✨ Welcome to the World of Linear Graphs!
Hello mathematicians! This chapter, Equations of Linear Graphs, is fundamental to Coordinate Geometry. It’s all about describing straight lines using simple algebraic rules. Think of it as giving directions to a machine so it can draw a perfect straight path.
Why is this important? Because linear relationships are everywhere! From calculating the cost of a taxi ride based on distance, to predicting temperature changes over time, linear equations help us model simple, consistent change.
Section 1: The Cartesian Plane and Coordinates (C4.1 / E4.1)
Before we find equations, we must remember where we are working: the Cartesian plane (also known as the coordinate grid).
What are Coordinates?
Coordinates are an ordered pair \((x, y)\) that tells you the exact location of a point:
- The \(x\)-coordinate is the horizontal distance from the origin (0, 0).
- The \(y\)-coordinate is the vertical distance from the origin (0, 0).
Remember: We walk along the corridor (\(x\)) before going up the stairs (\(y\)).
Key Takeaway:
Every point on a straight line has coordinates \((x, y)\) that satisfy the line's equation.
Section 2: The Concept of Gradient (C4.2 / E4.2)
The gradient is the measure of how steep a line is, and in which direction it slopes.
Imagine the gradient \(m\) as the pitch of a roof or the steepness of a hill. It tells you the rate of change of \(y\) compared to \(x\).
Calculating the Gradient from a Grid
If you have a grid, you can find the gradient using the simple ratio:
\[m = \frac{\text{Vertical Change (Rise)}}{\text{Horizontal Change (Run)}}\]
If the line goes up from left to right, the gradient is positive. If it goes down, the gradient is negative.
Calculating the Gradient from Two Points
If you are given two points, \((x_1, y_1)\) and \((x_2, y_2)\), use the formula:
\[m = \frac{y_2 - y_1}{x_2 - x_1}\]
🧠 Memory Aid: The Gradient Formula
Think of it as the Difference in Y divided by the Difference in X.
Example: Find the gradient of the line passing through (1, 5) and (3, 11).
\(m = \frac{11 - 5}{3 - 1} = \frac{6}{2} = 3\)
The line has a steepness of 3.
Special Gradients
There are two special types of straight lines:
- Horizontal Lines: These lines are perfectly flat. The rise is zero.
- Gradient \(m = 0\).
- Equation form: \(y = c\) (where \(c\) is the value of the \(y\)-intercept). Example: \(y=5\).
- Vertical Lines: These lines go straight up and down. The run is zero, and you cannot divide by zero.
- Gradient is Undefined.
- Equation form: \(x = k\) (where \(k\) is the value of the \(x\)-intercept). Example: \(x=-2\).
Quick Review: The gradient \(m\) tells us how steep and directional the line is.
Section 3: The Equation of a Straight Line (C4.4 / E4.4)
The most common and important way to write the equation of a straight line is the gradient-intercept form:
\[y = mx + c\]
Understanding the Components
-
\(m\) is the Gradient:
As discussed, this is the slope or steepness of the line.
-
\(c\) is the \(y\)-intercept:
This is the point where the line crosses the \(y\)-axis. At this point, \(x\) is always 0. The coordinates of the \(y\)-intercept are \((0, c)\).
Did you know? The term 'intercept' means to cut or cross. The y-intercept is where the line cuts the y-axis.
Interpreting the Equation
If you are given an equation in the form \(y = mx + c\), you can immediately determine its properties.
Example: Find the gradient and y-intercept of \(y = -2x + 7\).
- Gradient \(m = -2\). (The line slopes downwards.)
- \(y\)-intercept \(c = 7\). (The line crosses the \(y\)-axis at \((0, 7)\).)
Alternative Forms of Linear Equations (Extended Content)
Sometimes you will see linear equations written differently, for example, the general form \(ax + by = c\). You must be able to convert these back into the \(y = mx + c\) form to find the gradient and intercept.
Example: Find the gradient of the line \(5x + 4y = 8\). (As per syllabus E4.4)
Step 1: Isolate the \(y\) term.
\(4y = 8 - 5x\)
Step 2: Divide by the coefficient of \(y\).
\(y = \frac{8}{4} - \frac{5}{4}x\)
Step 3: Rearrange into standard form.
\(y = -\frac{5}{4}x + 2\)
Therefore, the gradient \(m = -\frac{5}{4}\) and the \(y\)-intercept \(c = 2\).
Quick Review: Always aim to rearrange equations into \(y=mx+c\) to easily identify the gradient \(m\) and the \(y\)-intercept \(c\).
Section 4: Finding the Equation of a Line (C4.4 / E4.4)
This is the main skill in this chapter. You will be asked to find the equation given different pieces of information. We will use the model \(y = mx + c\).
Method 1: Given Gradient (\(m\)) and a Point \((x, y)\)
If you know the gradient, you just need to find the \(y\)-intercept, \(c\).
Step-by-Step Process:
- Start with the template: \(y = mx + c\).
- Substitute the gradient \(m\).
- Substitute the coordinates \((x, y)\) of the given point into the equation.
- Solve the resulting equation to find \(c\).
- Write the final equation using the known \(m\) and calculated \(c\).
Example: Find the equation of a line with gradient 4 that passes through the point (2, 9).
- \(m = 4\), so \(y = 4x + c\).
- Substitute (2, 9): \(9 = 4(2) + c\).
- \(9 = 8 + c\).
- \(c = 1\).
- Final Equation: \(y = 4x + 1\).
Method 2: Given Two Points (C4.2 & C4.4 / E4.2 & E4.4)
If you are given two points, you need to calculate \(m\) first, then find \(c\).
Step-by-Step Process:
- Calculate \(m\): Use the gradient formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\).
- Find \(c\): Substitute the calculated \(m\) and the coordinates of EITHER of the two points into \(y = mx + c\).
- Write the final equation.
Example: Find the equation of the line passing through A(4, 1) and B(6, 5).
- 1. Find \(m\): \(m = \frac{5 - 1}{6 - 4} = \frac{4}{2} = 2\).
- 2. Find \(c\): Use \(m=2\) and point A(4, 1).
\(1 = 2(4) + c\) \(1 = 8 + c\) \(c = -7\). - Final Equation: \(y = 2x - 7\).
Common Mistake Alert! Always use a fully simplified value for \(m\) in the equation, otherwise, your value for \(c\) will be incorrect.
Section 5: Parallel and Perpendicular Lines
The relationship between gradients helps us describe lines that never meet (parallel) or lines that cross at a 90° angle (perpendicular).
5.1 Parallel Lines (C4.5 / E4.5)
Parallel lines travel in the exact same direction and therefore have the same gradient.
If Line 1 has gradient \(m_1\) and Line 2 has gradient \(m_2\):
\[\text{If lines are parallel, then } m_1 = m_2\]
Example: Find the equation of a line parallel to \(y = 4x - 1\) that passes through \((1, -3)\).
- The gradient of the given line is \(m = 4\).
- Since the new line is parallel, its gradient is also \(m=4\).
- Now, use Method 1 (from Section 4) with \(m=4\) and point \((1, -3)\):
\(-3 = 4(1) + c\) \(-3 = 4 + c\) \(c = -7\). - Final Equation: \(y = 4x - 7\).
5.2 Perpendicular Lines (E4.6) ***Extended Content Only***
Perpendicular lines meet at a right angle (90°).
The gradients of perpendicular lines have a special relationship: they are negative reciprocals of each other.
If Line 1 has gradient \(m_1\), and Line 2 has gradient \(m_2\):
\[m_1 \times m_2 = -1 \quad \text{or} \quad m_2 = -\frac{1}{m_1}\]
📐 The Negative Reciprocal Trick
To find the perpendicular gradient, simply:
- Flip the fraction (find the reciprocal).
- Change the sign (make it negative if it was positive, or positive if it was negative).
If \(m_1 = \frac{2}{3}\), then \(m_2 = -\frac{3}{2}\).
If \(m_1 = -5\), then \(m_2 = \frac{1}{5}\).
Example: Find the equation of the line perpendicular to \(y = 2x + 5\) that passes through (4, 1).
- The gradient of the given line is \(m_1 = 2\).
- The perpendicular gradient is \(m_2 = -\frac{1}{2}\).
- Now, use \(m = -\frac{1}{2}\) and point \((4, 1)\):
\(1 = (-\frac{1}{2})(4) + c\) \(1 = -2 + c\) \(c = 3\). - Final Equation: \(y = -\frac{1}{2}x + 3\).
Key Takeaway:
Parallel lines share the gradient. Perpendicular lines use the negative reciprocal gradient.
Section 6: Tools of Coordinate Geometry (C4.3 / E4.3)
In addition to finding equations, the syllabus requires you to calculate the length and midpoint of a line segment. These often appear in questions involving geometric shapes plotted on a grid.
6.1 Calculating the Length of a Line Segment
To find the length (distance) between two points \((x_1, y_1)\) and \((x_2, y_2)\), we use a formula derived directly from Pythagoras’ theorem (\(a^2 + b^2 = c^2\)).
Length \(L\) is the hypotenuse, and the differences in \(x\) and \(y\) form the other two sides.
\[L = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]
Example: Find the length of the line segment joining (1, 2) and (5, 5).
\(L = \sqrt{(5 - 1)^2 + (5 - 2)^2}\)
\(L = \sqrt{(4)^2 + (3)^2}\)
\(L = \sqrt{16 + 9}\)
\(L = \sqrt{25}\)
\(L = 5\) units.
6.2 Finding the Midpoint of a Line Segment
The midpoint is the exact center of the line segment. To find it, you simply calculate the average of the \(x\)-coordinates and the average of the \(y\)-coordinates.
Midpoint \(M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\)
Example: Find the midpoint of the line segment joining (1, 2) and (5, 5).
\(M = \left( \frac{1 + 5}{2}, \frac{2 + 5}{2} \right)\)
\(M = \left( \frac{6}{2}, \frac{7}{2} \right)\)
\(M = (3, 3.5)\)
***Extended Note: Perpendicular Bisector (E4.6)***
A common Extended question combines the Midpoint and Perpendicular line concepts. A perpendicular bisector is a line that cuts a segment exactly in half (using the midpoint) and meets it at a right angle (using the negative reciprocal gradient).
To find the equation of a perpendicular bisector:
- Find the Midpoint \((x_M, y_M)\) of the segment.
- Find the Gradient \(m_1\) of the segment.
- Determine the Perpendicular Gradient \(m_2\) (\(-1/m_1\)).
- Use the perpendicular gradient \(m_2\) and the midpoint \((x_M, y_M)\) to find the equation \(y = m_2x + c\).
Key Takeaway: Length uses squares and roots (Pythagoras); Midpoint uses averages.