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Coordinate Geometry

Distance Formula

What is the distance formula? The distance formula calculates the straight-line distance between any two points in a coordinate plane. If P(\(x_{1}\), \(y_{1}\)) and Q(\(x_{2}\), \(y_{2}\)) are two points, the distance PQ is given by:

PQ = √[(\(x_{2} - x_{1}\))\(^{2}\) + (\(y_{2} - y_{1}\))\(^{2}\)]

This formula comes from the Pythagorean theorem — the distance is the hypotenuse of a right triangle whose legs are the horizontal and vertical differences between the points.

Real-life analogy: Imagine you are at point A (your home) and want to know how far away point B (your school) is. Instead of walking along streets (horizontal and vertical), the distance formula gives you the "as the crow flies" straight-line distance. GPS navigation uses this to calculate shortest paths!

Key properties:

  • Distance is always non-negative
  • Distance from P to Q = Distance from Q to P (symmetric)
  • If points lie on a horizontal line (\(y_{1}\) = \(y_{2}\)), distance = |\(x_{2} - x_{1}\)|
  • If points lie on a vertical line (\(x_{1}\) = \(x_{2}\)), distance = |\(y_{2} - y_{1}\)|
┌─────────────────────────────────────────────────────────────┐
│              DISTANCE FORMULA - PYTHAGORAS IN ACTION         │
└─────────────────────────────────────────────────────────────┘

VISUAL DERIVATION:

    y
    │
    │    Q(x₂, y₂)
    │     /|
    │    / |
    │   /  | (y₂ - y₁)
    │  /   |
    │ /    |
    │/_____|
    │P(x₁,y₁) (x₂ - x₁)
    │
    └──────────────────► x

    PQ = √[(x₂-x₁)² + (y₂-y₁)²] (hypotenuse)


STEP-BY-STEP CALCULATION:

    Points: A(2, 3) and B(5, 7)
    
    Step 1: x₂ - x₁ = 5 - 2 = 3
    Step 2: y₂ - y₁ = 7 - 3 = 4
    Step 3: Square them: 3² = 9, 4² = 16
    Step 4: Add: 9 + 16 = 25
    Step 5: Square root: √25 = 5
    
    ✓ Distance AB = 5 units


REAL-LIFE APPLICATION (CITY MAP):

        City Grid (1 unit = 1 km)
    
        (0,5)   (3,5)
           ┌─────┐
           │     │
       (0,2)     │(3,2)
           │     │
           └─────┘
        (0,0)   (3,0)
    
    Distance from (0,2) to (3,5):
    = √[(3-0)² + (5-2)²] = √[9 + 9] = √18 ≈ 4.24 km


DISTANCE FROM ORIGIN:

    For point P(x, y), distance from origin O(0,0):
    
    OP = √[(x-0)² + (y-0)²] = √(x² + y²)
    
    Example: P(4, 3) → OP = √(16+9) = √25 = 5
Example 1: Find the distance between points A(2, 3) and B(5, 7).
  1. Step 1: \(x_{1}\) = 2, \(y_{1}\) = 3; \(x_{2}\) = 5, \(y_{2}\) = 7
  2. Step 2: \(x_{2} - x_{1}\) = 5 − 2 = 3
  3. Step 3: \(y_{2} - y_{1}\) = 7 − 3 = 4
  4. Step 4: Apply formula: AB = \(\sqrt{\(3^{2} + 4^{2}\)}\) = \(\sqrt{9 + 16}\) = \(\sqrt{25}\)

Answer: AB = 5 units

Example 2: Check whether points P(1, 2), Q(4, 6), and R(7, 10) are collinear.
  1. Step 1: Find PQ = √[(4−1)\(^{2}\) + (6−2)\(^{2}\)] = \(\sqrt{\(3^{2} + 4^{2}\)}\) = \(\sqrt{25}\) = 5
  2. Step 2: Find QR = √[(7−4)\(^{2}\) + (10−6)\(^{2}\)] = \(\sqrt{\(3^{2} + 4^{2}\)}\) = \(\sqrt{25}\) = 5
  3. Step 3: Find PR = √[(7−1)\(^{2}\) + (10−2)\(^{2}\)] = \(\sqrt{\(6^{2} + 8^{2}\)}\) = \(\sqrt{36+64}\) = \(\sqrt{100}\) = 10
  4. Step 4: Check if PQ + QR = PR: 5 + 5 = 10 ✓

Answer: Yes, points are collinear

Example 3: Find the point on the x-axis which is equidistant from (5, 4) and (−2, 3).
  1. Step 1: Let point on x-axis be P(x, 0)
  2. Step 2: Distance from P to A(5,4) = √[(x−5)\(^{2}\) + (0−4)\(^{2}\)] = √[(x−5)\(^{2} + 16\)]
  3. Step 3: Distance from P to B(−2,3) = √[(x+2)\(^{2}\) + (0−3)\(^{2}\)] = √[(x+2)\(^{2} + 9\)]
  4. Step 4: Set equal: (x−5)\(^{2} + 16\) = (x+2)\(^{2} + 9
  5. Step\) 5: Expand: \(x^{2} - 10x + 25 + 16\) = \(x^{2} + 4x + 4 + 9
  6. Step\) 6: Simplify: \(x^{2} - 10x + 41\) = \(x^{2} + 4x + 13
  7. Step\) 7: Cancel \(x^{2}\): −10x + 41 = 4x + 13 → −14x = −28 → x = 2

Answer: P(2, 0)

Quick recap
  • Distance formula: d = √[(\(x_{2}-x_{1}\))\(^{2}\) + (\(y_{2}-y_{1}\))\(^{2}\)]
  • Formula comes from the Pythagorean theorem
  • Distance is always a non-negative real number
  • For collinearity check, sum of two distances equals the third
  • Distance from origin: d = \(\sqrt{\(x^{2} + y^{2}\)}\)
  • Points on x-axis have y = 0; on y-axis have x = 0
✓ Quick check
The area of the triangle with vertices (0, 0), (4, 0) and (0, 3) is:
xyOO(4,0)(0,3)
It is right-angled with legs 4 and 3, so area = ½ × 4 × 3 = 6.
The centroid of a triangle is the:
The centroid has coordinates equal to the mean of the three vertices.

Section and Midpoint Formulae

What is the section formula? The section formula gives the coordinates of a point that divides a line segment joining two points A(\(x_{1}\), \(y_{1}\)) and B(\(x_{2}\), \(y_{2}\)) in a given ratio m : n (internally). The coordinates of point P dividing AB internally in the ratio m : n are:

P(x, y) = ( [m \(x_{2} + n\) \(x_{1}\)] / [m + n] , [m \(y_{2} + n\) \(y_{1}\)] / [m + n] )

Understanding the formula:

  • If P divides AB such that AP : PB = m : n, then P is closer to A if m < n, closer to B if m > n
  • When m = n (equal ratio 1:1), P becomes the midpoint: P = ((\(x_{1}+x_{2}\))/2, (\(y_{1}+y_{2}\))/2)

Important notes:

  • This formula is for internal division (P lies between A and B)
  • For external division (P outside AB), a modified formula is used (not in Grade 10 syllabus)
  • The ratio m : n can be written as AP : PB
┌─────────────────────────────────────────────────────────────┐
│         SECTION FORMULA (INTERNAL DIVISION) - VISUAL        │
└─────────────────────────────────────────────────────────────┘

INTERNAL DIVISION ON A LINE SEGMENT:

    A──────────P──────────B
    │◄── m ──►│◄── n ──►│
    
    A(x₁, y₁)          B(x₂, y₂)
         \                 /
          \               /
           \             /
            \           /
             \         /
              \       /
               \ P  /
                \ /
                 
    P divides AB in ratio m : n internally


MIDPOINT FORMULA (SPECIAL CASE m = n = 1):

    A(2, 3)────────────────────B(8, 7)
                       │
                       │
                  P((2+8)/2, (3+7)/2)
                       │
                       ▼
                    P(5, 5)


STEP-BY-STEP EXAMPLE:

    A(1, 2) and B(7, 10). Find point P dividing AB in ratio 1:3
    
        A────P──────────────B
        1    3
        (1,2)    (7,10)
    
    x = (m·x₂ + n·x₁)/(m+n) = (1×7 + 3×1)/(1+3) = (7+3)/4 = 10/4 = 2.5
    y = (m·y₂ + n·y₁)/(m+n) = (1×10 + 3×2)/4 = (10+6)/4 = 16/4 = 4
    
    P = (2.5, 4)


RATIO COMPARISON TABLE:

┌────────────────┬──────────────────┬─────────────────────────┐
│    RATIO m:n   │    POSITION OF P  │     COORDINATES         │
├────────────────┼──────────────────┼─────────────────────────┤
│      1:1       │    Midpoint       │   ((x₁+x₂)/2, (y₁+y₂)/2) │
│      1:2       │    Closer to A    │   Weighted towards A    │
│      2:1       │    Closer to B    │   Weighted towards B    │
│      m > n     │    Nearer to B    │   m·x₂ term dominates   │
│      m < n     │    Nearer to A    │   n·x₁ term dominates   │
└────────────────┴──────────────────┴─────────────────────────┘
Example 1: Find the coordinates of the point dividing the line joining A(2, 3) and B(6, 7) in the ratio 1 : 3 internally.
  1. Step 1: Identify \(x_{1}\)=2, \(y_{1}\)=3, \(x_{2}\)=6, \(y_{2}\)=7, m=1, n=3
  2. Step 2: x-coordinate = (m·\(x_{2} + n\)·\(x_{1}\))/(m+n) = (1×6 + 3×2)/(1+3) = (6 + 6)/4 = 12/4 = 3
  3. Step 3: y-coordinate = (m·\(y_{2} + n\)·\(y_{1}\))/(m+n) = (1×7 + 3×3)/4 = (7 + 9)/4 = 16/4 = 4

Answer: P(3, 4)

Example 2: Find the midpoint of the line segment joining P(3, 5) and Q(9, 13).
  1. Step 1: Midpoint divides in ratio 1:1, so m=n=1
  2. Step 2: x = (1×9 + 1×3)/(1+1) = (9+3)/2 = 12/2 = 6
  3. Step 3: y = (1×13 + 1×5)/2 = (13+5)/2 = 18/2 = 9

Answer: M(6, 9)

Example 3: In what ratio does the point P(3, 4) divide the line segment joining A(1, 2) and B(5, 6)?
  1. Step 1: Let ratio be m : n. Using x-coordinate: 3 = (m×5 + n×1)/(m+n)
  2. Step 2: Multiply: 3(m+n) = 5m + n → 3m + 3n = 5m + n
  3. Step 3: Rearrange: 3n − n = 5m − 3m → 2n = 2m → m/n = 1 → m:n = 1:1
  4. Step 4: Verify with y: (m×6 + n×2)/(m+n) = (6+2)/2 = 4 ✓

Answer: P divides AB in ratio 1:1 (midpoint)

Quick recap
  • Section formula (internal): P = ((m·\(x_{2} + n\)·\(x_{1}\))/(m+n), (m·\(y_{2} + n\)·\(y_{1}\))/(m+n))
  • For midpoint (1:1), P = ((\(x_{1}+x_{2}\))/2, (\(y_{1}+y_{2}\))/2)
  • The point is closer to A if m < n, closer to B if m > n
  • Always verify with both x and y coordinates
  • The denominator is always the sum m + n
  • This formula only works for internal division (point between A and B)
✓ Quick check
The distance between (2, 3) and (2, 8) is:
Same x-coordinate, so the distance is |8 − 3| = 5.
The distance between the points (−5, 7) and (−1, 3) is:
√((−5 + 1)² + (7 − 3)²) = √(16 + 16) = 4√2.

Area of a Triangle and Collinearity

What is the area of a triangle using coordinates? For a triangle with vertices A(\(x_{1}\), \(y_{1}\)), B(\(x_{2}\), \(y_{2}\)), and C(\(x_{3}\), \(y_{3}\)), the area is:

Area = (1/2) | \(x_{1}\)(\(y_{2} - y_{3}\)) + \(x_{2}\)(\(y_{3} - y_{1}\)) + \(x_{3}\)(\(y_{1} - y_{2}\)) |

The vertical bars | | mean absolute value (area is always positive). If the result is 0, the three points are collinear (lie on a straight line).

What is the reflection of a point in coordinate axes? Reflection means flipping a point over a line (like a mirror). The key rules:

Reflection inOriginal (x, y)Reflected point
x-axis(x, y)(x, −y)
y-axis(x, y)(−x, y)
Origin(x, y)(−x, −y)
Line y = x(x, y)(y, x)

Real-life analogy: Reflection is like looking at your face in a mirror — the mirror is the axis. When you raise your left hand, your reflection raises its right hand!

┌─────────────────────────────────────────────────────────────┐
│         AREA OF TRIANGLE & REFLECTION OF POINTS              │
└─────────────────────────────────────────────────────────────┘

AREA OF TRIANGLE FORMULA:

    A(x₁, y₁)
       /\
      /  \
     /    \
    /      \
   /________\
B(x₂, y₂)   C(x₃, y₃)

Area = ½ |x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)|

EXAMPLE: A(1,1), B(4,1), C(4,5)

Area = ½ |1(1-5) + 4(5-1) + 4(1-1)|
     = ½ |1(-4) + 4(4) + 4(0)|
     = ½ |-4 + 16 + 0| = ½ × 12 = 6 sq units ✓


REFLECTION OF POINTS IN AXES:

    ORIGINAL POINT P(3, 4)
    
    Reflection in x-axis:
        P'(3, -4)
        
    Reflection in y-axis:
        P''(-3, 4)
        
    Reflection in origin:
        P'''(-3, -4)


REFLECTION VISUAL ON GRID:

          y
          │
     (-3,4)│   (3,4)
         \ │ /
          \│/
    ────────┼──────── x
          /│\
         / │ \
    (-3,-4)│  (3,-4)
          │
          
    Each point's reflection across axes forms a rectangle!


MIDPOINT AS AVERAGE COORDINATES:

    A(2, 3) and B(10, 7)
    
    Midpoint M = ( (2+10)/2, (3+7)/2 ) = (6, 5)
    
    This is the "average" of the two endpoints!
Example 1: Find the area of triangle with vertices A(1, 2), B(4, 5), and C(7, 8).
  1. Step 1: Use area formula: ½ |\(x_{1}\)(\(y_{2}-y_{3}\)) + \(x_{2}\)(\(y_{3}-y_{1}\)) + \(x_{3}\)(\(y_{1}-y_{2}\))|
  2. Step 2: Substitute: ½ |1(5−8) + 4(8−2) + 7(2−5)|
  3. Step 3: Simplify: ½ |1(−3) + 4(6) + 7(−3)| = ½ |−3 + 24 − 21|
  4. Step 4: ½ |0| = 0

Answer: Area = 0 → points are collinear

Example 2: Find the reflection of point P(5, −3) in the x-axis and in the y-axis.
  1. Step 1: Reflection in x-axis: (x, y) → (x, −y)
  2. Step 2: P(5, −3) → \(P_{1}\)(5, 3)
  3. Step 3: Reflection in y-axis: (x, y) → (−x, y)
  4. Step 4: P(5, −3) → \(P_{2}\)(−5, −3)

Answer: In x-axis: (5, 3); in y-axis: (−5, −3)

Example 3: If the midpoint of a line segment joining A(2a, 4) and B(−2, 3b) is C(1, 2a+1), find a and b.
  1. Step 1: Midpoint formula: ((2a + (−2))/2, (4 + 3b)/2) = (1, 2a+1)
  2. Step 2: x-coordinate: (2a − 2)/2 = 1 → 2a − 2 = 2 → 2a = 4 → a = 2
  3. Step 3: y-coordinate: (4 + 3b)/2 = 2a + 1 = 2(2) + 1 = 5
  4. Step 4: 4 + 3b = 10 → 3b = 6 → b = 2

Answer: a = 2, b = 2

Quick recap
  • Area of triangle = ½ |\(x_{1}\)(\(y_{2}-y_{3}\)) + \(x_{2}\)(\(y_{3}-y_{1}\)) + \(x_{3}\)(\(y_{1}-y_{2}\))|
  • If area = 0, points are collinear
  • Reflection in x-axis: (x, y) → (x, −y)
  • Reflection in y-axis: (x, y) → (−x, y)
  • Reflection in origin: (x, y) → (−x, −y)
  • Midpoint = average of coordinates = ((\(x_{1}+x_{2}\))/2, (\(y_{1}+y_{2}\))/2)
✓ Quick check
A drone flies from (0, 0) to (8, 6) in a straight line. The distance flown is:
√(8² + 6²) = √100 = 10 units.
On a city map, a school is at (2, 3) and a park at (5, 7) (units in km). The straight-line distance between them is:
xyOSchoolPark
√((5 − 2)² + (7 − 3)²) = √(9 + 16) = 5 km.
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