Question 1 – Distance Between Points
Find the distance between the following pairs of points. This question demonstrates the application of the distance formula in coordinate geometry.
📑 Question Parts
Part (i)
Find the distance between the points \( (2, 3) \) and \( (4, 1) \).
📋 Given
- First point: \( P(2, 3) \) where \( x_1 = 2 \) and \( y_1 = 3 \)
- Second point: \( Q(4, 1) \) where \( x_2 = 4 \) and \( y_2 = 1 \)
🎯 To Find
Distance between points P and Q using the distance formula.
📐 Formula
📝 Solution
Substituting \( x_1 = 2, y_1 = 3, x_2 = 4, y_2 = 1 \):
\[ PQ = \sqrt{(4 – 2)^2 + (1 – 3)^2} \]Horizontal distance: \( 4 – 2 = 2 \)
Vertical distance: \( 1 – 3 = -2 \)
\[ PQ = \sqrt{(2)^2 + (-2)^2} \]Note: \( (2)^2 = 4 \) and \( (-2)^2 = 4 \)
\[ PQ = \sqrt{4 + 4} = \sqrt{8} \]Decimal approximation: \( 2\sqrt{2} \approx 2.83 \) units
💡 Alternative Method: Direct Application of Pythagoras Theorem
Step 1: Draw a right triangle with PQ as hypotenuse
- Base (horizontal leg) = \( |4 – 2| = 2 \) units
- Height (vertical leg) = \( |1 – 3| = 2 \) units
Step 2: Apply Pythagoras theorem
\[ PQ^2 = 2^2 + 2^2 = 4 + 4 = 8 \]Step 3: Take square root
\[ PQ = \sqrt{8} = 2\sqrt{2} \text{ units} \]This confirms our answer using the distance formula!
Part (ii)
Find the distance between the points \( (-5, 7) \) and \( (-1, 3) \).
📋 Given
- First point: \( A(-5, 7) \) where \( x_1 = -5 \) and \( y_1 = 7 \)
- Second point: \( B(-1, 3) \) where \( x_2 = -1 \) and \( y_2 = 3 \)
🎯 To Find
Distance between points A and B (both in the second quadrant).
📐 Formula
📝 Solution
Substituting \( x_1 = -5, y_1 = 7, x_2 = -1, y_2 = 3 \):
\[ AB = \sqrt{(-1 – (-5))^2 + (3 – 7)^2} \]For x-coordinates: \( -1 – (-5) = -1 + 5 = 4 \)
For y-coordinates: \( 3 – 7 = -4 \)
\[ AB = \sqrt{(4)^2 + (-4)^2} \]Both \( (4)^2 = 16 \) and \( (-4)^2 = 16 \)
\[ AB = \sqrt{16 + 16} = \sqrt{32} \]Decimal approximation: \( 4\sqrt{2} \approx 5.66 \) units
💡 Alternative Method: Component Analysis
Horizontal displacement:
\[ \Delta x = |-1 – (-5)| = |4| = 4 \text{ units} \]Vertical displacement:
\[ \Delta y = |3 – 7| = |-4| = 4 \text{ units} \]Apply Pythagoras theorem:
\[ d = \sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \text{ units} \]Interesting observation: Since both displacements are equal (4 units), the line AB makes a 45° angle with the coordinate axes!
Part (iii)
Find the distance between the points \( (a, b) \) and \( (-a, -b) \).
📋 Given
- First point: \( P(a, b) \) where \( x_1 = a \) and \( y_1 = b \)
- Second point: \( Q(-a, -b) \) where \( x_2 = -a \) and \( y_2 = -b \)
- These are general algebraic coordinates (variables, not specific numbers)
🎯 To Find
A general formula for the distance between two points that are symmetric about the origin.
📐 Formula
📝 Solution
Substituting \( x_1 = a, y_1 = b, x_2 = -a, y_2 = -b \):
\[ PQ = \sqrt{(-a – a)^2 + (-b – b)^2} \]For x-coordinates: \( -a – a = -2a \)
For y-coordinates: \( -b – b = -2b \)
\[ PQ = \sqrt{(-2a)^2 + (-2b)^2} \]Remember: \( (-2a)^2 = 4a^2 \) and \( (-2b)^2 = 4b^2 \)
\[ PQ = \sqrt{4a^2 + 4b^2} \]🔍 Verification with Example
Let’s verify with specific values: If \( a = 3 \) and \( b = 4 \)
Using our formula:
\[ PQ = 2\sqrt{3^2 + 4^2} = 2\sqrt{9 + 16} = 2\sqrt{25} = 2 \times 5 = 10 \text{ units} \]Direct calculation: Points are P(3, 4) and Q(-3, -4)
\[ PQ = \sqrt{(-3-3)^2 + (-4-4)^2} = \sqrt{(-6)^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ units} \]✓ Both methods give the same result!
Geometric Meaning: The distance from origin to P(a, b) is \( \sqrt{a^2 + b^2} \). Since Q is the reflection of P through the origin, the distance PQ is exactly twice the distance from origin to P.
💡 Alternative Approach: Using Origin as Midpoint
Since (a, b) and (-a, -b) are symmetric about the origin, O(0, 0) is the midpoint of PQ.
Distance from origin to P(a, b):
\[ OP = \sqrt{a^2 + b^2} \]Distance from origin to Q(-a, -b):
\[ OQ = \sqrt{(-a)^2 + (-b)^2} = \sqrt{a^2 + b^2} \]Since P, O, Q are collinear with O as midpoint:
\[ PQ = OP + OQ = \sqrt{a^2 + b^2} + \sqrt{a^2 + b^2} = 2\sqrt{a^2 + b^2} \]This elegant approach highlights the beautiful symmetry of the problem!
⚠️ Common Mistakes to Avoid
❌ Mistake 1: Not squaring the differences before adding
Incorrect: \( d = \sqrt{x_2 – x_1 + y_2 – y_1} \)
✓ Correct: \( d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2} \)
❌ Mistake 2: Sign errors with negative coordinates
Writing \( -1 – (-5) = -6 \) instead of \( -1 + 5 = 4 \)
✓ Correct: Remember: \( a – (-b) = a + b \)
❌ Mistake 3: Not simplifying radicals
Leaving \( \sqrt{8} \) instead of simplifying to \( 2\sqrt{2} \)
✓ Correct: Always simplify: \( \sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2} \)
❌ Mistake 4: Forgetting that squaring eliminates negative signs
Worrying about order of subtraction unnecessarily
✓ Correct: \( (4)^2 = (-4)^2 = 16 \), so order doesn’t matter
💡 Key Points to Remember
- Distance Formula: \( d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2} \) works for any two points in any quadrant
- Derived from Pythagoras: The formula comes from the Pythagoras theorem applied to coordinate geometry
- Distance is always positive: The square root ensures the result is non-negative
- Order doesn’t matter: Distance from P to Q equals distance from Q to P
- Simplify radicals: Express answers in simplest form (e.g., \( 2\sqrt{2} \) not \( \sqrt{8} \))
- Symmetric points: Points (a, b) and (-a, -b) are \( 2\sqrt{a^2 + b^2} \) units apart
- Works with variables: The formula applies to algebraic expressions, not just numbers
📝 Practice Similar Problems
Strengthen your understanding by solving:
