🎯 IGCSE International Mathematics (0607) Study Notes: Coordinates
👋 Introduction: Why Coordinate Geometry Matters
Welcome to the world of Coordinate Geometry! This chapter is all about locating points and lines precisely on a 2D plane. Think of it like being an air traffic controller or using a map with GPS—you need exact instructions to know where things are and where they are going.
In this section, you will learn the essential tools to describe, measure, and define straight lines, which forms a vital foundation for graphing functions and tackling more complex geometry problems later on. Don't worry if the formulas look complicated; they are just efficient shortcuts for calculations!
1. Cartesian Coordinates: Finding Your Location
C4.1 / E4.1: Use and interpret Cartesian coordinates in two dimensions.
The Cartesian Coordinate System (or coordinate plane) uses two perpendicular lines (axes) to define every location.
Key Definitions
- The Origin: This is the starting point, where the two axes cross. Its coordinates are \((0, 0)\).
- The x-axis: The horizontal line. Movement along this axis is the first number in the coordinate pair.
- The y-axis: The vertical line. Movement along this axis is the second number.
- A Coordinate is always written as \((x, y)\).
🧠 Memory Aid: "Walk, then Climb"
Always move along the horizontal x-axis (walk along the corridor) first, and then move up or down the vertical y-axis (climb the stairs) second.
Example: The point \((4, -2)\) is 4 units right and 2 units down from the origin.
Key Takeaway
Coordinates \((x, y)\) are the precise address of a point on the plane. Always start at the origin and read \(x\) first, then \(y\).
2. Gradient of a Straight Line (Steepness)
C4.2 / E4.2: Find the gradient of a straight line.
The Gradient (\(m\)) measures the steepness and direction of a line. A high gradient means a very steep line, and a zero gradient means a flat line.
The Basic Concept: Rise over Run
The gradient is calculated as the change in the vertical distance (\(y\)) divided by the change in the horizontal distance (\(x\)).
$$ \text{Gradient } (m) = \frac{\text{Change in } y}{\text{Change in } x} = \frac{\text{Rise}}{\text{Run}} $$
Calculation Method 1: From a Grid (Core Content)
If you have a graph drawn on a grid, you can count the squares:
- Choose two clear points on the line.
- Count how many units you go up or down (Rise).
- Count how many units you go across (Run).
- Calculate \(m = \text{Rise} / \text{Run}\).
Calculation Method 2: From Two Points (Extended Content)
If you are only given the coordinates of two points, \((x_1, y_1)\) and \((x_2, y_2)\), you must use the gradient formula:
$$ m = \frac{y_2 - y_1}{x_2 - x_1} $$
Example: Find the gradient between \(A(1, 5)\) and \(B(4, -1)\).
$$ m = \frac{-1 - 5}{4 - 1} = \frac{-6}{3} = -2 $$
Interpreting the Gradient Sign
- Positive Gradient: The line goes uphill (from left to right).
- Negative Gradient: The line goes downhill (from left to right).
- Zero Gradient (\(m=0\)): A perfectly flat, horizontal line (\(y = k\)).
- Undefined Gradient: A perfectly straight, vertical line (\(x = k\)). (You cannot divide by zero!)
⚠️ Common Mistake Alert: Always ensure you subtract the coordinates in the same order! If you start with \(y_2\), you must start with \(x_2\) in the denominator.
Key Takeaway
Gradient (\(m\)) tells you the line's slope. Calculate it using \(\text{Rise}/\text{Run}\) (counting squares for Core) or the formula (for Extended, using coordinates).
3. Measuring Line Segments: Length and Midpoint
C4.3 / E4.3: Calculate the length and find the midpoint of a line segment.
3.1 Length (Distance Formula)
The distance between two points, \(A(x_1, y_1)\) and \(B(x_2, y_2)\), can be found by essentially creating a right-angled triangle and using Pythagoras' Theorem.
The horizontal distance is \((x_2 - x_1)\), and the vertical distance is \((y_2 - y_1)\).
$$ \text{Length } (d) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$
Did you know? Since you square the distances, it doesn't matter which point you label \((x_1, y_1)\) or \((x_2, y_2)\). The length will always be positive!
Example: Find the length between \((1, 2)\) and \((4, 6)\).
$$ d = \sqrt{(4 - 1)^2 + (6 - 2)^2} $$ $$ d = \sqrt{(3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ units} $$
3.2 Midpoint Formula
The Midpoint is the point exactly halfway between two endpoints. To find the halfway point, you simply find the average of the \(x\)-coordinates and the average of the \(y\)-coordinates.
$$ \text{Midpoint } (M) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) $$
💡 Analogy: Sharing the journey. If you meet a friend halfway, you average your starting position with their starting position!
Example: Find the midpoint between \((1, 2)\) and \((5, 10)\).
$$ M = \left( \frac{1 + 5}{2}, \frac{2 + 10}{2} \right) = \left( \frac{6}{2}, \frac{12}{2} \right) = (3, 6) $$
Quick Review Box: Measurement Formulas
- Length: \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) (Pythagoras)
- Midpoint: \(\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\) (Averages)
4. Equations of Linear Graphs
C4.4 / E4.4: Interpret and obtain the equation of a straight-line graph in the form \(y = mx + c\).
The equation of a straight line allows you to predict any point on that line. The most important form is the Gradient-Intercept Form:
$$ y = mx + c $$
- \(m\) is the Gradient (steepness).
- \(c\) is the y-intercept (where the line crosses the y-axis, i.e., the point \((0, c)\)).
Step-by-Step: Finding the Equation
To find the equation of a line given two points, or a point and the gradient, follow these steps:
- Find \(m\): Calculate the gradient using the formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\) (or count squares if on a grid).
- Find \(c\): Substitute a known point \((x, y)\) and the calculated gradient \(m\) into the equation \(y = mx + c\). Solve for \(c\).
- Write the Final Equation: Write the final answer in the form \(y = mx + c\), replacing \(m\) and \(c\) with their numerical values.
Example: Find the equation of the line with gradient \(m=3\) that passes through \((2, 7)\).
1. \(m=3\) (Given).
2. Use \((2, 7)\) in \(y = 3x + c\):
\(7 = 3(2) + c\)
\(7 = 6 + c\)
\(c = 1\)
3. Equation is: \(\mathbf{y = 3x + 1}\)
Special Cases: Horizontal and Vertical Lines
- Horizontal Lines: The gradient is zero (\(m=0\)). The equation is always:
$$ y = k $$
(Where \(k\) is the y-intercept. Example: \(y=5\))
- Vertical Lines: The gradient is undefined. The equation is always:
$$ x = k $$
(Where \(k\) is the x-intercept. Example: \(x=-3\))
(Extended Only) Alternative Equation Forms
Sometimes, a linear equation may be presented in the form \(\mathbf{ax + by = c}\). You must be able to rearrange this into the \(y=mx+c\) form to easily identify the gradient (\(m\)) and y-intercept (\(c\)).
Example: Find \(m\) and \(c\) for \(5x + 4y = 8\) (Syllabus E4.4 example)
\(4y = -5x + 8\)
\(y = -\frac{5}{4}x + \frac{8}{4}\)
\(y = -\frac{5}{4}x + 2\)
Therefore, \(m = -\frac{5}{4}\) and \(c = 2\).
Key Takeaway
The core linear equation is \(y=mx+c\). Use your calculated gradient (\(m\)) and substitute a point to find the y-intercept (\(c\)).
5. Parallel and Perpendicular Lines
These concepts link the gradient of one line to the gradient of another, allowing you to find equations for related lines.
C4.5 / E4.5: Parallel Lines
If two lines are parallel, they never meet, meaning they have exactly the same steepness (the same gradient).
If Line 1 has gradient \(m_1\) and Line 2 has gradient \(m_2\):
$$ m_1 = m_2 $$
Example: A line parallel to \(y = 4x - 1\) must also have a gradient of \(m = 4\).
Step-by-Step: Finding the Equation of a Parallel Line
Problem: Find the equation of the line parallel to \(y = 4x - 1\) that passes through \((1, -3)\). (Syllabus E4.5 example)
- The gradient of the given line is \(m = 4\). The new parallel line must also have \(m = 4\).
- Substitute \(m=4\) and the point \((1, -3)\) into \(y = mx + c\):
\(-3 = 4(1) + c\) - Solve for \(c\): \(-3 = 4 + c\), so \(c = -7\).
- The equation is \(\mathbf{y = 4x - 7}\).
E4.6: Perpendicular Lines (Extended Content Only)
If two lines are perpendicular, they meet at a right angle (\(90^\circ\)). Their gradients are related in a specific way: they are negative reciprocals of each other.
If Line 1 has gradient \(m_1\), the perpendicular gradient \(m_2\) is:
$$ m_2 = -\frac{1}{m_1} $$
This also means that when you multiply the two gradients together, you must get \(-1\):
$$ m_1 \times m_2 = -1 $$
Example Gradients:
- If \(m_1 = 5\), then \(m_2 = -\frac{1}{5}\).
- If \(m_1 = \frac{2}{3}\), then \(m_2 = -\frac{3}{2}\).
- If \(m_1 = -4\), then \(m_2 = \frac{1}{4}\).
Key Applications of Perpendicular Lines (Extended)
You may be asked to find the equation of a perpendicular bisector (E4.6). This combines two skills:
- Finding the Midpoint (bisect means cut in half).
- Finding the Perpendicular Gradient.
- Using the midpoint and the perpendicular gradient to form the equation \(y=mx+c\).
Example: Line L1 joins \((1, 1)\) and \((3, 5)\). Find the equation of the line perpendicular to L1 passing through \((3, 0)\).
1. Find \(m_1\): \(m_1 = \frac{5 - 1}{3 - 1} = \frac{4}{2} = 2\)
2. Find \(m_2\) (perpendicular gradient): \(m_2 = -\frac{1}{2}\)
3. Substitute \(m_2 = -\frac{1}{2}\) and the point \((3, 0)\) into \(y = mx + c\):
\(0 = (-\frac{1}{2})(3) + c\)
\(0 = -\frac{3}{2} + c\)
\(c = \frac{3}{2}\) or 1.5
4. The equation is: \(\mathbf{y = -\frac{1}{2}x + \frac{3}{2}}\)
Key Takeaway
Parallel lines share the same gradient (\(m_1 = m_2\)). Perpendicular lines have negative reciprocal gradients (\(m_1 \times m_2 = -1\)).