Linear Graphs (Cambridge O Level Additional Maths)

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Equations of a Straight Line

How do I find the gradient of a straight line?

  • Find two points that the line passes through with coordinates (x1, y1) and (x2, y2)
  • The gradient between these two points is calculated by

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

  • The gradient of a straight line measures its slope
    • A line with gradient 1 will go up 1 unit for every unit it goes to the right
    • A line with gradient -2 will go down two units for every unit it goes to the right

Finding the gradient of a straight line graph

What are the equations of a straight line?

  • space y equals m x plus c
    • This is the gradient-intercept form
    • It clearly shows the gradient m and the y-intercept (0, c)
  • space y minus y subscript 1 equals m open parentheses x minus x subscript 1 close parentheses
    • This is the point-gradient form
    • It clearly shows the gradient m and a point on the line (x1, y1)
  • space a x plus b y plus d equals 0
    • This is the general form
    • You can quickly get the x-interceptspace stretchy left parenthesis negative d over a comma space 0 stretchy right parenthesis and y-interceptspace stretchy left parenthesis 0 comma space minus d over b stretchy right parenthesis

Equations of a straight line graph

How do I find an equation of a straight line?

  • You will need the gradient
    • If you are given two points then first find the gradient
  • It is easiest to start with the point-gradient form
    • then rearrange into whatever form is required
      • multiplying both sides by any denominators will get rid of fractions

Finding the equation of a straight line graph using the point-gradient form

Examiner Tip

  • Quickly sketching the graph of the straight line(s) can be helpful if you are struggling with an exam question
  • Ensure you state equations of straight lines in the format required
    • Usually  y equals m x plus c  or  a x plus b y plus d equals 0
    • Check whether coefficients need to be integers (they usually are for a x plus b y plus d equals 0)

Worked example

(a)

Find the equation of the straight line with gradient 3 that passes through (5, 4).


We know that the gradient is 3 so the line takes the form 

y equals 3 x plus c

To find the value of c, substitute (5, 4) into the equation


4 equals 3 open parentheses 5 close parentheses plus c
4 equals 15 plus c
c equals negative 11


Replace c with −11 to complete the equation of the line

y = 3x − 11

(b)
Find the equation of the straight line that passes through (-2, 6) and (8, 1).

You may find it helpful to sketch the information given
XXg8k5ry_2-13-1-finding-equations-of-straight-lines
First find m, the gradient
table attributes columnalign right center left columnspacing 0px end attributes row m equals cell fraction numerator 6 minus 1 over denominator negative 2 minus 8 end fraction end cell row blank equals cell fraction numerator 5 over denominator negative 10 end fraction end cell row blank equals cell negative 1 half end cell end table
We know that the line takes the form 
table attributes columnalign right center left columnspacing 0px end attributes row y equals cell negative 1 half x plus c end cell end table

To find the value of c, substitute either of the given points into this equation. Here we will pick (8, 1) as it is doesn't contain negative numbers so is easier to work with


1 equals negative 1 half open parentheses 8 close parentheses plus c
1 equals negative 4 plus c
c equals 5


Replace c with −11 to complete the equation of the line

bold italic y bold equals bold minus bold 1 over bold 2 bold italic x bold plus bold 5

We can check against our sketch that this equation looks correct- it has a negative gradient and it crosses the y-axis between 1 and 6

Midpoint of a Straight Line

How do I find the midpoint of a line (segment)?

  • The midpoint of a line will be the same distance from both endpoints
  • You can think of a midpoint as being the average (mean) of two coordinates
  • The midpoint of open parentheses x subscript 1 comma space y subscript 1 close parentheses and open parentheses x subscript 2 comma space y subscript 2 close parentheses is

open parentheses fraction numerator x subscript 1 plus x subscript 2 over denominator 2 end fraction space comma space fraction numerator y subscript 1 plus y subscript 2 over denominator 2 end fraction close parentheses

Midpoint of a line segment

 

  • M is often used as the midpoint between two points (x1, y1) (x2, y2)
  • It is the average of both the x and y coordinates

Worked example

The coordinates of A are (−4, 3) and the coordinates of B are (8, −12).

Find M, the midpoint of AB.

The midpoint can be found using Mopen parentheses fraction numerator x subscript 1 plus x subscript 2 over denominator 2 end fraction space comma space fraction numerator y subscript 1 plus y subscript 2 over denominator 2 end fraction close parentheses

Fill in the values of x and y  from each coordinate

open parentheses fraction numerator negative 4 plus 8 over denominator 2 end fraction space comma space fraction numerator 3 plus negative 12 over denominator 2 end fraction close parentheses equals open parentheses 4 over 2 comma space fraction numerator negative 9 over denominator 2 end fraction close parentheses

Simplify

M = (2, −4.5)

Length of a Straight Line

How do I calculate the length of a line?

  • The distance between two points with coordinates open parentheses x subscript 1 space comma space y subscript 1 close parentheses and open parentheses x subscript 2 space comma space y subscript 2 close parentheses can be found using the formula

d equals square root of open parentheses x subscript 1 minus x subscript 2 close parentheses squared plus open parentheses y subscript 1 minus y subscript 2 close parentheses squared end root

  • This formula is really just Pythagoras’ Theorem  a squared equals b squared plus c squared, applied to the difference in the x-coordinates and the difference in the y-coordinates;

Distance between two points on a straight line graph

  • You may be asked to find the length of a diagonal in 3D space
    • This can be answered using 3D Pythagoras

Examiner Tip

  • Work with the square of a distance for as long as possible as this avoids early rounding errors
    • In the non-calculator paper your answer may need to be left as a surd
  • Only square root when forced to or for a final answer, and use the ANS button (and other memory features) on your calculator

Worked example

Point A has coordinates (3, -4) and point B has coordinates (-5, 2).

Calculate the distance of the line segment AB.

Using the formula for the distance between two points, d equals square root of open parentheses x subscript 1 minus x subscript 2 close parentheses squared plus open parentheses y subscript 1 minus y subscript 2 close parentheses squared end root 

Substituting in the two given coordinates:

d equals square root of open parentheses 3 minus negative 5 close parentheses squared plus open parentheses negative 4 minus 2 close parentheses squared end root

Simplify: 

d equals square root of open parentheses 8 close parentheses squared plus open parentheses negative 6 close parentheses squared end root space equals space square root of 64 plus 36 end root equals square root of 100 equals 10

Answer = 10 units

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Amber

Author: Amber

Expertise: Maths

Amber gained a first class degree in Mathematics & Meteorology from the University of Reading before training to become a teacher. She is passionate about teaching, having spent 8 years teaching GCSE and A Level Mathematics both in the UK and internationally. Amber loves creating bright and informative resources to help students reach their potential.