Lesson video

Hello, my name is Miss Parnham.

In this lesson we're going to learn how to write the equation of a straight line if parallel to a line and passing through the coordinate.

Here we have two lines y= 2X + 1 and y= 2X- 2 drawn on the same grid.

So let's look at these lines and think about first what's the same and we notice that they are parallel.

And looking at the equations, they have the same x-coefficient of two.

So that is the gradient of these lines.

So what we're saying is that when two or more lines have the same gradient then they are parallel.

And now what is different these lines? Well, y= 2x + 1 intercepts the y axis at one because the y-axis is when x equals zero, and therefore y = 2x-2 intercepts the y-axis at negative two.

So if we draw a third line parallel to these two lines and passing through the point.

What would be the equation of that line? Well, it is parallel so has the same gradient.

So the similarity is it will start with y = x, but this intercepts the y-axis at positive three, which means the constant on the end of the equation is positive three so putting that all together, draw that line in and this is y = 2x + 3.

The last example showed us that parallel lines have the same gradient and we identified the gradient in the equation of a line as the x-coefficient.

So if we want to write down the equation of a line parallel to y= 5x -7, we know that it will also start with y = 5x, but it passes through the point.

and in the last example we saw that this was the y- intercept when x has the value zero, y has the value negative nine for our line.

So we know it starts with 5x and the c must be negative nine.

So our answer is y= 5x -9.

In this example, we have to find the equation of a line parallel to y = 12- 11 x.

Now that's not written in y= mx + c form because m the gradient, is negative and the constant is positive we sometimes see it written like this.

And we're looking for a line that's parallel to this line and goes to the point.

So when we look at our equation that we've been given, we're looking for the x-coefficient, and this is negative 11.

So the gradient is negative 11.

So the gradient on our line will also be negative 11.

So it starts with negative 11x, and our constant is 17.

So that gives us y =- 11 x + 17 or if you prefer, y = 17 -11 x.

Here's a question for you to try.

Pause the video to complete the task and then restart the video once you're finished.

Here are the answers.

In part a the word origin is used instead of and so if you had y = -4x + 0, then this is also a correct solution for that question.

In this example we have to find the equation of a line parallel to x + 2y= 12.

And this is not written in y = mx + c form, so it's not immediately obvious what the gradient is.

So let's rearrange that.

So let's first subtract x from both sides.

And this gives us 2y = -x + 12 and then to isolate y we're going to need to divide by two and that gives us that y =-1/2 x + 6.

So we can see that this has a gradient of negative a half.

Now our line is going to pass through the point.

So we know c will be 31 and therefore the equation of the line parallel to x + 2y =12 is y =- 1/2x + 31.

Here are some questions for you to try.

Pause the video to complete the task, and then restart the video when you're finished.

Here are the answers.

Rewriting the given equation in the form y = mx + C will give you for part a, y =2x + 4, c is y = -2x -3, d re- arranges y = 2x -1.

5 and the final one y =-1/2x -1.

So all your answers should start in exactly the same way because they are parallel and have the same gradient.

So the x-coefficient is the same, but the constant value is taken from the y-coordinate of the pair of coordinates that your given.

Now let's find the equation of a line that's parallel to this line L and passes through the point marked B.

Right we first need to find the equation of our line L.

We can see that the horizontal change is one and the vertical change is three.

Therefore the gradient is three divided by one.

So we know that L is going to start with y = 3x.

Now this line intersects the y axis at positive one.

So the constant on the end of that is going to be plus one.

So the equation of L is y = 3x + 1.

Now our equation is parallel to this so it starts with y = 3x in exactly the same way.

But it intercepts at B, and we can see B is marked at negative four.

So our final equation for the line that is parallel to L and passing through the point B is y = 3x - 4.

Here's a question for you to try.

Pause the video to complete the task, and then restart the video when you're finished.

Here are the answers.

The line we can see here has got a gradient of four, because the vertical change is four when the horizontal change is one.

And we can see that using the coordinates and scale on the axes.

So anything parallel has the same gradient.

So our equation must start with y = 4x.

And with a y intercept marked at , that's got to be y = 4x -4.

Here's a further question for you to try, pause the video to complete the task, and then restart the video when you're finished.

Here are the answers.

This line has gradient negative three, because as the x-coordinate increases by one, so the y-coordinate decreases by three, and therefore our answer must start with y = -3x.

And since the line that we are talking about intercepts at positive one the equation must be y = -3x + 1.

Now sometimes when we have a negative x-coefficient and have a positive constant we can see that written as y = 1- 3x, which is also acceptable for this question.

So that's all for this lesson.

Thank you for watching.