Maths · Graphs
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Equations of Straight Lines
Finding the equation of a line from two points, midpoints, and parallel and perpendicular lines.
Learning Objectives
- 1Find the equation of a straight line from its gradient and a point, or from two points.
- 2Find the midpoint of a line segment, and its length (Higher tier).
- 3Write the equation of a line parallel to a given line.
- 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.
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Step 1: gradient
Work out \(m = \dfrac{y_2 - y_1}{x_2 - x_1}\).
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Step 2: find c
Substitute \(m\) and one point into \(y = mx + c\), and solve for \(c\).
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Step 3: write it
Put \(m\) and \(c\) back into \(y = mx + c\).
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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 Gradient \(\dfrac{9 - 1}{6 - 2} = \dfrac{8}{4} = 2\).
- 2 Substitute a point \(y = 2x + c\) with \((2, 1)\) gives \(1 = 4 + c\), so \(c = -3\).
- 3 Write the equation \(y = 2x - 3\).
- 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.
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Formula
The midpoint of \((x_1, y_1)\) and \((x_2, y_2)\) is \(\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)\).
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Example
The midpoint of \((2, 1)\) and \((6, 9)\) is \(\left(\dfrac{2 + 6}{2}, \dfrac{1 + 9}{2}\right) = (4, 5)\).
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Finding an end point
If the midpoint and one end are known, double the midpoint and subtract the end you know.
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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.
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Same m, different c
A line parallel to \(y = 3x + 1\) has the form \(y = 3x + c\).
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Through a point
Put the point in to find \(c\). Through \((2, 9)\): \(9 = 6 + c\), so \(c = 3\), and the line is \(y = 3x + 3\).
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Rearranging first
\(2y - 4x = 6\) becomes \(y = 2x + 3\), so its gradient is 2. Lines with different-looking equations can still be parallel.
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Same line?
If \(m\) and \(c\) are both the same, it is the same line, not a parallel one.
Perpendicular lines
When two lines meet at a right angle, one gradient is the negative reciprocal of the other. Turn the fraction upside down and change the sign.
Perpendicular gradients (Higher tier)
- The rule If two lines are perpendicular, \(m_1 \times m_2 = -1\).
- Negative reciprocal The perpendicular to gradient 2 has gradient \(-\dfrac{1}{2}\). The perpendicular to gradient \(-\dfrac{3}{4}\) has gradient \(\dfrac{4}{3}\).
- Horizontal and vertical A horizontal line and a vertical line are perpendicular, but the rule does not apply to them.
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 Perpendicular gradient The gradient of \(y = 2x + 3\) is 2, so the new gradient is \(-\dfrac{1}{2}\).
- 2 Find c \(1 = -\dfrac{1}{2} \times 4 + c\), so \(1 = -2 + c\) and \(c = 3\).
- 3 Write the equation \(y = -\dfrac{1}{2}x + 3\).
- 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.
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Form ax + by = c
\(3x + 2y = 12\) rearranges to \(2y = -3x + 12\) and then \(y = -\dfrac{3}{2}x + 6\), so the gradient is \(-\dfrac{3}{2}\).
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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 \((4, 0)\).
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Equal gradients
To check two lines are parallel, rearrange both and compare \(m\).
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Fractions
Keep gradients as fractions, not decimals, unless the question asks.
Test yourself
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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 \(y = mx + c\).
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2
What is the midpoint of \((0, 4)\) and \((6, 10)\)?
Show answerHide answer
\((3, 7)\).
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3
What gradient does a line parallel to \(y = 5x - 2\) have?
Show answerHide answer
5.
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4
What is the gradient of a line perpendicular to \(y = 4x\) (Higher tier)?
Show answerHide answer
\(-\dfrac{1}{4}\).
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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.
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Write the gradient calculation
Show the change in \(y\) over the change in \(x\).
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Substitute a point
Write "\(1 = 2 \times 2 + c\)", not just \(c = -3\).
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Finish with a full equation
The answer is \(y = 2x - 3\), not just "\(m = 2\)".
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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.
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