Year 9 Mathematics · Number and Algebra · 18m read

Linear Graphs — Equations and Forms

Ever wondered why your phone bill stays the same price per gigabyte, or why a car's distance from home increases at a constant rate? That's a linear relationship, and you can draw it as a straight line on a graph. By mastering the equation of a line, you'll understand what the gradient (slope) and intercept tell you about real situations — like how fast something is changing or where it starts. This skill unlocks everything from VCE maths to predicting costs, trends, and motion in the real world.

Learning objective: Graph linear relationships and interpret gradient and intercept forms.

By the end of this lesson you can:

I find gradient from a graph or two points.
I use y = mx + c form.
I convert between forms.

Gradient is the steepness of a ski run. Bunny slopes have small gradients; black runs have big ones. Downhill runs (left to right) have negative gradients.

Line essentials

GRADIENT (slope) — the measure of steepness.

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

| Gradient | Meaning |
|----------|---------|
| $m > 0$ | Line rises left to right |
| $m < 0$ | Line falls left to right |
| $m = 0$ | Horizontal line |
| Undefined | Vertical line (run = 0) |

### Gradient-Intercept Form (most useful)
$$y = mx + c$$
• $m$ = gradient
• $c$ = y-intercept (where line crosses y-axis, i.e. when $x = 0$)

To SKETCH: plot the y-intercept, then use rise/run to find a second point.

### Point-Gradient Form
$$y - y_1 = m(x - x_1)$$
Use when you know one point $(x_1, y_1)$ and the gradient $m$.

### General Form
$$ax + by + c = 0$$
Rearrange to find $m$ and intercepts.

Example: $2x - 3y + 6 = 0$
$3y = 2x + 6$, so $y = \frac{2}{3}x + 2$. Thus $m = \frac{2}{3}$, $c = 2$.

### X-intercept
Set $y = 0$ in the equation and solve for $x$.
For $y = 2x - 4$: $0 = 2x - 4 \Rightarrow x = 2$. X-intercept: $(2, 0)$.

### Parallel and Perpendicular Lines

| Relationship | Gradient condition |
|--------------|--------------------|
| Parallel | $m_1 = m_2$ (same gradient) |
| Perpendicular | $m_1 \times m_2 = -1$ (negative reciprocal) |

Example: Line with $m = 3$.
• Parallel line has $m = 3$.
• Perpendicular line has $m = -\frac{1}{3}$.

### Finding an Equation from Two Points
1. Calculate gradient $m = \frac{y_2 - y_1}{x_2 - x_1}$.
2. Substitute one point into $y - y_1 = m(x - x_1)$.
3. Rearrange to $y = mx + c$.

### Australian context
A taxi in Melbourne charges \$3.40 flag fall plus \$2.20/km. The cost equation $C = 2.20d + 3.40$ is a straight line with $m = 2.20$ and $c = 3.40$.

Linear Equation Forms

Gradient from two points: m = \dfrac{y_2 - y_1}{x_2 - x_1}
Gradient-intercept: y = mx + c
Point-gradient: y - y_1 = m(x - x_1)
Perpendicular gradients: m_1 \times m_2 = -1

Line reading

Gradient sign: Positive: line goes up left-to-right. Negative: goes down.
Steeper = bigger |m|: Absolute value of $m$ shows steepness.
Equal gradient: Lines are parallel — they never meet.

Worked example (Medium)

Find the equation of the line through $(1, 2)$ and $(4, 11)$.

Gradient. $m = \dfrac{11 - 2}{4 - 1} = \dfrac{9}{3} = 3$
Point-gradient. $y - 2 = 3(x - 1)$
Expand. $y - 2 = 3x - 3$
Rearrange. $y = 3x - 1$

Answer: $y = 3x - 1$

Worked example (Hard)

Line $L_1$ has equation $3x + 2y = 12$. Find the equation of line $L_2$ that is perpendicular to $L_1$ and passes through $(6, -1)$.

Gradient of L1. Rearrange: $2y = -3x + 12$, so $y = -\frac{3}{2}x + 6$. Thus $m_1 = -\frac{3}{2}$.
Perpendicular gradient. $m_2 = -\frac{1}{m_1} = -\frac{1}{-3/2} = \frac{2}{3}$
Point-gradient. $y - (-1) = \frac{2}{3}(x - 6)$
Simplify. $y + 1 = \frac{2}{3}x - 4$, so $y = \frac{2}{3}x - 5$

Answer: $y = \dfrac{2}{3}x - 5$

Common mistakes

Computing gradient as $\frac{x_2 - x_1}{y_2 - y_1}$ (rise over run flipped) Gradient is change in $y$ OVER change in $x$, i.e. vertical over horizontal. Fix: Remember: gradient $= \frac{\text{rise}}{\text{run}} = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}$. Rise (vertical) is always the numerator.
Saying two lines with $m_1 = 2$ and $m_2 = -2$ are perpendicular $m_1 \times m_2 = 2 \times (-2) = -4 \neq -1$. Perpendicular requires the PRODUCT to equal $-1$. Fix: Perpendicular to $m = 2$ has $m = -\frac{1}{2}$. Check: $2 \times (-\frac{1}{2}) = -1$ ✓.
Confusing x-intercept and y-intercept when reading $y = mx + c$ $c$ is the y-intercept (value of $y$ when $x = 0$). The x-intercept requires setting $y = 0$. Fix: In $y = 3x - 6$: $c = -6$ is the y-intercept. X-intercept: $0 = 3x - 6 \Rightarrow x = 2$.

Practice questions

In $y = -2x + 5$, the y-intercept is:
1. $-2$
2. $2$
3. $5$
4. $-5$
(1 marks)
Find a line perpendicular to $y = 2x + 3$ that passes through $(0, 1)$.
(2 marks)
Find the gradient of the line through $(-2, 5)$ and $(3, -5)$.
(2 marks)
Find the x-intercept and y-intercept of $4x - 3y = 12$.
(3 marks)
A plumber charges a $\$60$ call-out fee plus $\$90$ per hour. Write the cost equation, identify the gradient and y-intercept, and find the cost for 2.5 hours.
(3 marks)

Going Deeper — Midpoint and Distance

Two other fundamental formulas for working with line segments:

MIDPOINT formula — the point halfway between $(x_1, y_1)$ and $(x_2, y_2)$:
$$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$

DISTANCE formula — the length of the segment from $(x_1, y_1)$ to $(x_2, y_2)$:
$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

Note: this is just Pythagoras' theorem in coordinate form! The horizontal distance is $|x_2 - x_1|$ and vertical distance is $|y_2 - y_1|$.

Example: Midpoint and distance from $(1, 3)$ to $(7, 11)$:
$M = \left(\frac{1+7}{2}, \frac{3+11}{2}\right) = (4, 7)$
$d = \sqrt{(7-1)^2 + (11-3)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$

Summary

Gradient $m = \dfrac{y_2 - y_1}{x_2 - x_1}$; positive = rising, negative = falling.
Gradient-intercept form: $y = mx + c$ where $c$ is the y-intercept.
Point-gradient form: $y - y_1 = m(x - x_1)$.
Parallel lines share the same gradient.
Perpendicular lines: $m_1 \times m_2 = -1$.