Mathematics (0580) Study Notes: Coordinate Geometry

Welcome to the Coordinate Geometry chapter! This topic is all about linking Algebra and Geometry by using numbers (coordinates) to describe exact locations in a 2D space. Mastering these tools will allow you to find distances, calculate steepness, and describe lines using simple equations. It’s essential for understanding graphs, physics, and map reading!


1. The Basics: Cartesian Coordinates (C4.1 / E4.1)

What are Cartesian Coordinates?

The Cartesian coordinate system uses two axes—the horizontal axis (\(x\)-axis) and the vertical axis (\(y\)-axis)—which intersect at the origin, \((0, 0)\). We use an ordered pair \((x, y)\) to define the exact location of a point.

  • The \(x\)-coordinate is the distance along the horizontal axis.
  • The \(y\)-coordinate is the distance along the vertical axis.

Memory Aid: The Airplane Rule

Think of plotting a point like taking off from an airport:

Walk Across (\(x\)) before you Climb Up/Down (\(y\)). Always state the \(x\) value first, then the \(y\) value: \((x, y)\).

Example: The point \((3, -5)\) is 3 units right of the origin and 5 units down.

Quick Review: Coordinates

A coordinate pair is always written \((x, y)\). The axes divide the plane into four regions called quadrants.


2. Gradient of a Straight Line (C4.2 / E4.2)

What is Gradient?

The gradient (\(m\)) of a straight line measures its steepness and direction. Think of it as the slope of a roof or a hill.

We calculate the gradient using the ratio of the vertical change (the rise) to the horizontal change (the run).

$$m = \frac{\text{Vertical Change (Rise)}}{\text{Horizontal Change (Run)}}$$

Finding Gradient from Two Points (E4.2)

If you have two points, \((x_1, y_1)\) and \((x_2, y_2)\), the formula for the gradient is:

$$m = \frac{y_2 - y_1}{x_2 - x_1}$$

Did you know? It doesn't matter which point you label as \((x_1, y_1)\) and which as \((x_2, y_2)\), as long as you are consistent across the numerator and the denominator!

Types of Gradient

  • Positive Gradient (\(m > 0\)): The line slopes upwards from left to right (like climbing a hill).
  • Negative Gradient (\(m < 0\)): The line slopes downwards from left to right (like skiing downhill).
  • Zero Gradient (\(m = 0\)): A horizontal line (flat ground). This line has the equation \(y = k\).
  • Undefined Gradient: A vertical line (a cliff face). This line has the equation \(x = k\).
Key Takeaway: Gradient

Gradient (\(m\)) is Rise over Run. Consistency is key when using the formula: subtract the coordinates in the same order.


3. Distance and Midpoint (C4.3 / E4.3)

We can use coordinates to find the physical distance between two points and the exact centre point of the line segment connecting them.

Finding the Midpoint (C4.3 / E4.3)

The midpoint (\(M\)) is simply the average of the \(x\)-coordinates and the average of the \(y\)-coordinates.

If the end points are \((x_1, y_1)\) and \((x_2, y_2)\), the midpoint \(M\) is:

$$M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$$

Example: Find the midpoint of \(A(2, 8)\) and \(B(6, 4)\).
$$M = \left( \frac{2 + 6}{2}, \frac{8 + 4}{2} \right) = \left( \frac{8}{2}, \frac{12}{2} \right) = (4, 6)$$

Calculating the Length (Distance) (C4.3 / E4.3)

To find the length (\(D\)) of the line segment, we use Pythagoras' Theorem (\(a^2 + b^2 = c^2\)). Imagine the line segment is the hypotenuse of a right-angled triangle. The distance formula is:

$$D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

Step-by-Step Distance Calculation:

  1. Find the difference in \(x\) values: \((x_2 - x_1)\).
  2. Find the difference in \(y\) values: \((y_2 - y_1)\).
  3. Square both differences.
  4. Add the squared differences together.
  5. Take the square root of the result.
Key Takeaway: Midpoint and Distance

Midpoint uses addition (averaging). Distance uses subtraction and squares (like Pythagoras).


4. Equations of Linear Graphs (C4.4 / E4.4)

The Gradient-Intercept Form

The equation of a straight line is most commonly written in the gradient-intercept form:

$$y = mx + c$$

  • \(m\) is the gradient (steepness).
  • \(c\) is the \(y\)-intercept (the point where the line crosses the \(y\)-axis, i.e., where \(x=0\)).

Example: For the line \(y = 6x + 3\), the gradient is \(m=6\), and the line crosses the \(y\)-axis at \((0, 3)\).

Obtaining the Equation of a Line

If you are given the gradient (\(m\)) and a point \((x_1, y_1)\) on the line, you can find \(c\):

Step 1: Find \(m\) (use the gradient formula if given two points).

Step 2: Substitute \(m\), \(x_1\), and \(y_1\) into the equation \(y = mx + c\).

Step 3: Solve the resulting equation to find \(c\).

Step 4: Write the final equation using \(m\) and \(c\).

Example: A line has a gradient \(m = 4\) and passes through \((1, -3)\).
Substitute into \(y = mx + c\):
\(-3 = 4(1) + c\)
\(-3 = 4 + c\)
\(c = -7\)
The equation is: \(y = 4x - 7\).

Special Lines (C4.4 / E4.4)

The syllabus requires you to recognize the equations of horizontal and vertical lines:

  • Horizontal Line: \(y = k\). The gradient is zero. (e.g., \(y = 5\))
  • Vertical Line: \(x = k\). The gradient is undefined. (e.g., \(x = -2\))
Key Takeaway: Linear Equations

Every non-vertical straight line can be defined by its steepness (\(m\)) and its starting point (\(c\)) using \(y = mx + c\). Always ensure your final equation is in a fully simplified form.


5. Relationships Between Lines

Parallel Lines (C4.5 / E4.5)

Parallel lines are lines that run next to each other and never intersect. This means they must have the exact same steepness.

If line 1 has gradient \(m_1\) and line 2 has gradient \(m_2\), then for them to be parallel:

$$m_1 = m_2$$

Example: Find the equation of the line parallel to \(y = 4x - 1\) that passes through \((1, -3)\).
The gradient of the given line is \(m = 4\). The parallel line must also have \(m = 4\).
Use \(m=4\) and point \((1, -3)\) to find \(c\):
\(-3 = 4(1) + c\)
\(c = -7\)
Equation: \(y = 4x - 7\).

Perpendicular Lines (E4.6 - Extended Content Only)

Perpendicular lines meet at a right angle (\(90^\circ\)). Their gradients are related in a special way: they are negative reciprocals of each other.

If line 1 has gradient \(m_1\), the gradient of the perpendicular line, \(m_2\), is found by:

$$m_2 = -\frac{1}{m_1}$$

or

$$m_1 \times m_2 = -1$$

Memory Trick: "Flip it and Change the Sign"

  • If \(m_1 = 2\), then \(m_2 = -\frac{1}{2}\)
  • If \(m_1 = -\frac{3}{4}\), then \(m_2 = +\frac{4}{3}\)
  • If \(m_1 = -5\), then \(m_2 = \frac{1}{5}\)

Step-by-Step Finding a Perpendicular Line:

  1. Find the gradient (\(m_1\)) of the original line (rearrange to \(y=mx+c\) if needed).
  2. Calculate the negative reciprocal to find the new gradient (\(m_2\)).
  3. Use the new gradient (\(m_2\)) and the given point \((x_1, y_1)\) to find the \(y\)-intercept \(c\).
  4. Write the final equation in the form \(y = m_2x + c\).
Quick Review: Parallel and Perpendicular

Parallel: Same gradient (\(m_1 = m_2\)).
Perpendicular (Extended): Negative reciprocal gradient (\(m_1 \times m_2 = -1\)).