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Equations of Straight Lines

Finding the equation of a line from two points, midpoints, and parallel and perpendicular lines.

  • 9 key terms
  • All boards

Learning Objectives

  1. 1Find the equation of a straight line from its gradient and a point, or from two points.
  2. 2Find the midpoint of a line segment, and its length (Higher tier).
  3. 3Write the equation of a line parallel to a given line.
  4. 4Use the fact that perpendicular lines have gradients with product \(-1\) (Higher tier).

From a picture to an equation

The last lesson went from an equation to a graph. This one goes the other way: given a gradient and a point, or two points, write down the equation of the line. The method is always the same three steps, find the gradient, put a point in to find \(c\), then write \(y = mx + c\), so it is worth learning as a routine. Exam questions often add a second part, such as the midpoint, a parallel line or a perpendicular line, so each of those gets a short method here too.

The equation of a line through two points

Follow the same three steps every time and you will not need to remember a formula.

  • Step 1: gradient

    Work out \(m = \dfrac{y_2 - y_1}{x_2 - x_1}\).

  • Step 2: find c

    Substitute \(m\) and one point into \(y = mx + c\), and solve for \(c\).

  • Step 3: write it

    Put \(m\) and \(c\) back into \(y = mx + c\).

  • Check

    Put the other point in. If it does not fit, there is a slip.

Through two points

Find the equation of the line through \((2, 1)\) and \((6, 9)\).

Show the solutionHide the solution
  1. 1 Gradient \(\dfrac{9 - 1}{6 - 2} = \dfrac{8}{4} = 2\).
  2. 2 Substitute a point \(y = 2x + c\) with \((2, 1)\) gives \(1 = 4 + c\), so \(c = -3\).
  3. 3 Write the equation \(y = 2x - 3\).
  4. 4 Check with the other point \(2 \times 6 - 3 = 9\). Correct.

Answer\(y = 2x - 3\)

Midpoints

The midpoint of a line segment is halfway along it, so its coordinates are the averages of the end points.

  • Formula

    The midpoint of \((x_1, y_1)\) and \((x_2, y_2)\) is \(\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)\).

  • Example

    The midpoint of \((2, 1)\) and \((6, 9)\) is \(\left(\dfrac{2 + 6}{2}, \dfrac{1 + 9}{2}\right) = (4, 5)\).

  • Finding an end point

    If the midpoint and one end are known, double the midpoint and subtract the end you know.

  • Length

    The length of the segment uses Pythagoras: \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) (Higher tier).

Parallel lines

Parallel lines have the same gradient, so you can copy it.

  • Same m, different c

    A line parallel to \(y = 3x + 1\) has the form \(y = 3x + c\).

  • Through a point

    Put the point in to find \(c\). Through \((2, 9)\): \(9 = 6 + c\), so \(c = 3\), and the line is \(y = 3x + 3\).

  • Rearranging first

    \(2y - 4x = 6\) becomes \(y = 2x + 3\), so its gradient is 2. Lines with different-looking equations can still be parallel.

  • Same line?

    If \(m\) and \(c\) are both the same, it is the same line, not a parallel one.

A perpendicular line

(Higher tier) Find the equation of the line perpendicular to \(y = 2x + 3\) that passes through \((4, 1)\).

Show the solutionHide the solution
  1. 1 Perpendicular gradient The gradient of \(y = 2x + 3\) is 2, so the new gradient is \(-\dfrac{1}{2}\).
  2. 2 Find c \(1 = -\dfrac{1}{2} \times 4 + c\), so \(1 = -2 + c\) and \(c = 3\).
  3. 3 Write the equation \(y = -\dfrac{1}{2}x + 3\).
  4. 4 Check At \(x = 4\), \(y = -2 + 3 = 1\). Correct.

Answer\(y = -\dfrac{1}{2}x + 3\)

Awkward forms

Lines are not always given as \(y = mx + c\), so you may need to rearrange.

  • Form ax + by = c

    \(3x + 2y = 12\) rearranges to \(2y = -3x + 12\) and then \(y = -\dfrac{3}{2}x + 6\), so the gradient is \(-\dfrac{3}{2}\).

  • Where it crosses the axes

    Put \(x = 0\) to find the \(y\)-intercept and \(y = 0\) to find the \(x\)-intercept. For \(3x + 2y = 12\) these are \((0, 6)\) and \((4, 0)\).

  • Equal gradients

    To check two lines are parallel, rearrange both and compare \(m\).

  • Fractions

    Keep gradients as fractions, not decimals, unless the question asks.

Test yourself

  1. 1

    What are the three steps for the equation of a line through two points?

    Show answerHide answer

    Find the gradient, find \(c\) with one point, then write \(y = mx + c\).

  2. 2

    What is the midpoint of \((0, 4)\) and \((6, 10)\)?

    Show answerHide answer

    \((3, 7)\).

  3. 3

    What gradient does a line parallel to \(y = 5x - 2\) have?

    Show answerHide answer

    5.

  4. 4

    What is the gradient of a line perpendicular to \(y = 4x\) (Higher tier)?

    Show answerHide answer

    \(-\dfrac{1}{4}\).

  5. 5

    What is the gradient of \(2y = 6x + 8\)?

    Show answerHide answer

    3.

Exam technique: equations of lines

Show every step, because the answer has several parts that can each earn a mark.

  • Write the gradient calculation

    Show the change in \(y\) over the change in \(x\).

  • Substitute a point

    Write "\(1 = 2 \times 2 + c\)", not just \(c = -3\).

  • Finish with a full equation

    The answer is \(y = 2x - 3\), not just "\(m = 2\)".

  • Check with the second point

    It takes ten seconds and catches most slips.

Summary and exam focus

  • For the line through two points, find \(m\), then find \(c\), then write \(y = mx + c\).
  • The midpoint is the average of the \(x\)-coordinates and the average of the \(y\)-coordinates.
  • Parallel lines have the same gradient.
  • Perpendicular lines have gradients that multiply to \(-1\) (Higher tier).

Exam focus

Find an equation of the line that passes through \((0, 5)\) and \((4, 13)\). (3 marks) (3 marks)

The point \((0, 5)\) is on the \(y\)-axis, so \(c = 5\) straight away. The gradient is \(\dfrac{13 - 5}{4 - 0} = 2\), so \(y = 2x + 5\). Spotting a point with \(x = 0\) saves a step.

Key terms

The words this lesson expects you to use. Each one is linked from the first place it appears above.

Equation of a line
A rule such as \(y = 2x + 1\) that is true for every point on the line.
Midpoint
The point halfway along a line segment.
Line segment
The part of a line between two end points.
Perpendicular
At a right angle to each other.
Negative reciprocal
A number turned upside down with its sign changed, such as \(-\dfrac{1}{2}\) for 2.
Substitute
Replace letters in an equation with numbers.
Gradient-intercept form
The form \(y = mx + c\).
Parallel
Lines with the same gradient.
Reciprocal
One divided by a number, such as \(\dfrac{1}{4}\) for 4.

Questions and answers

15 questions set on this lesson, with the mark schemes and model answers open.

1. Exam question Calculate 2 marks Easier

Calculate the midpoint of \((-3, -2)\) and \((5, 4)\).

Mark scheme — 2 marks available

  • One coordinate correct — M1
  • \((1, 1)\) — A1

Model answer

\(\left(\dfrac{-3 + 5}{2}, \dfrac{-2 + 4}{2}\right) = (1, 1)\).

2. Exam question Calculate 5 marks Core

The diagram shows a straight line through the points \(A\) and \(B\). (a) Calculate the gradient of \(AB\). [2 marks] (b) Find the equation of \(AB\). [2 marks] (c) Find the coordinates of the midpoint of \(AB\). [1 mark]

A straight line passing through the points A (0, 6) and B (4, 0).

Mark scheme — 5 marks available

  • (a) \(\dfrac{0 - 6}{4 - 0}\) — M1
  • (a) \(-\dfrac{3}{2}\) — A1
  • (b) \(y = -\dfrac{3}{2}x + c\) or \(y = mx + 6\) — M1
  • (b) \(y = -\dfrac{3}{2}x + 6\) — A1
  • (c) \((2, 3)\) — B1

Model answer

(a) \(\dfrac{0 - 6}{4 - 0} = -\dfrac{3}{2}\). (b) The line crosses the \(y\)-axis at 6, so \(y = -\dfrac{3}{2}x + 6\). (c) \(\left(\dfrac{0 + 4}{2}, \dfrac{6 + 0}{2}\right) = (2, 3)\).

3. Exam question Find 3 marks Core

Find the equation of the line that is parallel to \(y = 4x + 1\) and passes through \((1, 9)\).

Mark scheme — 3 marks available

  • \(y = 4x + c\) — M1
  • \(9 = 4 \times 1 + c\) — M1
  • \(y = 4x + 5\) — A1

Model answer

The gradient is 4, so \(y = 4x + c\). Putting in \((1, 9)\) gives \(9 = 4 + c\), so \(c = 5\) and \(y = 4x + 5\).

4. Exam question Find 3 marks Core

A line has equation \(2x - 3y = 12\). Find the coordinates of the points where the line crosses the \(x\)-axis and the \(y\)-axis.

Mark scheme — 3 marks available

  • \(y = 0\) or \(x = 0\) used — M1
  • \((6, 0)\) — A1
  • \((0, -4)\) — A1

Model answer

When \(y = 0\), \(2x = 12\) and \(x = 6\), so the point is \((6, 0)\). When \(x = 0\), \(-3y = 12\) and \(y = -4\), so the point is \((0, -4)\).

5. Exam question Find 3 marks Stretch

A line has equation \(3y + x = 6\). Find the gradient of a line that is perpendicular to it.

Mark scheme — 3 marks available

  • \(y = -\dfrac{1}{3}x + 2\) or the gradient \(-\dfrac{1}{3}\) — M1
  • Negative reciprocal used — M1
  • 3 — A1

Model answer

Rearranging, \(3y = -x + 6\), so \(y = -\dfrac{1}{3}x + 2\) and the gradient is \(-\dfrac{1}{3}\). The perpendicular gradient is 3.

6. Exam question Calculate 4 marks Stretch

\(A\) is the point \((3, -2)\) and \(B\) is the point \((-2, 10)\). (a) Calculate the coordinates of the midpoint of \(AB\). [2 marks] (b) Calculate the length of \(AB\). [2 marks]

Mark scheme — 4 marks available

  • (a) One coordinate correct — M1
  • (a) \((0.5, 4)\) — A1
  • (b) \(\sqrt{5^2 + 12^2}\) — M1
  • (b) 13 — A1

Model answer

(a) \(\left(\dfrac{3 + (-2)}{2}, \dfrac{-2 + 10}{2}\right) = (0.5, 4)\). (b) The differences are 5 and 12, so \(AB = \sqrt{5^2 + 12^2} = \sqrt{169} = 13\).

7. Multiple choice 1 mark Core

What is the equation of the line through \((2, 1)\) and \((6, 9)\)?

  1. A \(y = 2x + 3\)
  2. B \(y = \dfrac{1}{2}x - 3\)
  3. C \(y = 2x - 3\) Correct
  4. D \(y = -2x - 3\)

Why: The gradient is \(\dfrac{8}{4} = 2\). Then \(1 = 2 \times 2 + c\) gives \(c = -3\).

8. Multiple choice 1 mark Easier

What is the midpoint of \((2, 1)\) and \((6, 9)\)?

  1. A \((8, 10)\)
  2. B \((4, 5)\) Correct
  3. C \((2, 4)\)
  4. D \((4, 8)\)

Why: \(\left(\dfrac{2 + 6}{2}, \dfrac{1 + 9}{2}\right) = (4, 5)\).

9. Multiple choice 1 mark Easier

A line is parallel to \(y = 5x - 2\). What is its gradient?

  1. A \(5\) Correct
  2. B \(-5\)
  3. C \(-\dfrac{1}{5}\)
  4. D \(\dfrac{1}{5}\)

Why: Parallel lines have equal gradients.

10. Multiple choice 1 mark Core

A line has gradient 3 and passes through \((2, 9)\). What is its equation?

  1. A \(y = 3x + 9\)
  2. B \(y = 3x - 3\)
  3. C \(y = 3x + 6\)
  4. D \(y = 3x + 3\) Correct

Why: \(9 = 3 \times 2 + c\) gives \(c = 3\).

11. Multiple choice 1 mark Core

What is the gradient of the line \(2y - 4x = 6\)?

  1. A \(4\)
  2. B \(3\)
  3. C \(2\) Correct
  4. D \(-2\)

Why: Rearranged, \(2y = 4x + 6\), so \(y = 2x + 3\).

12. Multiple choice 1 mark Easier

What is the midpoint of \((0, 4)\) and \((6, 10)\)?

  1. A \((6, 14)\)
  2. B \((3, 7)\) Correct
  3. C \((3, 3)\)
  4. D \((6, 7)\)

Why: \(\left(\dfrac{0 + 6}{2}, \dfrac{4 + 10}{2}\right) = (3, 7)\).

13. Multiple choice 1 mark Core

Where does the line \(3x + 2y = 12\) cross the axes?

  1. A \((0, 6)\) and \((4, 0)\) Correct
  2. B \((0, 4)\) and \((6, 0)\)
  3. C \((0, 12)\) and \((12, 0)\)
  4. D \((0, 3)\) and \((2, 0)\)

Why: Put \(x = 0\) to get \(y = 6\), and \(y = 0\) to get \(x = 4\).

14. Multiple choice 1 mark Stretch

What is the gradient of a line perpendicular to a line with gradient 4?

  1. A \(\dfrac{1}{4}\)
  2. B \(-4\)
  3. C \(4\)
  4. D \(-\dfrac{1}{4}\) Correct

Why: Perpendicular gradients multiply to \(-1\), so the new gradient is \(-\dfrac{1}{4}\).

15. Multiple choice 1 mark Stretch

What is the equation of the line perpendicular to \(y = 2x + 3\) through \((4, 1)\)?

  1. A \(y = -2x + 9\)
  2. B \(y = \dfrac{1}{2}x - 1\)
  3. C \(y = -\dfrac{1}{2}x + 3\) Correct
  4. D \(y = -\dfrac{1}{2}x + 1\)

Why: The gradient is \(-\dfrac{1}{2}\). Then \(1 = -2 + c\), so \(c = 3\).