Work out the gradient of the line.

Work out the gradient of the line passing through the points (-5, -9) and (1, 3).

Find the gradient of a line

﻿Hello! My name is Mr. Clasper, and today we are going to be finding the equation of a straight line by using y equals mx plus c. Let's take a look at the general equation of a straight line. This is also known as the form. So if we have a line in the form of y equals mx plus c, the value of m tells us the gradient, and the value of c tells us the y-intercept. This is why we often see lines written in this form. If we take the straight line y equals three x minus five, this has a gradient of three, and it crosses the y-axis at zero, minus five. Let's find the equation of this line. We can find the gradient, as for every unit across, we go two units up. This means our gradient or the value of m is two. We can also see that our line crosses the y-axis at zero, minus one. This means the value of c is negative one. Therefore, the equation of my line is y equals two x minus one. If I look at my next graph, my gradient is negative one over two, as for every two units across, I go one unit down. And the value of c is three, as it crosses the y-axis at zero, three. Therefore, my equation is y equals negative one over two x plus three. Here's a question for you to try. Pause the video to complete your task, and click resume once you're finished. And here is your solution. So if we look at question one, our line has a gradient of one over two, and it crosses the y-axis at zero, one. And that gives us the equation y equals one over two x plus one. If you've written 0.5 x plus one, this is also fine. Here's another question for you to try. Pause the video to complete your task, and click resume once you're finished. And here is your solution for question two. So our line has a gradient of negative two, and it crosses the y-axis at zero, minus one. So therefore, we get the equation y equals negative two x minus one. Let's have a look at this question. The first thing we're going to need to do is to calculate the gradient of the line. We can do this if we take a change in y and divide it by a change in x. So if we take the y-coordinate from our second pair of coordinates and subtract the y coordinate from the first pair of coordinates, this is our change in y. And if we take the x-coordinate from our second pair of coordinates and subtract the x-coordinate from the first pair of coordinates, this will be our change in x. This will give us a gradient of negative two over two, which is equivalent to negative one. This means that our equation has a gradient or a value of m of negative one. Now we need to find our y-intercept. Well, because we know that our graph is going to cross at zero, one, we know that the y-intercept must be one. Therefore, the equation of the line given must be y equals negative x plus one. You could also write y equals negative one x plus one. Let's have a look at a slightly different example. Work out the equation of the straight line which passes through the points one, seven and four, one. Well again, we can work out our gradient by finding our change in y and dividing it by our change in x. And this would give us a gradient of negative six over three, which is equivalent to negative two. I then need to find my y-intercept, but this example is slightly trickier, as I don't have a coordinate which has an x-coordinate of zero. So let's have a look at this. I know that my equation is y equals negative two x plus c, but I don't know what c is. What I can do is look at my coordinates. Because the coordinate one, seven is on the line, this means that when x is equal to one, y must be equal to seven. Therefore, we can substitute these into our equation. And if we tidy this up and solve this equation and add two to both sides, we find that c must be equal to nine. Now we know the value of m, and we know the value of c. The equation of our line must be y equals negative two x plus nine. Here are some questions for you to try. Pause the video to complete your task, and click resume once you're finished. And here are your solutions. So for question three, we could rearrange this equation and write it as y equals negative two x plus three instead, and this would give us our gradient of negative two and a y-intercept of zero, three. And if we look at question four, we need to find our difference in y and our difference in x, and divide these. So our difference in y was 20, and our difference in x was four. Therefore, 20 divided by four would give us a negative gradient of five. And from here, once we've got our gradient of negative five, we can substitute either of our two coordinates, a or b, into the equation, and we should get our final answer. Here is a question for you to try. Pause the video to complete your task, and click resume once you're finished. And here is your solution. So if we take our coordinate two, nine, if we substitute a value of two for x, we are aiming to find a value of nine. And this happens when we substitute two into the equation for line a and the equation for line c. Therefore, these two lines must both have the coordinates two, nine on them. And that brings us to the end of our lesson. So you've been using y equals mx plus c to find the equation of a straight line. Why not show off your skills with the exit quiz? I'll hopefully see you soon.

Work out the equation of the line.

Find the equation of the straight line that has a gradient of 4 and passes through the point (1, 10)

Find the equation of a straight line using y=mx+c

In this lesson, we will find the equation of a straight line using y=mx+c. We will use coordinates taken from a plotted straight line to help us calculate the gradient, then use a method of substitution to find the equation of the line.

Maths

Y11 Maths (Higher)

Key Stage 4

Year 11

Revise - Linear and Quadratic Graphs

Algebra (A)

THEME

UNIT

Find the intercept and gradient from a line given in any form

Using gradient to solve problems with parallel lines

Plot simple quadratic equations

Plot other quadratic equations

Solving Quadratic Equations Graphically

Identify and interpret roots, intercepts and turning points of quadratic graphs

Draw quadratic graphs (a > 1)

Draw quadratic graphs (negative x squared)

Solve quadratic graphs = 0, = a and = ax + b

Solving quadratic equations, given a different quadratic, using a sketch

Which of these equations of a line is parallel to y = 5x− 8?

Which of these equations of the line is parallel to x + y = 8 and passes through (0, -4).

What is the equation of the line labelled P?

What is the equation of the line labelled Q?

What is the equation of the line labelled R?

What is the gradient of the line y=4x-5

What is the y intercept of the line y=7x-2

Work out the gradient of the line 2y + x = 20

Here are four quadratic graphs. The red curve shows y = x². Which equation is the yellow curve labelled i?

Lesson overview:Find the equation of a straight line using y=mx+c

Revise - Linear and Quadratic Graphs: