How do I find the equation of a straight line?

Let's draw a line. It crosses the point (0,1), intercepting the y-axis (remember, y to the sky!) at 1 and the x-axis at the point (-2,0).

An equation tells us what a line looks like, it's a way of describing a line in a mathematical way. If we do it right, we can draw a line just by looking at the equation. Equations for straight lines are ususally given in the form y=mx+c - this tells us what y values do for different values of x. But what are m and c?

So, in order to find the equation for any straight line, we need two things: the gradient and the y-intercept. We call the y-intercept c and the gradient m. As soon as we know these two things, the rest is easy!

First step: finding the y-intercept - this is the simple bit, all you need to do is look at the y-axis and see where the line crosses it. In this case it is 1, so c=1.

Second step: finding the gradient - this requires a little bit more thought. All gradient means is the steepness of the line, so for every place the line moves to the right, how many places up does it go? In this case the line is not very steep at all and, in fact, you can see that for every place you move to the right it only goes up half a place up. Therefore m=0.5.

The tricky bit is over! Now comes the simple bit. All you have to do now is sub the values you've found for m and c back into the template equation, y=mx+c. So, if c=1 and m=0.5 then y=0.5x+1 - problem solved!

Answered by Dominique G. Maths tutor

4888 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

What is the correct answer if you rearrange the following, making "c" the subject? (3c+b)/2 = c + a


Rearrange the following to make m the subject: 4(m - 2) = t (5m +3)


Use factorisation to simplify the following expression (x^2-9)/(x^2-4x+3)


Simplify (14y^2)/y + 5y x 2


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2024

Terms & Conditions|Privacy Policy
Cookie Preferences