Linear Graphs

Mathematics (Core) revision notes · Cambridge (CIE) · 0580 · Core

1. Coordinates on the grid

Reading and plotting a point

  • The grid used throughout this course has a horizontal x-axis and a vertical y-axis, meeting at the origin
  • The origin is the point (0, 0), and every other point is described by how far it lies from there in each direction
  • A pair of coordinates is written in brackets with the horizontal value first: (x, y)
  • Positive x moves right and negative x moves left; positive y moves up and negative y moves down
    • (4, 3) sits four to the right and three up
    • (−2, −5) sits two to the left and five down
  • Plot a point with a small cross or a solid dot, and write its letter beside it when the question names it, so it is clear which point is which
A squared grid with both axes running from −6 to 6 and a single labelled point A marked with a dot at three to the left and four up, the position written as (−3, 4)
Source: 2D coordinates by Save My Exams
  • Check the scale on the axes before counting: one square is frequently worth more than one unit, and assuming otherwise puts every answer out by a factor

2. Gradient

What the number tells you

  • The gradient measures how steep a line is and which way it slopes
  • A gradient of 3 means the line rises 3 for every 1 step to the right, and a gradient of −4 means it falls 4 for every 1 step to the right
  • An uphill line, running from bottom left to top right, has a positive gradient; a downhill line has a negative one
  • Comparing steepness ignores the sign, so a gradient of −5 is steeper than one of 2
  • A horizontal line has gradient 0, and a vertical line has no gradient at all because there is no horizontal step to divide by

Measuring it

  • On a printed grid, pick two points that sit on the line and complete a right-angled triangle between them
A line drawn on a grid through (0, 2) and (1, 5), with a dashed right-angled triangle between those two points marking a rise of 5 − 2 = 3 against a run of 1 − 0 = 1, so the gradient is 3
Source: Gradient of a line by Save My Exams
  • The gradient is the vertical change divided by the horizontal change
  • Choose two points that sit exactly on grid intersections and are far apart, because a short triangle magnifies any reading error
  • Read the scales before counting squares: the two axes often use different scales, and one square is frequently not one unit
  • Put the minus sign back in yourself for a downhill line, since the counting gives only the size
A downhill line on a grid through (0, 1) and (1, −1), with a dashed triangle showing 2 down against 1 across, giving a gradient of −2
Source: Gradient of a line by Save My Exams

3. The straight-line equation

Reading a line from its equation

  • Any straight line that is not vertical can be written y = mx + c: the number m sets its steepness and the number c fixes where it meets the y-axis
  • The line therefore passes through (0, c)
    • y = 4x − 9 has gradient 4 and meets the y-axis at −9
    • y = 11 − 2x has gradient −2 and meets the y-axis at 11, so take the number attached to x, not the one written first
  • A horizontal line has the form y = c, since every point on it has the same y value
  • A vertical line has the form x = k, and it cannot be written as y = mx + c at all
  • Answers must be given fully simplified, so tidy fractions and collect any like terms before writing the final line

Worked example

Reading a line from its equation

Write down the gradient and the y-axis crossing of y = 3x − 4.

Solution:

  • The equation is already in the form y = mx + c
  • The number in front of x is the gradient, so the gradient is 3
  • The constant is where the line crosses the y-axis, at (0, −4)

Worked example

Where a line meets a horizontal line

The lines y = 25 − 4x and y = −3 cross at the point P. Find the coordinates of P.

Solution:

  • Every point on y = −3 has a y-coordinate of −3, so P does too
  • Substitute y = −3 into the other equation: −3 = 25 − 4x
  • Subtract 25 from both sides: −28 = −4x
  • Divide both sides by −4: x = 7
  • P is (7, −3)
  • Check in y = 25 − 4x: 25 − 4 × 7 = 25 − 28 = −3
  • Give both coordinates when the point is asked for: the −3 comes straight from the horizontal line

4. Drawing a straight line

From an equation

  • Build a small table of values by substituting three x values that are spread across the axes you are given
  • Three points, not two: a slip in one of three points shows up because it misses the line through the other two, whereas two points always look convincing
  • Plot the points and join them with a ruled line that runs the full width of the grid
Two straight lines drawn on the same grid: a steep uphill line through (0, −1) and (1, 2), stepping 1 across and 3 up, and a shallow downhill line through (0, 3) and (5, 0)
Source: Drawing linear graphs by Save My Exams
  • Alternatively start at (0, c) and step out the gradient, going one right and m up each time, down instead if m is negative
  • For a gradient written as a fraction, step across by the bottom number and up by the top number
A shallow uphill line on a grid through (0, −1) and (3, 1), with a dashed triangle showing 3 across against 2 up, giving a gradient of two thirds
Source: Gradient of a line by Save My Exams

Worked example

Drawing a line from its equation

Draw y = 2x + 3 for x from −1 to 2.

Solution:

  • Build a short table of values
  • x = −1 gives y = 1; x = 0 gives 3; x = 1 gives 5; x = 2 gives 7
  • Plot (−1, 1), (0, 3), (1, 5) and (2, 7), and join them with a ruled line
A grid from x = −1 to 3 and y = −2 to 8 with y = 2x + 3 ruled through the crosses at (−1, 1), (0, 3), (1, 5) and (2, 7)

5. Finding the equation of a line

From a graph

A downhill line crossing the y-axis at 5 and the x-axis at 10, with a red right-angled triangle marking 4 down against 8 across between (2, 4) and (10, 0), so the gradient is −0.5
Source: Finding equations of straight lines by Save My Exams
  • Work out the gradient from a triangle, then read where the line crosses the y-axis
  • Put both numbers into y = mx + c and write the equation out
  • If the crossing point is off the edge of the grid, use a point the line does pass through instead

Exam tip

The equation is marked in pieces

Finding the equation of a line from a grid is marked in pieces. A right-angled triangle drawn on the grid, with the correct rise and run written on it, earns a method mark on its own. In the answer, the right gradient scores even with a wrong c, and the right c scores even with a wrong gradient, so always finish with a whole equation, y = mx + c.

Worked example

The equation of a line from a grid

Line L is drawn on the grid below. Find its equation in the form y = mx + c.

A grid from x = −1 to 5 and y = −5 to 5 with a straight line labelled L ruled across it, sloping down from the top left to the bottom right; no points on it are marked

Solution:

  • Use two points where L crosses the grid exactly, as far apart as the grid allows: (0, 3) and (4, −3)
  • Draw a right-angled triangle between them and read its sides off the grid
  • The run is 4 to the right and the line drops 6, so the rise is −6 because L slopes downhill
  • Gradient = rise ÷ run = −6 ÷ 4 = −1.5
  • L crosses the y-axis at 3, so c = 3
  • The equation is y = −1.5x + 3
  • Check with the other point: −1.5 × 4 + 3 = −6 + 3 = −3

From a gradient and a point

  • Write y = mx + c with the gradient filled in and c left as a letter
  • Substitute the coordinates of the known point and solve the resulting equation for c
  • Rewrite the equation with that value in place

Worked example

A gradient and one point

Find the equation of the line with gradient 5 that passes through (3, 7). Give your answer in the form y = mx + c.

Solution:

  • Start from y = 5x + c
  • Substitute x = 3 and y = 7: 7 = 5 × 3 + c
  • So 7 = 15 + c
  • Subtract 15: c = −8
  • The equation is y = 5x − 8

6. Parallel lines

Same gradient, different intercept

  • Two different lines are parallel exactly when their gradients are equal (or both are vertical), and they then never meet
  • y = 7x + 2 and y = 7x − 5 are parallel; y = 7x + 2 and y = 4x + 2 are not
  • Check by rearranging both equations into y = mx + c first, since a shared gradient can be hidden by the form
  • To find the equation of a line parallel to a given one, copy the gradient across and then find the new intercept from the point you are given

Worked example

Parallel through a given point

A line is parallel to y = 7x + 2 and passes through (3, 8). Find its equation.

Solution:

  • Parallel means the same gradient, so the new line is y = 7x + c
  • Substitute x = 3 and y = 8: 8 = 21 + c
  • Subtract 21: c = −13
  • The line is y = 7x − 13