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CIE IGCSE Maths Study Notes

3.4.1 Length of a Line Segment

Understanding the calculation of the length of a line segment in coordinate geometry is vital. This knowledge enables us to find the distance between two points on a Cartesian plane, a fundamental concept in many mathematical and real-world applications.

Introduction to the Distance Formula

The distance between any two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) on a Cartesian plane is calculated using the distance formula, which is an application of Pythagoras' theorem:

d=(x2x1)2+(y2y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}Distance formula

Image courtesy of Chili Math

Calculating the Length of a Line Segment

Example 1: Basic Calculation

Find the length of the line segment between A(1,2)A(1, 2) and B(4,6)B(4, 6).

Solution:

1. Identify Coordinates:

x1=1x_1 = 1, y1=2y_1 = 2, x2=4x_2 = 4, y2=6y_2 = 6.

2. Apply the Formula:

d=(41)2+(62)2d = \sqrt{(4 - 1)^2 + (6 - 2)^2}d=32+42d = \sqrt{3^2 + 4^2}d=9+16d = \sqrt{9 + 16}d=25d = \sqrt{25}d=5d = 5

The length of the line segment AB is 5 units.

Example 2: Negative Coordinates

Calculate the length of the line segment between C(3,1)C(-3, -1) and D(1,5)D(1, -5).

Solution:

1. Coordinates:

x1=3,y1=1,x2=1,y2=5x_1=-3, y_1=-1, x_2=1, y_2=-5

2. Apply the Formula:

d=(1(3))2+(5(1))2d = \sqrt{(1 - (-3))^2 + (-5 - (-1))^2}d=(1+3)2+(5+1)2d = \sqrt{(1 + 3)^2 + (-5 + 1)^2}d=42+(4)2d = \sqrt{4^2 + (-4)^2}d=16+16d = \sqrt{16 + 16}d=32d = \sqrt{32}d=42d = 4\sqrt{2}

The length of CD is 424\sqrt{2} units.

Practice Problems

Problem 1

Find the length of the line segment between E(2,3)E(2, -3) and F(4,1)F(-4, 1).

Solution:

1. Coordinates:

x1=2,y1=3,x2=4,y2=1x_1 = 2, y_1 = -3, x_2 = -4, y_2 = 1

2. Apply the Formula:

d=(42)2+(1(3))2d = \sqrt{(-4 - 2)^2 + (1 - (-3))^2}d=(6)2+(4)2d = \sqrt{(-6)^2 + (4)^2}d=36+16d = \sqrt{36 + 16}d=52d = \sqrt{52}d=213d = 2\sqrt{13}

Problem 2

Calculate the distance between G(2,2)G(-2, -2) and H(3,3)H(3, 3).

Solution:

1. Coordinates:

x1=2,y1=2,x2=3,y2=3x_1 = -2, y_1 = -2, x_2 = 3, y_2 = 3

2. Apply the Formula:

d=(3(2))2+(3(2))2d = \sqrt{(3 - (-2))^2 + (3 - (-2))^2}d=(5)2+(5)2d = \sqrt{(5)^2 + (5)^2}d=25+25d = \sqrt{25 + 25}d=50d = \sqrt{50}d=52d = 5\sqrt{2}

Key Points

  • Accuracy in Coordinates: Ensure coordinates are correctly placed within the formula.
  • Handling Negative Values: Pay attention to negative signs to avoid calculation errors.
  • Simplification: Simplify under the square root before calculating the final distance.
  • Units of Measurement: Lengths are always positive and depend on the context for units.

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