Equation of a Straight Line (AQA GCSE Further Maths)

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Jamie W

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Jamie W

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Equation of a Straight Line

What is the equation of a straight line?

  • y = mx + c is the equation for any straight line
  • m is gradient given by “difference in y” ÷ “difference in x” or fraction numerator bold d bold italic y over denominator bold d bold italic x end fraction
  • c is the y-axis intercept
  • An alternative form is ax + by + c = 0
    • where a, b and c are integers

 

Equation of a Straight Line Notes Diagram 1, A Level & AS Level Pure Maths Revision Notes

How do I find the equation of a straight line?

Equation of a Straight Line Notes Diagram 2, A Level & AS Level Pure Maths Revision Notes

  •  Two features of a straight line are needed
    • gradient, m
    • a point the line passes through, (x1, y1)
  • The equation can then be found using yy1 = m(x - x1)
  • This can be arranged into either y = mx + c or axby + c = 0

Equation of a Straight Line Notes Diagram 6, A Level & AS Level Pure Maths Revision Notes 

How do I find the gradient of a straight line?

  • There are lots of ways to find the gradient of a line

  • Using two points on a line to find the change in y divided by change in x

bold italic m bold equals fraction numerator bold italic y subscript bold 2 bold minus bold italic y subscript bold 1 over denominator bold italic x subscript bold 2 bold minus bold italic x subscript bold 1 end fraction

Equation of a Straight Line Notes Diagram 3, A Level & AS Level Pure Maths Revision Notes  

  • Using the fact that lines are parallel or perpendicular to another line
    • see Gradients - Parallel and Perpendicular Gradients

 Equation of a Straight Line Notes Diagram 4, A Level & AS Level Pure Maths Revision Notes 

  • Using Tangents and Normals - Differentiation
    • see Tangents & Normals

 

Equation of a Straight Line Notes Diagram 5, A Level & AS Level Pure Maths Revision Notes

 

  • Other ways
    • Collinear lines are the same straight line so gradients are equal
    • Angle facts and circle theorems
      e.g. a radius and tangent are perpendicular

Worked example

Equation of a Straight Line Example Diagram, A Level & AS Level Pure Maths Revision Notes

Intersection of Two Lines

How do I find where two lines intersect?

  • The coordinates of the point of intersection of two lines, open parentheses x comma space y close parentheses, is the solution for x and y when the pair of equations are solved simultaneously
    • A pair of parallel lines will not have any intersections, and therefore no solutions to the simultaneous equation, so it is useful to check if the lines are parallel before trying to solve them
  • We can find the solutions using either substitution or elimination
  • The most straight-forward method for finding where two straight lines intersect is to
    • Rearrange each line's equation to y equals...
    • Substitute one expression for y into the other equation, in place of y
    • Solve to find x
    • Substitute the value of x into either line's equation to find y
  • The method is exactly the same when finding where a line and a curve, or two curves, intersect
    • When finding where a quadratic curve and a line intersect, you will end up with a quadratic equation to solve, and there could be either 2, 1, or 0 solutions
      • This corresponds to the number of intersections of the line and the curve
  • The same idea can be used to find the intersection of a line or curve with the x or y axis
    • The equation of the y-axis is x equals 0
    • The equation of the x-axis is y equals 0
    • Either of these can be substituted into the equation of the line or curve and solved

Solving Equations Graphically Notes Diagram 1, A Level & AS Level Pure Maths Revision Notes

Solving Equations Graphically Notes Diagram 2, A Level & AS Level Pure Maths Revision Notes

Exam Tip

  • Consider the types of graphs (straight line, quadratic, cubic) that you are finding the intersections of
  • Sketch a graph to help you consider the number of intersections there will be, and roughly where those will be

Worked example

Find the points of intersection of the following

(a)

2 x plus 3 y equals 24 with the x-axis and y-axis


The equation of the x-axis is y equals 0
Substitute this in to the other (given) equation

2 x plus 3 open parentheses 0 close parentheses equals 24

Solve for x

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x end cell equals 24 row x equals 12 end table

There is no need to solve for y, since y equals 0

The point of intersection with the x-axis is (12, 0)

The equation of the y-axis is x equals 0
Substitute this in to the other (given) equation

2 open parentheses 0 close parentheses plus 3 y equals 24

Solve for y

table row cell 3 y end cell equals 24 row y equals 8 end table

There is no need to solve for x, since x equals 0

The point of intersection with the y-axis is (0, 8)

 
(b)

y equals 2 x minus 1 and 4 x plus y equals 26


These are the equations of two straight lines
The easiest way to find their intersection is to start by writing both in the form y equals...
The first equation already is but the second equation becomes

y equals 26 minus 4 x

Substitute this for y in the first equation and solve for x

table row cell 26 minus 4 x end cell equals cell 2 x minus 1 end cell row cell 26 plus 1 end cell equals cell 2 x plus 4 x end cell row cell 6 x end cell equals 27 row x equals cell 4.5 end cell end table

Substitute this value of x into either of the original equations and solve for y

table row y equals cell 2 open parentheses 4.5 close parentheses minus 1 end cell row y equals 8 end table

The point of intersection between the two lines is (4.5, 8)

 
(c)

y equals x squared minus 5 x plus 5 and 3 x minus y equals 7


One of these equations is quadratic (and already in the form y equals...)
The second (straight line) equation needs rearranging to y equals... form

y equals 3 x minus 7

Substitute this for y into the first equation and rearrange to a quadratic equation

table row cell 3 x minus 7 end cell equals cell x squared minus 5 x plus 5 end cell row cell x squared minus 5 x plus 5 minus 3 x plus 7 end cell equals 0 row cell x squared minus 8 x plus 12 end cell equals 0 end table

Solve (if possible) this equation to find the x-coordinates of any points of intersection
(Factorising, quadratic formula and completing the square are possible techniques or, if allowed, a calculator)

table row cell open parentheses x minus 6 close parentheses open parentheses x minus 2 close parentheses end cell equals 0 row x equals 6 row x equals 2 end table

Substitute these into either of the original equations to find the y-coordinates
(Using the straight line equation is usually easiest)

x equals 6 comma space space y equals 3 open parentheses 6 close parentheses minus 7 equals 11
x equals 2 comma space space y equals 3 open parentheses 2 close parentheses minus 7 equals negative 1

Write your final answers as coordinates but as there are two intersections be careful to match x's and y's correctly

The points of intersection between the curve and line are (6, 11) and (2, -1)

Finding Parallel & Perpendicular Lines

What are parallel lines?

  • Parallel lines are lines that have the same gradient, but are not the same line
  • Parallel lines do not intersect with each other
  • You can easily spot that two lines are parallel when they are written in the form y equals m x plus c, as they will have the same value of m (gradient)
    • y equals 3 x plus 7 and y equals 3 x minus 4 are parallel
    • y equals 2 x plus 3 and y equals 3 x plus 3 are not parallel
    • y equals 4 x plus 9 and y equals 4 x plus 9 are not parallel; they are the exact same line

How do I find the equation of a line parallel to another line?

  • As parallel lines have the same gradient, a line of the form y equals m x plus c will be parallel to a line in the form y equals m x plus d, where m is the same for both lines
    • If c equals d then they would be the same line and therefore not parallel
  • If you are asked to find the equation of a line parallel to y equals m x plus c, you will also be given some information about a point that the parallel line, y equals m x plus d passes through; open parentheses x subscript 1 space comma space y subscript 1 close parentheses
  • You can then substitute this point into y equals m x plus d and solve to find d

What are perpendicular lines?

  • You should already know that parallel lines have equal gradients
  • Perpendicular lines meet each other at right angles
    • i.e. they meet at 90°

What’s the deal with perpendicular gradients (and lines)?

  • Before you start trying to work with perpendicular gradients and lines, make sure you understand how to find the equation of a straight line – that will help you do the sorts of questions you will meet
  • Gradients m1 and m2 are perpendicular if m1 × m2 = −1
    • For example
      • 1 and -1
      • 1 third and -3
      • negative 2 over 3 and 3 over 2
  • We can use m2 = −1 ÷ m1 to find a perpendicular gradient. This is called the negative reciprocal.
  • If in doubt, SKETCH IT!

Exam Tip

  • Working with straight lines can involve lots of algebra, but sketching a diagram will always help
  • Use a sketch to check if answers seem about right

Worked example

(a)
Find the equation of the line that is parallel to y equals 3 x plus 7 and passes through (2,1).
 

As the gradient is the same, the line that is parallel will be in the form: 

y equals 3 x plus d

Substitute in the coordinate that the line passes through: 

1 equals 3 open parentheses 2 close parentheses plus d

Simplify: 

1 equals 6 plus d

Subtract 6 from both sides: 

negative 5 equals d

Final answer: 

bold italic y bold equals bold 3 bold italic x bold minus bold 5

 

(b)
Find the equation of the line that is perpendicular to y equals 2 x minus 2 and passes through (2, -3).
Leave your answer in the form a x plus b y plus c equals 0 where a comma space b comma space c are integers.
 

L is in the form y equals m x plus c so we can see that its gradient is 2

m subscript 1 equals 2

Therefore the gradient of the line perpendicular to L will be the negative reciprocal of 2

m subscript 2 equals negative 1 half

Now we need to find c for the line we're after
Do this by substituting the point open parentheses 2 comma space minus 3 close parentheses into the equation y equals negative 1 half x plus c and solving for c

table row cell negative 3 end cell equals cell negative 1 half cross times 2 plus c end cell row cell negative 3 end cell equals cell negative 1 plus c end cell row c equals cell negative 2 end cell end table

Now we know the line we want is 

table attributes columnalign right center left columnspacing 0px end attributes row y equals cell negative 1 half x minus 2 end cell end table

But this is not in the form asked for in the question. So rearrange into the form a x plus b y plus c equals 0 where ab and c are integers

table row cell y plus 1 half x plus 2 end cell equals 0 row cell 2 y plus x plus 4 end cell equals 0 end table

Write the final answer

bold italic x bold plus bold 2 bold italic y bold plus bold 4 bold equals bold 0

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Jamie W

Author: Jamie W

Jamie graduated in 2014 from the University of Bristol with a degree in Electronic and Communications Engineering. He has worked as a teacher for 8 years, in secondary schools and in further education; teaching GCSE and A Level. He is passionate about helping students fulfil their potential through easy-to-use resources and high-quality questions and solutions.