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Well done for making the decision to learn with this video today.

My name is Ms. Davies and I'm gonna be helping you as you work through this lesson.

Let's get started then.

Welcome to this lesson on parallel linear graphs.

By the end of the lesson you'll be able to identify, from their equations or graphs, whether two lines are parallel.

The most important key word for today is the idea of parallel.

Two lines are parallel if they're straight lines that are always the same, non-zero distance apart.

So let's start by identifying parallel lines from a graph.

The image below shows the graph of three linear equations.

Which one is the odd one out? And I'd like you to justify your answer.

So it doesn't matter what you say as long as you can justify your answer.

The more mathematical language you can use the better, off you go.

Now if you haven't already done so, can you give a justification for why each line could be the odd one out? Is it possible to reason that each one could be the odd one out and for what reason? Give this a go.

So for A, you might have said A is the only one with a negative gradient and that's why it's the odd one out.

You might have gone further and said B and C are parallel and A is not, so A has a different gradient to the other two and the other two's gradient are exactly the same.

For B, you might have said B was the odd one out because it has a different x-intercept.

The other two cross the x-axis at (2,0) but B doesn't.

So B would be the odd one out.

And then C would be the odd one out if we were looking at y-intercepts, 'cause C has a different y-intercept to the other two.

Well done, if you managed to convince someone or yourself or me that each of those could be the odd one out.

Okay, which line is the odd one out this time? Just like before, can you justify your answer? It doesn't matter which one you pick as long as you can explain why.

Well done, I think some of these were more obvious to see why they might be the odd one out than others.

So A could be the odd one out because it's the only one with a dashed line.

It's also the only one with a negative y-intercept.

A little bit more unusual that one.

We can see the y-intercepts for B and C are both positive values, whereas A is gonna have a negative value.

So you might've said that.

Now, B being the odd one out was a little tricky.

B is the only one that does not intercept one of the axes on a coordinate containing the value two, you can see that C intercepts the y-axis at (0,2) and A intercepts the x-axis at (2,0).

However, B does not intercept either axis at a point with coordinate two, but that was a very niche idea, so don't worry if you didn't think about that one.

C could be the odd one out because it has a different gradient to the other two.

You might have noticed that A and B are parallel and C has a different gradient.

Don't worry if you didn't come up with all of those ideas, but hopefully you had a chance to just explore what you already know about linear graphs.

So let's talk about parallel then.

Two lines are parallel if they're straight lines that are always the same, non-zero distance apart.

With that in mind, which of those lines are parallel? Pause the video, can you see it? Well done If you spotted that it's C and D.

If we look at those two in more detail, they are always the same distance apart and you can see that they will never intersect.

So Sam has drawn the line with equation y = 3x + 1, has a gradient of three and goes through the y-axis at (0,1).

Izzy says, I wonder if the line with equation y = 2x + 1 will be parallel? I wonder if you know already whether she's gonna be right or not.

There you go, was that what you thought? See if you can explain then why Izzy's line is not parallel to Sam's.

Well you can see that the lines are not always the same distance apart.

Parallel lines will not intersect and you can see clearly that those two lines intersect at (0,1).

Sofia says, well let's try y = 1x + 3.

There you go.

And we can see again the lines are not always the same distance apart.

If the lines were parallel they would not intersect and they do clearly-intercept there at (1,4).

So let's see if we can see what is happening.

Andeep now has drawn a parallel line.

So can you work out the equation of Andeep's line? Sam's is y = 3x + 1.

What would Andeep's be? Well we have a gradient of three and a y-intercept of five.

So y = 3x + 5.

Why is this parallel to Sam's? Can you explain it this time? Well, the lines are always the same distance apart.

We know that for certain 'cause they both go right one, up three, right one, up three, so they're always staying exactly the same distance apart.

Both lines have a gradient of three.

This time Sam has drawn the line with equation y = -2x - 1.

Izzy has drawn some parallel lines.

How could we work out the equations of Izzy's lines? What's gonna be the most efficient way? What do you think? If you look at Izzy's lines, they're all gonna have the same gradient as each other and the same gradient as Sam's.

So all we need to do really is look at the y-intercepts.

So the blue line A is y = -2x - 3.

Sam's line was y = -2x - 1.

The green line B is y = -2x and the purple line C is y = -2x + 4.

So we have seen how they all have the same gradient but different y-intercepts.

And you can see then in the equations A, B and C, the similarities that we have.

Sofia has drawn the line y = 2x.

Izzy says parallel lines will have the same gradient, so you can add any constant and get a parallel line.

Sam uses that idea and says all of these would be parallel to y = 2x.

Have a read of those four.

Do you agree with all of these statements? Okay, well Sofia was correct, that is the line with equation y = 2x.

Izzy is almost correct with her statement.

There's a little bit of a problem with what she said but she's close.

Three of Sam's lines will be parallel to y = 2x.

Y = 2x + 1 will be parallel.

y = 2x - 8 will be parallel, they're both drawn on the graph.

Y = 2x + 128.

25 will be parallel, I've had to zoom my graph out a little bit to find it, but there's a problem with y = 2x + 0.

There's y = 2x + 0.

It cannot be considered parallel to y = 2x 'cause they're actually the exact same line.

Our definition of parallel is two straight lines that are always the same non-zero distance apart, two lines sitting on top of each other cannot be called parallel.

So Izzy has adjusted her definition.

Parallel lines will have the same gradient but a different y-intercept.

So which two lines are parallel to each other? Have a look, what do you think? Well done if you forgot that it was C and E.

If you look at B and C, you can see that the distance between them is getting narrower as the graph moves right.

So they're not parallel.

If you look at B, it goes along four.

up one, whereas with C it goes along two, up one.

So they've not got the same gradient, you can see that they're not the same distance apart.

So then your only choices are B and E or C and E.

And again you can count the squares to help you with E, we go along two, up one, along two, up one, and that's the same as C.

So C and E must be parallel and that means that B and E cannot be parallel.

Time to have a practise then.

So for question one, I'd like you to draw any two lines which are parallel to the line already plotted.

Use the squares to help you, absolutely use a ruler.

For question two, I'd like you to explain whether those two lines are parallel or not.

Can you convince me that they are parallel or convince me that they are not parallel? For question three, I've drawn a line y = x - 3 I'd like you to draw a parallel line and then tell me the equation of your line.

For four, I'd like you to draw a parallel line to the one already drawn, but I want it to go through the point that I've marked on, the point with coordinate (0,8).

Once you've done that, can you tell me the equation of your line? And finally we're practising a few different skills here.

I'd like you to draw the line with equation y = -x - 2.

Think about two coordinates that satisfy that relationship and then you can draw your line.

You might wanna check it with a third coordinate as well.

Can you also draw the line with equation y = 4 - x? Again, pick a couple of x coordinates, use the equation to work out the y coordinates and then maybe a pick a third one to make sure you're definitely right.

And then are your lines parallel? For six, I'd like you to do the same.

I've picked some slightly trickier equations, but all you need do again is pick a couple of x values and work out the y values.

I suggest you pick even numbers for your x values so that when you halve them you get integers.

I'd like you to do the same with equation y = 2 - 1/2x.

And then tell me if your lines are parallel.

Give those a go, I know there's quite a lot of thinking to do for that last one.

So take your time and come back when you're ready for the answers.

Let's have a look then.

There's all sorts of lines you could have drawn for A and B.

All your lines should go right one, up three.

So check that they're parallel to the ones that I've drawn.

For, B, same again.

A little bit trickier to draw the gradient this time.

Two squares right, three squares down and so on.

Right then your explanation.

I've drawn some gradient triangles on my lines to help with my explanation.

Drawing things on a graph can be part of your explanation, that's absolutely fine.

So I've shown that these lines do not stay the same distance apart, they do not have the same gradient.

The black line goes right three, up two, whereas the purple line goes right five, up four.

They're actually getting closer together as the graphs move right.

For question 3A, there's many different lines you could have drawn.

They should all have a gradient of one, right one, up one.

And then your equation is gonna depend whereabouts on the graph you drew it.

So you could have had y = x + 4, y = x + 3, y = x - 1.

You could have had the one going through the origin, which is just y = x.

So it should be y = X + c.

But of course C cannot be -3 because otherwise it'd be exactly the same equation as we already have.

And then for question 4A.

The gradient should be -4.

Notice that the step on the y-axis is two.

So the equation of your line should be y = -4x + 8 'cause I needed it to go through eight on the y-axis.

And finally for question 5A, you might want some coordinates on that line to help you draw them and you can see that they are parallel.

For 6A, again, you want to pick a couple of coordinates to help you plot those, the y-intercept and the x-intercept might be the easiest ones to go for.

And then you can clearly see that they are not parallel, they intercept, one has a gradient of a half, one has a gradient of -1/2, they're going in completely different directions.

So now we're gonna explore this idea of parallel lines and see if we can identify them algebraically.

So we have seen how parallel lines are always the same non-zero distance part and we can identify whether lines are gonna be parallel by looking at their equations and without drawing them.

Izzy said earlier that parallel lines have the same gradient and a different y-intercept.

What that means is when we write equations in the form, y = mx + c, it's gonna be easy to see whether they're parallel or not.

So let's have a look at y = 8x + 1 and y = 8x + 9.

They will be parallel 'cause they have the same gradient.

They both have a gradient of eight.

It's really easy to see that because they're written in the form y = mx + c.

Here are the lines y = 2x - 4 and y = 4 - 2x.

How could we tell they were not parallel without drawing them? We can see they're not parallel on the diagram, but how could we see they were not parallel if we hadn't drawn the lines? What do you think? Well they have different gradients.

The first one has a gradient of two and the second one has a gradient of -2.

What about these two? y = 4x + 3 and y = 3x + 4, why are they not parallel? Of course it's the same thing, they have different gradients.

They're both in the form y = mx + c and they have different coefficients of x, and the coefficient of x is telling us the gradient.

What'd you think about these two? y = 3/5X and y = 0.

6x, will they be parallel? I dunno if you spotted that 3/5 is equivalent to 0.

6, so these two lines have the same gradient, however they have the same y-intercept, so they're not parallel lines, they're actually the exact same line, but y = 3/5x and y = 0.

6x + 0.

01 would be parallel.

Yeah, they have the same gradient but a different y-intercept.

So which of these lines would be parallel to y = x - 2? What do you think? Yeah, it's the one with the same gradient.

So it should be y = x + 18, both of those have a gradient of one.

Which of these would be parallel to y = 7x? Both B and C are gonna be parallel.

They both have a gradient of seven.

A has a different gradient, it has a negative gradient.

Well I dunno if you spotted that D has the same gradient but also the same y-intercept.

So they're exactly the same line, so we don't call them parallel.

Okay, so what about if they're not in the form y = mx + c? So let's look at 2y = 4x + 6 and y + 2x = 5.

At the moment it's hard to see if they're going to be parallel.

All we need to do is rearrange them into the form y = mx + c.

So for that first one, that just involves dividing every terms by two.

So y = 2x + 3.

For the second one, it's just gonna involve subtracting 2x from both sides.

So y = 5 - 2x.

Now it's a lot easier to see that they're not parallel as the gradients are different.

These equations are both in the form axe + by = c.

Again, we can rearrange them to see if they're parallel.

It's absolutely fine to have fractions in our answers, we're just gonna need to compare it with the other one.

They have the same gradient and different y-intercepts so they are parallel.

Ooh, Jun's been thinking.

Without rearranging, we can see the variables in the second equation of both double those in the first equation, but the constant has not been doubled.

Right, pause the video, what is he on about? Let's have a look, in the second equation, x has been doubled and y has been doubled, but the constant hasn't been doubled.

So these equations are not exactly the same equation.

Well that's good 'cause we know that if they're exactly the same equation, we don't call them parallel.

Doubling both the variables have kept the gradient the same, essentially because the variables are still in the same ratio, which means they're parallel.

So which of these would be parallel to y - 3x = 8? You might wanna do some rearranging to help you.

So we're looking for any equation of a line with a gradient of three.

So not the first one that has a gradient of -3.

Yes, the second one.

Yes, the third one, and not the last one, that has a gradient of 1/3.

Well done if you use Jun's idea to see that B is definitely parallel.

You'll see that the y and the x variables have both been multiplied by three so they stay in the same ratio but the constant has not been multiplied by three, so it is a different line.

So let's find the equation of a line parallel to y = 3x which goes through the point (5,10).

Well if it's parallel to y = 3x, the gradient must be three.

And then we can use the point (5,10) to find the y-intercept.

You can either count back to get to the y-intercept or you can substitute the coordinate in.

I'm gonna show you that countback method first.

So the gradient's three, we know we have to go through the point (5,10).

So I need to count back three five times, which is subtracting 15.

So (0,-5) must be my y-intercept.

Alternatively, once we know the gradient is three, we can substitute.

So we know our equation must be in the form y = 3x + c.

And then if we substitute our point in.

10 three lots of five + c.

So 10 = 15 + c, so c must be -5.

Okay, we're gonna have a practise of that skill then.

I'm gonna show you on the left hand side, you are gonna do one on the right hand side.

So why I want an equation of the line parallel to 2y = 3x + 6, which passes through the point (6,4).

So the first thing, to find the gradient, I'd like it in the form y = mx + c.

If I divide every term by two, I can see that y is going to be 3/2x + 3.

A parallel line will have equation y = 3/2x + something that's not three.

I can substitute my coordinate in to help me.

So four has got to equal 3/2 times six + c.

So four has got to equal nine + c, so c must be -5.

So my equation must be y = 3/2x - 5.

Your turn, have a look at what I did, see if you can do the same thing.

Okay, so rearranging, if we subtract x from both sides, we've got y = -x + 3.

5.

A parallel line will have equation y = -x + something else.

We substitute the coordinate we have.

So seven has got to equal -2 + c, therefore c must be nine.

So y = -x + 9.

Well done, time to put that all into practise.

I would like you to match up any equations for parallel lines.

There may be multiple lines which are parallel.

If you have a look at the grid on the right hand side, there's space for three different lines which could be parallel to each other.

There may be some equations of lines which are not parallel to any of the others.

So I've told you to help you out that A and M parallel and that's why I've written them in that first row of the grid.

So when you find another two or three which are parallel, you can write that in the next row of the grid.

Give that a go and then we'll have a look at the next bit.

Well done.

For question two, I'd like you to find the equation of the line parallel to y = 2x - 4, which passes through the point (0,7).

And for 2B, similar idea, I'd like you to find the equation of the line parallel to y = 5x - 1 which passes through the (3,8).

Feel free to sketch things out to help you.

And for 2C, I'd like you to find a line parallel to y = 5 - 3x, but it needs to go through the point with coordinates (7,-15).

And for 2D, you're probably gonna want to rearrange that equation into the form y = mx + c, so that you can then get a parallel line which passes through (-2,2).

For 3A, I'd like you to explain why those lines will be parallel, the same for 3B.

And for 3C, can you write an equation of a third line parallel to the two above? But I want your answer in the form y = mx + c.

And finally, Jun has drawn a line which goes through the points (-4,2) and (-4,7).

I'd like you to write the equation of a line parallel to Jun's.

And then Sam has drawn a line which goes through the points (-3,-1) and (3,3).

Can you write down the equation of a line parallel to Sam's, which passes through (15,26).

Off you go.

Right, there was loads to do in that activity, so well done for really pushing yourself and doing your absolute best.

So let's have a look at these parallel lines.

So B, P and T were all parallel.

They had a gradient of one.

C and I were both parallel, they had a gradient of three.

D, R and S all had a gradient of -2.

E, H and J all had a gradient of -1.

F, G and Q all had a gradient of -1/2.

And for K and N, well I dunno if you noticed, they were parallel.

For K, you might have written the gradient as 4/5ths but that's the same as 0.

8.

L and O were not parallel to any of the other lines.

For 2A, we know that to be parallel, it'll have to also have a gradient of two and it has a y-intercept of seven.

So y = 2x + 7.

For 2B to be parallel, it also has to have a gradient of five.

So y = 5x + c.

Five threes are 15 and we need it to be eight, so we need to subtract seven, so y = 5x - 7.

For C, we have a gradient of -3, so y = -3x + c.

<v ->3 times seven is -21, so we'd need to add six.

</v> So y = -3x + 6 or y = 6 - 3x.

For 2D we want to rearrange to get our gradient first.

So we've got y = -4x + 4/3.

Not too interested in the constant, it's the gradient I'm interested in.

So parallel line will have an equation of y = -4x + c.

<v ->4 times -2 is eight.

</v> Two subtract eight is -6.

So I need y = -4x - 6.

Well done if you got this far, that was quite a lot to do already.

So explain why these lines will be parallel.

So you could have rearranged them into the form y = mx + c and shown that they have the same gradient, you might have decided to look at it without rearranging and use Jun's idea.

So without rearranging, both variables have been doubled, and the constant is not doubled.

So that means the variables are still in the same ratio, so they'll be the same gradient but a different y-intercept, so parallel.

That's definitely the easiest way to do this.

You could rearrange them, that's gonna be a little bit of a pain.

The easiest thing to spot is that variables in the second equation have been multiplied by three, but the constant has not.

If you did rearrange, you can see that the gradients of both is 2.

5 or 5/2, but this is definitely the easiest way this time.

For a parallel line, you're gonna need any equation in the form y = 5/2x + c, as long as c is not -1/2 or -2/3.

And for the last one, well done for persevering this far.

So if you sketch the pointss out, it would've been easy to see that Jun's line is a vertical line.

So any equation of the form x = a, where a is a constant, not equal to -4 would be parallel.

So you might have gone with x = one, x = zero, x = -3.

Any equation for a vertical line which is not x = -4.

There was a lot to do for Sam's, so we needed the gradient first.

So the change in x is six and the change in y is four, so the gradient is two thirds, but then our equation must have a gradient of two thirds, but also go through the point (15,26).

So if we substitute that coordinate in to y = 2/3x + c, rearrange to get c, you'll see c is 16, y = 2/3x + 16.

Well done for bringing all your graphing skills together for that final question.

If you didn't quite get to that final answer, don't panic, go back through and see if you can work out which stage you found difficult.

Fantastic work today.

We've done lots of work with parallel linear graphs.

Even if you found that last bit a little bit tricky, you have shown that you can identify parallel lines on a graph.

We all know that lines will be parallel if they have the same gradient and therefore we can write equations of parallel lines in the form y = mx + c.

Well done if you stretched yourself even further today to have a look at those trickier questions, which involved rearranging, they involved substitution skills, they're all things that we've done before, but sometimes we just need a little bit of a reminder of.

Fantastic effort and I look forward to seeing you again.