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Cambridge IGCSE0580

Equations of linear graphs

Mathematics 0580 Chapter Notes

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Equations of linear graphs
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In this chapter

  1. 1The equation y = mx + c
  2. 2Finding the equation from a graph
  3. 3Finding c from a point and a gradient
  4. 4Lines parallel to the axes: x = k and y = k
  5. 5Lines in the form ax + by = c

1. The equation y = mx + c

A straight-line graph has an equation of the form y = mx + c. The number m is the gradient. The number c is the y-intercept: where the line crosses the y-axis, at (0, c).

Read m as the number multiplying x, with its sign. In y = 3x + 7 the gradient is 3 and the y-intercept is 7. Give the equation of a line in a fully simplified form.

y = mx + c

m = gradient, c = y-intercept

Examiner insight

“Many candidates clearly did not know the meanings of ‘m’ and ‘c’ in the equation y = mx + c.” “Most candidates recognised that the gradient was the coefficient of x, but some gave 5x.”

Worked example 14 marks

Line L has the equation y = 3x + 7.

(a) Write down the gradient of line L.

(b) Point P and point W lie on line L. P has coordinates (4,p) and W has coordinates (w, −11).

(i) Work out the value of p.

(ii) Work out the value of w.

  1. 1

    (a) the gradient is the coefficient of x: 3 [1]

  2. 2

    (b)(i) put x = 4: p = 3 × 4 + 7 = 19 [1]

  3. 3

    (ii) put y = -11: -11 = 3w + 7 [1]

  4. 4

    3w = -18, so w = -6 [1]

2. Finding the equation from a graph

To find the equation of a line, find m and c separately.

For the gradient, take two points on the line. Divide the change in y by the change in x, always in that order. A line that falls as you move right has a negative gradient.

For c, read the y-intercept: where the line crosses the y-axis. If (0,k) is on the line then c = k. No calculation is needed.

m = (y₂ - y₁)/(x₂ - x₁)

Examiner insight

“A significant number of candidates incorrectly used the co-ordinates (3, 0) instead of (0, 3) in their gradient calculation.” “This part on finding the equation of the given line was generally reasonably answered, although a significant number did not seem to appreciate that the intercept value could be read directly from the given graph.”

Worked example 13 marks

A straight line passes through the points (0, 5) and (4, −3). Find its equation in the form y = mx + c.

  1. 1

    gradient = (change in y)/(change inx) = (-3 - 5)/(4 - 0) [1]

  2. 2

    = (-8)/4 = -2 [1]

  3. 3

    (0, 5) is on the y-axis, so c = 5 and y = -2x + 5 [1]

3. Finding c from a point and a gradient

Sometimes the line is given by its gradient and one point that is not on the y-axis. Write y = mx + c with the known m. Then substitute the point's x and y and solve for c.

A line with the same gradient as a given line has the same m. Only c changes.

Give the final answer as a complete equation: it starts y = and keeps the x.

c = y - mx

Examiner insight

“Those that recognised that a parallel line had the same gradient gave an equation that started with y = 3x but finding the intercept was the most challenging.” “Some candidates omitted the y = and so did not have a complete equation as required.”

Worked example 12 marks

Find the equation of the line parallel to y = 4x + 3 that passes through (2, 9).

  1. 1

    same gradient as y = 4x + 3, so y = 4x + c [1]

  2. 2

    9 = 4 × 2 + c, so c = 1 and y = 4x + 1 [1]

4. Lines parallel to the axes: x = k and y = k

A vertical line has the same x at every point, so its equation is x = k. A horizontal line has the same y at every point, so its equation is y = k.

A horizontal line has gradient 0. A vertical line has no y = mx + c form, so do not try to write one.

x = k (vertical)

y = k (horizontal)

Examiner insight

“Most candidates thought the equation should be in the form y = mx + c, not remembering the simpler form of equation for lines parallel to the axes.”

Worked example 11 mark

Write down the equation of the line that passes through the point (0, 7) and has gradient 0.

  1. 1

    gradient 0 means a horizontal line, so every point has y = 7: y = 7 [1]

5. Lines in the form ax + by = c

At Extended level a line may be given as ax + by = c. To find its gradient and y-intercept, make y the subject.

Move the x term across, then divide every term by the number multiplying y. The gradient is the new coefficient of x.

ax + by = c ⇒ y = -a/bx + c/b

Examiner insight

“The rearrangement of the equation to find the equation of the straight line was found to be a challenge.”

Worked example 13 marks

Find the gradient and y-intercept of the graph with equation 5x + 4y = 8.

  1. 1

    make y the subject: 4y = -5x + 8 [1]

  2. 2

    divide every term by 4: y = -5/4x + 2 [1]

  3. 3

    gradient = -5/4, y-intercept = 2 [1]

Recap

  • y = mx + c: m is the gradient and c is the y-intercept.
  • Gradient = change in y ÷ change in x. A falling line has a negative gradient.
  • If the line crosses the y-axis at (0, k), then c = k.
  • Given m and a point, substitute the point and solve for c.
  • Vertical line: x = k. Horizontal line: y = k, with gradient 0.
  • Extended: for ax + by = c, make y the subject and divide every term.

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. Do not use a calculator. Write down the gradient and the y-intercept of the line y = -4x + 7.2 marks
  2. A line has gradient 3 and crosses the y-axis at (0, -5). Write down its equation.2 marks
  3. Write down the equation of (a) the vertical line through (4, -3), (b) the horizontal line through (4, -3).2 marks
  4. A line passes through the points (0, 3.5) and (6, 12.5). Find its equation in the form y = mx + c.3 marks
  5. Do not use a calculator. A straight line crosses the y-axis at (0, 9) and the x-axis at (3, 0). Find its equation in the form y = mx + c.3 marks
  6. A line has gradient 0.4 and passes through the point (5, -1). Find its equation in the form y = mx + c.3 marks
  7. Do not use a calculator. Points P (-2, 7) and Q (4, -5) lie on a straight line. (a) Find the gradient of the line. (b) Find the equation of the line in the form y = mx + c. (c) The line crosses the x-axis at R. Find the coordinates of R.6 marks
  8. Do not use a calculator. Find the gradient and the y-intercept of the line with equation 3x + 2y = 12.3 marks
  9. Do not use a calculator. Line L has equation 4x - 3y = 24. (a) Find the gradient of L. (b) A second line has the same gradient as L and passes through the point (6, -5). Find its equation in the form ax + by = c, where a, b and c are integers. (c) The point (t, 7) lies on the second line. Find the value of t.7 marks
  10. Line L has equation px + 4y = 30, where p is a constant. L passes through the point (-2, 9). (a) Find the value of p. (b) L crosses the y-axis at A and the x-axis at B. Find the coordinates of A and B. (c) Line M passes through A and through the point (5, -2.5). Find the equation of M in the form y = mx + c. (d) The point (k, -12) lies on M. Find the value of k.8 marks

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