Practice Maths

Graphing Linear Functions

← Back to Linear Equations and Their Graphs›General Maths

Key Terms

A linear function graphs as a straight line. The standard form is y = mx + c, where m is the gradient and c is the y-intercept.
The gradient (m) measures steepness: m = rise ÷ run = (y&sub2; − y&sub1;) ÷ (x&sub2; − x&sub1;). Positive m rises left-to-right; negative m falls left-to-right.
The y-intercept (c) is where the line crosses the y-axis (set x = 0). The x-intercept is where it crosses the x-axis (set y = 0).
General form
: ax + by + c = 0. To graph, rearrange to y = mx + c first.
Parallel lines
have the same gradient (same m, different c).
Perpendicular lines
: m&sub1; × m&sub2; = −1, so m&sub2; = −1/m&sub1; (negative reciprocal).
📚 Key Formulas
Gradient–intercept form: y = mx + c (m = gradient, c = y-intercept)
Gradient from two points: m = (y&sub2; − y&sub1;) / (x&sub2; − x&sub1;)
Equation from point and gradient: y − y&sub1; = m(x − x&sub1;)
Perpendicular gradient: m⊥ = −1/m   (negative reciprocal)
Parallel condition: same m   Perpendicular condition: m&sub1; × m&sub2; = −1
x y −1 −2 1 2 3 1 2 −1 −2 y-int (0, −1) run = 1 rise = 2 y = 2x − 1 m = 2

Worked Example 1 — Rearrange to gradient–intercept form

Find the gradient and y-intercept of: 3x − 2y + 6 = 0

Step 1 — Rearrange to y = mx + c: isolate y on the left.

   3x − 2y + 6 = 0

   −2y = −3x − 6   (subtract 3x and 6 from both sides)

   y = (3/2)x + 3   (divide both sides by −2)

Gradient m = 3/2 = 1.5   y-intercept c = 3

Interpretation: the line rises 1.5 units for every 1 unit across, and cuts the y-axis at (0, 3).

Worked Example 2 — Equation from two points

Find the equation of the line through (2, 5) and (6, 13).

Step 1 — Find gradient:

   m = (13 − 5) / (6 − 2) = 8/4 = 2

Step 2 — Use point–gradient form with (2, 5):

   y − 5 = 2(x − 2)

   y − 5 = 2x − 4

   y = 2x + 1

Check with (6, 13): y = 2(6) + 1 = 13 ✓

Hot Tip — Perpendicular gradient: “flip and negate.” If m = 2/3, then m⊥ = −3/2. If m = 4, then m⊥ = −1/4. Always multiply the two gradients to check: (2/3) × (−3/2) = −1 ✓. For horizontal lines (m = 0), the perpendicular is vertical (undefined gradient) — no formula needed.

Introduction

Linear functions are the foundation of all mathematical modelling. Their graphs — straight lines — are defined entirely by two numbers: gradient and y-intercept. Once you can read and write these two values, you can describe, compare, and apply any straight-line relationship.

The Gradient–Intercept Form: y = mx + c

The gradient m tells you how steep the line is and which direction it runs. A gradient of 3 means “go 1 right, go 3 up.” A gradient of −2 means “go 1 right, go 2 down.” The y-intercept c is the starting value when x = 0 — where the line hits the vertical axis.

Reading m and c directly

For y = −3x + 7: gradient m = −3, y-intercept c = 7.

For 2x + y − 5 = 0: rearrange → y = −2x + 5, so m = −2, c = 5.

Finding Gradient from Two Points

Use the gradient formula: m = (y&sub2; − y&sub1;) / (x&sub2; − x&sub1;). It does not matter which point you label (x&sub1;, y&sub1;) and which you label (x&sub2;, y&sub2;) — you will get the same answer either way.

Example: Gradient between two points

Points: (−3, 1) and (5, −7).

m = (−7 − 1) / (5 − (−3)) = −8/8 = −1

The line falls at 45° from left to right.

Writing the Equation of a Line

Three common scenarios:

1. Given m and c directly: write y = mx + c immediately.

2. Given m and one point (x&sub1;, y&sub1;): use y − y&sub1; = m(x − x&sub1;), then rearrange to y = mx + c.

3. Given two points: find m first (step 1), then apply method 2.

Example: Line through (3, −2) with m = 4

y − (−2) = 4(x − 3)

y + 2 = 4x − 12

y = 4x − 14

Parallel and Perpendicular Lines

Parallel lines never intersect because they have identical gradients. If you know one line is y = 3x + 5, any line parallel to it has the form y = 3x + c (different c value).

Perpendicular lines cross at exactly 90°. Their gradients satisfy m&sub1; × m&sub2; = −1. To find a perpendicular gradient, take the negative reciprocal: “flip and negate.”

Example: Perpendicular line through a point

Find the line perpendicular to y = ½x + 3, passing through (4, 1).

Original gradient: m = ½. Perpendicular gradient: m⊥ = −2.

Using point (4, 1): y − 1 = −2(x − 4) → y = −2x + 9

💡 Sketching a line from y = mx + c: plot the y-intercept first (0, c). Then use the gradient to find a second point — from the y-intercept, move “1 right, m up” (or if m is negative, m down). Draw a straight line through the two points and extend with arrows.

Summary

y = mx + c: m is gradient (rise/run), c is y-intercept. Find equation using: m and c directly; point–gradient form; or two-point method. Parallel lines share gradient; perpendicular gradients multiply to −1. Rearrange general form ax + by + c = 0 by isolating y.

Mastery Practice

See Answers ➔
  1. Fluency

    Identify the gradient m and y-intercept c for each line.

    1. (a) y = 3x − 4
    2. (b) y = −2x + 7
    3. (c) 2x + y = 5
    4. (d) 3x − 4y + 8 = 0
  2. Fluency

    Calculate the gradient of the line passing through each pair of points.

    1. (a) (1, 3) and (4, 9)
    2. (b) (−2, 5) and (3, −5)
    3. (c) (0, −3) and (6, 0)
  3. Fluency

    Write the equation of the line described.

    1. (a) Gradient m = 4 and y-intercept c = −2.
    2. (b) Gradient m = −½, passing through (4, 3).
    3. (c) Passing through (2, −1) and (5, 8).
  4. Understanding

    For the line y = −3x + 6:

    1. (a) Find the x-intercept and y-intercept.
    2. (b) Describe the key features needed to sketch this line.
    3. (c) State the gradient and explain what it means as a rate of change.
  5. Understanding

    Classify each pair of lines as parallel, perpendicular, or neither. Justify your answer.

    1. (a) y = 2x + 3   and   y = 2x − 5
    2. (b) y = 4x − 1   and   y = −¼x + 2
    3. (c) y = 3x + 1   and   y = −3x + 1
  6. Understanding

    Find the equation of each line.

    1. (a) Parallel to y = 3x − 2, passing through (1, 4).
    2. (b) Perpendicular to y = 2x + 1, passing through (6, 3).
  7. Understanding

    A line passes through (−3, 7) and (1, −1). Find:

    1. (a) The gradient of the line.
    2. (b) The y-intercept.
    3. (c) The x-intercept.
    4. (d) The equation in general form ax + by + c = 0.
  8. Understanding

    The line y = −2x + 8 represents the remaining phone credit ($) after making a number of calls (x).

    1. (a) What does the y-intercept represent in this context?
    2. (b) What does the gradient represent in this context?
    3. (c) After how many calls does the credit run out?
  9. Problem Solving

    The midpoint of line segment AB is M(3, 1). Point A is at (−1, 4). Find:

    1. (a) The coordinates of point B.
    2. (b) The gradient of AB.
    3. (c) The equation of the perpendicular bisector of AB (the line through M perpendicular to AB).
  10. Problem Solving

    The line through (k, 3) and (2k, −1) has a gradient of −2.

    1. (a) Use the gradient formula to write an equation in k and solve for k.
    2. (b) Write the coordinates of both points, then find the equation of the line in the form y = mx + c.