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Equation of a straight line: y = mx + c

Equation of a straight line: y = mx + c

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Summary

Equation of a straight line: y=mx+cy=mx+c​​

In a nutshell

The equation y=mx+cy=mx+c gives a straight line on a coordinate grid, where mm and cc are constants. It is the equation for almost any straight line, the exception being vertical lines, which have equations of the form x=dx=d where dd is a constant (the xx-intercept). You can use a straight line graph to work out the equation of the line.



The components of the equation y=mx+cy=mx+c

mm​ is the value of the gradient of the straight line and cc is the yy-intercept. xx and yy correspond to coordinates of points on the line. For any point (x,y)(x,y) on the line, multiplying the xx-coordinate by mm and adding cc, gives the yy-coordinate. If this doesn't work, then the point you are using is not actually on the line.



Finding points on a line

If you have a straight line graph y=mx+cy=mx+c then you will be given mm and cc to specify the line. Concurrently, xx and yy are not given as specific values, since they represent the xx- and yy-coordinates of any point on the line. If you insert an xx-coordinate into the equation of the line, it gives the corresponding yy-coordinate for the point on the line.


Example 1

Take the line with equation 

y=7x9y=7x-9


What are the coordinates of the point on this line that has xx-coordinate 33?


Substitute x=3x=3 into this equation:

y=7x9=7(3)9=219=12y=7x-9=7(3)-9=21-9=\underline{12}​​


Hence when x=3x=3 on the line, y=12y=12. So the point (3,12)\underline{(3,12)} is on the line with equation y=7x9y=7x-9.


Similarly, you can find a corresponding xx-coordinate if given a yy-coordinate of a point on a line. It just requires some algebraic rearranging to make xx the subject of the equation of the line.


Example 2

Take the line with equation

y=2x+6y=-2x+6


What are the coordinates of the point on this line that has yy-coordinate 88​?


Start by substituting y=8y=8 into the equation:

8=2x+68=-2x+6​​


Now rearrange to make xx the subject:​

8+2x=68+2x=6


2x=68=22x=6-8=-2


x=22=1x=\frac{-2}2=\underline{-1}


Hence when y=8y=8 on the line, x=1x=-1. So the point (1,8)\underline{(-1,8)} is on the line with equation y=2x+6y=-2x+6.



​​Using a graph to solve a linear equation

If you have the graph of a linear equation, you can solve a corresponding linear equation for a given xx- or yy-value. Simply read off the corresponding value from the graph.


Example 3

Consider the graph of y=2x4y=2x-4:

Maths; Graphs; KS4 Year 10; Equation of a straight line: y = mx + c

Using the graph, solve the equation 

6=2x46=2x-4​​

​​​​

The equation you have to solve is the equation of the line but with y=6y=6. Hence read off the graph to find the xx-coordinate when y=6y=6. You find that the xx-coordinate is 5\underline{5}.




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FAQs - Frequently Asked Questions

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