x – 2)^2 + 5^2 \] \[ PA^2 = (x – 2)^2 + 25 \]Using points \(P(x, 0)\) and \(B(-2, 9)\):
\[ PB^2 = (x – (-2))^2 + (0 – 9)^2 \] \[ PB^2 = (x + 2)^2 + (-9)^2 \] \[ PB^2 = (x + 2)^2 + 81 \]Expanding both sides:
\[ x^2 – 4x + 4 + 25 = x^2 + 4x + 4 + 81 \] \[ x^2 – 4x + 29 = x^2 + 4x + 85 \]Canceling \(x^2\) from both sides:
\[ -4x + 29 = 4x + 85 \]Bringing all x terms to one side:
\[ -4x – 4x = 85 – 29 \] \[ -8x = 56 \] \[ x = \frac{56}{-8} = -7 \]Since \(x = -7\) and the point lies on the x-axis (y = 0), the required point is:
\[ P(-7, 0) \]✓ Verification
Let’s verify that P(-7, 0) is indeed equidistant from A(2, -5) and B(-2, 9):
| Distance | Calculation | Result |
|---|---|---|
| PA | \(\sqrt{(-7-2)^2 + (0-(-5))^2} = \sqrt{81 + 25}\) | \(\sqrt{106}\) units |
| PB | \(\sqrt{(-7-(-2))^2 + (0-9)^2} = \sqrt{25 + 81}\) | \(\sqrt{106}\) units |
✅ Since PA = PB = \(\sqrt{106}\) units, our answer is verified!
💡 Alternative Method: Using Perpendicular Bisector Concept
Any point equidistant from two given points lies on the perpendicular bisector of the line segment joining those two points.
⚠️ Common Mistakes
❌ Mistake 1: Forgetting that a point on the x-axis has y-coordinate = 0.
✅ Correct Approach: Always write the point as \((x, 0)\) when it’s on the x-axis.
❌ Mistake 2: Not squaring both sides before equating PA and PB, leading to complex square root equations.
✅ Correct Approach: Square both sides first: \(PA^2 = PB^2\) to eliminate square roots.
❌ Mistake 3: Sign errors when expanding \((x + 2)^2\) or \((x – 2)^2\).
✅ Correct Approach: Remember: \((a \pm b)^2 = a^2 \pm 2ab + b^2\)
❌ Mistake 4: Calculation error: \((-9)^2 = -81\)
✅ Correct Approach: \((-9)^2 = 81\) (always positive)
💡 Key Points to Remember
- Any point on the x-axis has coordinates \((x, 0)\)
- Any point on the y-axis has coordinates \((0, y)\)
- If PA = PB, then \(PA^2 = PB^2\) (squaring eliminates square roots)
- The locus of points equidistant from two points is the perpendicular bisector
- Always verify your answer by calculating actual distances
- Use the distance formula correctly with proper signs
- When expanding \((a – b)^2\), remember the middle term: \(-2ab\)
- This concept is useful in finding circumcenters and other geometric constructions