Edwin Ogie Library
Coordinate Geometry E‑Book
Objectives:
• Determine the midpoint and gradient of a line segment.
• Find the distance between two points.
• Identify conditions for parallelism and perpendicularity.
• Find the equation of a line (two‑point, point‑slope, slope‑intercept, and general forms).
Page 1: Introduction
Coordinate geometry (analytic geometry) uses algebra to study geometric figures. In this e‑book, we learn to calculate midpoints, gradients, distances between points, and the equations of straight lines. We also examine conditions for parallelism and perpendicularity.
Page 2: Midpoint and Gradient of a Line Segment
The midpoint of a line segment joining (x1, y1) and (x₂, y₂) is given by:
((x1+x₂)/2, (y1+y₂)/2)
The gradient (slope) is given by:
(y₂ - y1) / (x₂ - x₁)
Page 3: Distance Between Two Points
The distance (d) between two points (x₁, y₁) and (x₂, y₂) is calculated by:
d = √[(x₂ - x₁)² + (y₂ - y₁)²]
Page 4: Parallel and Perpendicular Lines
Two lines are parallel if their gradients are equal. They are perpendicular if the product of their gradients is -1.
m₁ = m₂ (parallel) m₁ × m₂ = -1 (perpendicular)
Page 5: Equations of a Line – Two-Point Form
The two-point form of a line through (x₁, y₁) and (x₂, y₂) is:
(y - y₁) / (x - x₁) = (y₂ - y₁) / (x₂ - x₁)
This form is useful when two points on the line are known.
Page 6: Equations of a Line – Point-Slope and Slope-Intercept Forms
The point-slope form for a line with slope m through (x₁, y₁) is:
y - y₁ = m(x - x₁)
The slope-intercept form is:
y = mx + c
Page 7: Equation of a Line – General Form
The general form of a line is:
Ax + By + C = 0
This form is versatile and can be derived from other forms.
Page 8: Worked Examples – Midpoint, Gradient, and Distance
Example 1: Find the midpoint of the segment joining (2, 3) and (8, 11).
Solution:
Midpoint = ((2+8)/2, (3+11)/2) = (5, 7).
Example 2: Determine the gradient of the line passing through (1, 4) and (5, 12).
Solution:
Gradient, m = (12 - 4) / (5 - 1) = 8 / 4 = 2.
Example 3: Calculate the distance between the points (3, 4) and (7, 1).
Solution:
Distance = √[(7 - 3)² + (1 - 4)²] = √[16 + 9] = √25 = 5.
Page 9: Worked Examples – Parallel and Perpendicular Lines
Example 4: Determine if the lines with equations y = 2x + 3 and y = 2x - 4 are parallel or perpendicular.
Solution:
Both lines have the same gradient (2); hence, they are parallel.
Example 5: Check if the lines with equations y = -1/2x + 5 and y = 2x - 3 are perpendicular.
Solution:
Gradients are -1/2 and 2; since (-1/2)×2 = -1, they are perpendicular.
Page 10: Worked Examples – Equations of a Line (Two-Point Form)
Example 6: Find the equation of the line passing through (2, 3) and (6, 11) using the two‑point form.
Solution:
Gradient = (11-3)/(6-2)=8/4=2.
Two-point form: (y - 3)/(x - 2)=2, so y - 3 = 2(x - 2) → y = 2x - 4 + 3 → y = 2x - 1.
Page 11: Worked Examples – Equations of a Line (Point-Slope & Slope-Intercept Forms)
Example 7: Using the point (3, 4) and gradient 3, write the equation in point-slope form and convert it to slope-intercept form.
Solution:
Point-slope form: y - 4 = 3(x - 3).
Slope-intercept form: y = 3x - 9 + 4 → y = 3x - 5.
Example 8: Write the equation of the line in general form for a line with slope 2 and passing through (1, -3).
Solution:
Point-slope form: y + 3 = 2(x - 1) → y = 2x - 2 - 3 = 2x - 5.
General form: 2x - y - 5 = 0.
Page 12: Worked Examples – Additional Equations of a Line
Example 9: Determine the equation of a line in slope-intercept form that passes through (0, 2) and is parallel to y = -3x + 7.
Solution:
The gradient of y = -3x + 7 is -3. A parallel line has the same gradient. Using point (0, 2), the line is y = -3x + 2.
Example 10: Find the equation in two-point form for the line passing through (4, 5) and (8, 13).
Solution:
Gradient = (13 - 5)/(8 - 4) = 8/4 = 2.
Two-point form: (y - 5)/(x - 4) = 2 → y - 5 = 2(x - 4) → y = 2x - 8 + 5 = 2x - 3.
Page 13: Summary of Key Formulas and Concepts
- Midpoint: ((x₁+x₂)/2, (y₁+y₂)/2)
- Gradient: (y₂ - y₁)/(x₂ - x₁)
- Distance: √[(x₂ - x₁)² + (y₂ - y₁)²]
- Parallel lines: equal gradients; Perpendicular lines: product of gradients = -1
- Two-point form: (y - y₁)/(x - x₁) = (y₂ - y₁)/(x₂ - x₁)
- Point-slope form: y - y₁ = m(x - x₁)
- Slope-intercept form: y = mx + c
- General form: Ax + By + C = 0
Page 14: Extended Discussion and Applications
Coordinate geometry is used in many fields including engineering, navigation, and computer graphics. Being able to find midpoints, gradients, and distances allows you to model and solve real-world problems. Understanding line equations is essential for graphing and analysis.
Page 15: Final Summary and Quiz Introduction
- Midpoint, gradient, and distance formulas are fundamental.
- Parallelism: equal gradients; Perpendicularity: product of gradients equals -1.
- Equations of lines can be written in several forms: two-point, point-slope, slope-intercept, and general.
Review these concepts and then test your understanding with the quiz below.
30 CBT JAMB Quiz on Coordinate Geometry
Click the "Start Quiz" button to begin. You will have 15 minutes to answer 30 questions.
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