πŸ“ˆ Linear Graphs

Plot straight-line graphs, find gradients, and write equations in the form y = mx + c.

Reading y = mx + c
y = 3x + 2 β†’ gradient = 3, y-intercept = 2
Gradient from two points
gradient = (yβ‚‚ βˆ’ y₁) Γ· (xβ‚‚ βˆ’ x₁) = rise Γ· run
Parallel lines
y = 2x + 5 and y = 2x βˆ’ 3 are parallel β€” same gradient!

What you'll learn:

  • What m (gradient) and c (y-intercept) mean in y = mx + c
  • How to plot a straight-line graph using a table of values
  • How to calculate the gradient from two points (rise Γ· run)
  • How to find the equation of a line from a graph or two points
  • Horizontal lines (y = k) and vertical lines (x = k)
  • How to spot parallel lines (same gradient)

πŸ“– Linear Graphs

The equation y = mx + c

Every straight-line graph can be written in the form y = mx + c.

m = the gradient β€” how steep the line is (rise Γ· run)
c = the y-intercept β€” where the line crosses the y-axis
Example: y = 3x + 2 β†’ gradient = 3, y-intercept = 2
Example: y = βˆ’x + 5 β†’ gradient = βˆ’1, y-intercept = 5
Example: y = 4x β†’ gradient = 4, y-intercept = 0 (passes through origin)

What the gradient tells you

Positive gradient β†’ line slopes uphill (left to right)
Negative gradient β†’ line slopes downhill (left to right)
Larger gradient β†’ line is steeper
Gradient = 0 β†’ line is horizontal (flat)
Think of gradient as: "for every 1 step to the right, how many steps up or down?"

Plotting a line from its equation (Table of Values method)

To draw y = 2x + 1:

Step 1: Choose x values (e.g. x = βˆ’2, βˆ’1, 0, 1, 2)
Step 2: Substitute each x into y = 2x + 1
x = βˆ’2 β†’ y = 2(βˆ’2) + 1 = βˆ’3  |  x = 0 β†’ y = 1  |  x = 2 β†’ y = 5
Step 3: Plot the (x, y) coordinates and draw a straight line through them
Step 4: Label the line with its equation

Calculating gradient from two points

Use the formula: gradient = (yβ‚‚ βˆ’ y₁) Γ· (xβ‚‚ βˆ’ x₁)

Through (1, 3) and (4, 9):
gradient = (9 βˆ’ 3) Γ· (4 βˆ’ 1) = 6 Γ· 3 = 2
Through (0, 5) and (3, 2):
gradient = (2 βˆ’ 5) Γ· (3 βˆ’ 0) = βˆ’3 Γ· 3 = βˆ’1
Tip: It doesn't matter which point you call (x₁, y₁) β€” you get the same answer!

Finding the equation from two points

Step 1: Calculate the gradient (m) using the formula above
Step 2: Substitute m and one point into y = mx + c to find c
Through (1, 5) and (3, 11):
m = (11 βˆ’ 5) Γ· (3 βˆ’ 1) = 6 Γ· 2 = 3
Substitute (1, 5): 5 = 3(1) + c β†’ c = 2
Equation: y = 3x + 2

Special lines

Horizontal lines: y = k (e.g. y = 3 is a horizontal line at height 3)
Vertical lines: x = k (e.g. x = βˆ’2 is a vertical line at x = βˆ’2)
Horizontal lines have gradient = 0. Vertical lines have undefined gradient.

Parallel lines

Parallel lines have the same gradient (same value of m) but different y-intercepts (different c).
y = 4x + 1 and y = 4x βˆ’ 7 are parallel β€” both have gradient 4
y = 2x + 3 and y = 3x + 3 are NOT parallel β€” different gradients

πŸ’‘ Worked Examples

Example 1: Read gradient and y-intercept

What is the gradient and y-intercept of y = βˆ’3x + 7?

The equation is in the form y = mx + c
m = βˆ’3 (gradient β€” line slopes downhill)
c = 7 (y-intercept β€” line crosses y-axis at 7)
Example 2: Complete a table of values

Complete the table for y = 3x βˆ’ 2:

xβˆ’2βˆ’1012
yβˆ’8βˆ’5βˆ’214
x = βˆ’2: y = 3(βˆ’2) βˆ’ 2 = βˆ’6 βˆ’ 2 = βˆ’8
x = 0: y = 3(0) βˆ’ 2 = 0 βˆ’ 2 = βˆ’2
x = 2: y = 3(2) βˆ’ 2 = 6 βˆ’ 2 = 4
Example 3: Find the equation through two points

Find the equation of the line through (2, 3) and (5, 9).

Step 1 β€” Gradient: m = (9 βˆ’ 3) Γ· (5 βˆ’ 2) = 6 Γ· 3 = 2
Step 2 β€” Find c: substitute (2, 3) into y = 2x + c
3 = 2(2) + c β†’ 3 = 4 + c β†’ c = βˆ’1
Equation: y = 2x βˆ’ 1
Check with (5, 9): y = 2(5) βˆ’ 1 = 10 βˆ’ 1 = 9 βœ“
Example 4: Are these lines parallel?

y = 4x + 3    y = 4x βˆ’ 5    y = 2x + 3

y = 4x + 3: gradient = 4
y = 4x βˆ’ 5: gradient = 4
y = 2x + 3: gradient = 2
y = 4x + 3 and y = 4x βˆ’ 5 are parallel (both gradient = 4)
y = 2x + 3 is NOT parallel to the others (different gradient)
Example 5: Does a point lie on a line?

Does (3, 11) lie on y = 4x βˆ’ 1?

Substitute x = 3: y = 4(3) βˆ’ 1 = 12 βˆ’ 1 = 11
11 = 11 βœ“ β†’ Yes, (3, 11) lies on the line y = 4x βˆ’ 1

πŸ“Š Graph Visualiser

Two modes β€” build a line with sliders, or click two points to find the equation.

y = 2x + 1
Gradient = 2 β†’ go 1 right, go 2 up

Green = run (1 right)  |  Blue = rise (m up/down)

Click any two points on the grid to find the equation of the line through them.

🎯 Exercise 1: Fill in the Table

Substitute each x value into the equation to find y.

🎯 Exercise 2: Identify m and c

For each equation, write the gradient (m) and y-intercept (c).

🎯 Exercise 3: Match Equations to Descriptions

Click an equation, then click its matching description. All 4 pairs to find!

🎯 Exercise 4: Gradient from Two Points

Find the gradient, then the y-intercept, then write the full equation.

🎯 Exercise 5: Find the Parallel Pair

Click the TWO equations that are parallel to each other.

🎯 Exercise 6: Plot the Line

Click TWO points on the grid that lie on the given equation.

Click integer grid points. Both must satisfy the equation.

✏️ Practice Questions

Type your answers then click Check. Accept decimals or fractions where needed.

  1. What is the gradient of y = 5x βˆ’ 3?
  2. What is the y-intercept of y = 2x + 7?
  3. What is the gradient of y = βˆ’4x + 1?
  4. What is the y-intercept of y = 3x?
  5. Write the equation of a line with gradient 2 and y-intercept 5.
  6. For y = 3x + 1, what is y when x = 2?
  7. For y = βˆ’2x + 4, what is y when x = 3?
  8. What is the gradient of a horizontal line?
  9. What is the equation of the horizontal line through (0, 6)?
  10. Two points are (0, 3) and (1, 7). What is the gradient?
  11. Line through (0, 3) and (1, 7): what is the y-intercept?
  12. Line through (0, 3) and (1, 7): write the equation.
  13. Are y = 3x + 1 and y = 3x βˆ’ 5 parallel? (yes/no)
  14. Are y = 2x + 4 and y = 4x + 2 parallel? (yes/no)
  15. Does (2, 5) lie on y = 3x βˆ’ 1? (yes/no)
  16. What is the gradient of y = βˆ’x + 9?
  17. For y = 2x βˆ’ 5, what is y when x = 0?
  18. The gradient between (1, 4) and (3, 10) is:
  19. Rearrange: y = 8 βˆ’ 2x. What is the gradient?
  20. Write the equation of a line parallel to y = 5x + 1 that crosses the y-axis at βˆ’3.
Answers: 1) 5   2) 7   3) βˆ’4   4) 0   5) y=2x+5   6) 7   7) βˆ’2   8) 0   9) y=6   10) 4   11) 3   12) y=4x+3   13) yes   14) no   15) yes   16) βˆ’1   17) βˆ’5   18) 3   19) βˆ’2   20) y=5xβˆ’3

πŸ† Challenge Questions

Harder multi-step problems. Show your working on paper, then type answers.

  1. The line y = 3x + k passes through (2, 11). Find k.
  2. Find the equation of the line through (1, 2) and (4, 11).
  3. Find the equation of the line through (βˆ’1, 6) and (2, 0).
  4. A line has gradient βˆ’3 and passes through (2, 1). What is the y-intercept?
  5. A line crosses the x-axis at (4, 0) and y-axis at (0, 8). What is the gradient?
  6. A line crosses the x-axis at (4, 0) and y-axis at (0, 8). Write its equation.
  7. Do the lines y = 3x + 2 and y = 3x βˆ’ 4 ever meet? (yes/no)
  8. At what x-value do y = 2x + 1 and y = x + 4 intersect?
  9. A phone bill charges Β£10 per month plus 5p per minute. If C = total cost in pence and t = minutes, write the equation for C.
  10. The points (1, a), (3, 10), (5, b) all lie on y = 2x + 4. Find a and b. Enter a+b.
Answers:
1) k=5 (11=3Γ—2+k β†’ k=5)   2) y=3xβˆ’1 (m=3, 2=3(1)+c β†’ c=βˆ’1)   3) y=βˆ’2x+4 (m=(0βˆ’6)Γ·(2βˆ’(βˆ’1))=βˆ’6/3=βˆ’2; 6=βˆ’2(βˆ’1)+c β†’ c=4)   4) c=7 (1=βˆ’3(2)+c β†’ c=7)   5) m=βˆ’2 ((8βˆ’0)Γ·(0βˆ’4))   6) y=βˆ’2x+8   7) No β€” parallel lines never meet   8) x=3 (2x+1=x+4 β†’ x=3)   9) C=5t+1000 (Β£10=1000p fixed cost)   10) a=6 (y=2+4=6), b=14 (y=10+4=14), a+b=20