Lesson video

Hello there, welcome, and thank you for joining us for today's lesson.

My name is Ms. Davies, and I'm gonna be helping you as you work your way through this topic.

There's lots of exciting things gonna come up in today's lesson.

Make sure you've got all the things you need before you start and take your time.

Feel free to pause the video whenever necessary so that you can really think about what it is that you are doing.

Let's get ourselves started then.

Welcome to this lesson on identifying perpendicular linear graphs.

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

So a few keywords we're gonna use today.

Two lines are perpendicular if they meet at right angles.

And in the second part of the lesson, we're gonna explore some irrational numbers.

So an irrational number is one that cannot be written in the form a/b, where a and b are integers and b is not equal to 0.

So for example, pi is an irrational number.

It cannot be written as a fraction a/b, where a and b are integers.

A surd is an irrational number expressed as the root of a rational number.

So for example, the square root of 2 is a surd 'cause it's an irrational number.

You cannot write it as the fraction a/b, where a and b are integers, but it's also a root of a rational number.

So the square root of 2 would be a surd.

The root sign is called the radical symbol.

So when I use that term today, you'll know I'm referring to that root sign.

So let's get started with identifying gradients of perpendicular lines.

Two lines are perpendicular if they meet at right angles.

We can use the gradients of the lines to determine if they are perpendicular.

So Izzy says, "How can I tell if the lines with equations y = 2x + 4 and y = 1/2x - 4 are perpendicular?" Sofia says, "We can look at their gradients.

One has a gradient of 2, the other has a gradient of 1/2." Do you think these will be perpendicular? Let's have a look.

Well done if you said no.

I've drawn the lines with those equations, and we can see they are not perpendicular.

Lucas says, "In order to be perpendicular, one must have a positive gradient and one must have a negative gradient." He is correct.

Jacob's gone even further and said, "The gradients must be negative reciprocals of each other." We're gonna explore that in a moment.

Lucas says, "These two lines are perpendicular." Yes, one has a gradient of 2 and the other is -1/2.

So let's look at what Jacob was saying in more detail.

Negative reciprocals have a product of -1.

If two lines have gradients which multiply to -1 then they are perpendicular.

You can prove this always works by using Pythagoras.

For today, we're gonna assume that we are already happy with why two lines that have a gradient which multiply to -1 are perpendicular.

So there's some examples, 3 and -1/3, 2/3 and -3/2, 3/4 and -4/3.

I'm gonna use some gradient triangles to show that these are perpendicular.

If I rotate one of the gradient triangles through 90 degrees, I get the gradient triangle for the other line.

And that works for all three examples.

So which of these lines will be perpendicular to the line with equation y = -4x? What do you think? Well done if you said y = 1/4x.

<v ->4 and 1/4 are negative reciprocals.

</v> When we multiply those gradients together, we get -1.

At the moment, we're not worried about the constant.

So it does not matter what constant is added, any line with a gradient of 1/4 will be perpendicular to y = -4x.

Here are some examples.

Let's check then that we're happy with negative reciprocals.

Let's have a look.

The negative reciprocal of -2/5 is 5/2.

The negative reciprocal of 1/5 is -5, and 2/5 and -5/2 are negative reciprocals.

They need to multiply to give you -1.

Another one to have a go at.

Can you match the values to their negative reciprocals? So we should have positive 1/10.

This is an interesting one, 1 and -1 are negative reciprocals.

They're the only two integers which will be negative reciprocals of each other.

And 5 and -1/5.

So sometimes the gradient is not immediately obvious from the equation.

In that case, we can rearrange the equations into the form y = mx + c.

So what would the gradient of any line perpendicular to x + y = 7 be? Well, let's rearrange it.

So we can write this as y = 7 - x.

This has a gradient of -1 because -1 and 1 are negative reciprocals.

Any line which is perpendicular will have a gradient of 1.

So for example, y = 1x + 2 would be perpendicular.

So we found the gradient of the original line, found its negative reciprocal, and that's gonna be the gradient of our perpendicular line.

Sofia says, "I can ignore the constant as I'm only interested in the gradient." Do you agree with her statement? Let's have a look.

So we can rearrange.

It's taking me a couple of steps, and then here I need to divide by 3.

So I've got y = 5/3x.

And then Sofia says, "It doesn't matter about the constant," so let's just ignore it.

And then I can see that a perpendicular line will have a gradient of -3/5.

So yes, Sofia is correct.

We didn't need to work out the constant, which represents the y-intercept, as it's the gradient that determines whether lines are perpendicular.

However, it is a good habit to write the equation in full because it might be that we need the equation for a future calculation, and it's not difficult to write that as -8/3.

Izzy says, "I think the lines y = 2.

3x and y = -3.

2x might be perpendicular." Is Izzy correct? Let's draw them to find out.

No, they are not perpendicular, and that's because 2.

3 and -3.

2 are not negative reciprocals.

When the gradient is given as a decimal, the easiest way to find a reciprocal or a negative reciprocal is to convert it to a fraction.

So 2.

3 is the same as 2 and 3/10, or 23/10, and -3.

2 is the same as -3 and 1/5, or -16/5.

If you compare those values now, they're not negative reciprocals.

So which of these lines are perpendicular to the line with equation y = 4.

5x? Off you go.

Well done if you rewrote that as y = 9/2x.

Then we want the negative reciprocal, so our gradient needs to be -2/9.

So it must be B, 1 - 2/9x.

Have a go at this one.

Which of these lines would be perpendicular to this one? So we need to rearrange.

We get y = -1/3x + 4, and then we need a perpendicular gradient, which will be 3.

Now, you might need to rearrange some of these equations to help you.

That bottom one is the only one with a gradient of 3.

Time for you to have a go then.

I would like you to match up any equations for pairs of lines which are perpendicular to each other.

You might find that there's some lines which are not perpendicular to any of the others.

One has been done for you.

A and F are going to be perpendicular.

You might want to check to see why and then move on and see if you can find any more pairs.

Give that one a go.

Come back when you're ready for the next bit.

So Izzy has drawn a line which goes through the point -1.

52 and -1.

57.

What is the equation of Izzy's line? Can you write down the equations for two different lines perpendicular to Izzy's? When you're happy with that one, I'd like you to explain why the two lines in question 3 are perpendicular and the two lines in question 4 are not perpendicular.

Fantastic.

So we had a little bit of rearranging to do here to find our gradients and then see if there were negative reciprocals.

A and F we've given for you, B and H because B has a gradient of 1 and H has a gradient of -1, C and M 'cause C has a gradient of 4 and M has a gradient of -1/4, D and L 'cause D has a gradient of -2 two and L is 1/2, E and Q, E is 5/6 and Q is -6/5, G and S, G has a gradient of 3, S has a gradient of -1/3.

If you subtract 3x from both sides and then divide by 9, you get y = -1/3x + 6.

I and R, I has a gradient of 5/2 and R is -2/5.

And then K and N.

With a bit of rearranging, you can see that K has a gradient of 4/5 and N is -5/4.

<v ->5x/4 is the same as -5/4x.

</v> Any lines not mentioned there were not perpendicular to any others.

So Izzy's line is actually a vertical line with equation x = -1.

5, or x = -3/2.

This is a vertical line with an undefined gradient, so we can't use negative reciprocals to find a perpendicular line, but to be perpendicular to a vertical line must be a horizontal line.

So anything in the form y = a, where a is a constant, so y = 4, y = -2, y = 0, anything in that form.

The easiest way to show these were perpendicular was to put their gradients as fractions.

So the first one could be written as y = 5/4x - 4, and the second one, y = -4/5x - 4.

And we then need to say that the gradients are negative reciprocals, so the lines are perpendicular.

For 4, you could rearrange and show that the gradients are not negative reciprocals.

However, a quick way is to show that the variables in the first equation have both been multiplied by -1.

This means these lines are actually parallel because the variables have been multiplied by -1 but the constant hasn't.

So what that means is the variables are in the same ratio, so they are parallel.

Well done.

We're gonna now put those skills to the test to find some equations of perpendicular lines.

So we can use our understanding of perpendicular gradients to find equations of perpendicular lines with certain features.

So we are gonna find the equation of a line perpendicular to y = 2/3x + 4 for which passes through the point 6,2.

So the first thing you can identify is the gradient of this line.

That's 2/3.

So the gradient of a perpendicular line will be -3/2.

We've practised that loads now.

So a perpendicular line will have equation y = -3/2x + c, and all we need to do is find that value for c.

We have a coordinate of a point on that line, so we can use that to find c.

We can substitute the x and the y values into our equation so far and then rearrange to find c.

You may have done similar things in the past with parallel lines or finding lines from two coordinate pairs.

Sometimes we want our equation to be in a specific form.

So Izzy says, "Let's write this in the form axe + by = c, where a, b and c are integers." So if we multiply everything by 2, we will no longer have any fractional terms. So 2y = -3x + 22.

And then if we add 3x to both sides, we've got 3x + 2y = 22.

And that's in the form that Izzy wanted it.

Let's give this a go.

I'm gonna show you one on the left-hand side, and then I'd like you to have a go at a similar one on the right-hand side.

We have an equation of our line, and we want a perpendicular line which goes through the point 7, 8.

Well, to start with, we need to know the gradient of our original line.

To do that, I need to rearrange.

If I rearrange this equation, I get 2y = 7x, so y = 7/2x.

To get a perpendicular line, it'll have a gradient of -2/7.

I can then substitute the coordinate 7, 8 in for x and y and rearrange to get c.

My final equation must be y = -2/7x + 10.

You might wanna pause the video and read through those last steps just to see how I've rearranged, and then give this one a go.

Well done.

So we need to start by adding 8 to both sides and then divide in by 5.

So we get y = 1/5x + 8/5.

The constant's not particularly useful for me at the moment, but I am gonna use the gradient.

A perpendicular line will have gradient -5, so y = -5x + c.

Then we can use our coordinate to find our value for c.

And -12 = -5 X 2 + c, so c must be -2.

Final equation, y = -5x - 2.

Let's try another one.

A line L1 goes through the points 7, 6 and 11, 1.

Find an equation for a line which goes through the point 0, 8 and is perpendicular to L1.

Okay, well if it's perpendicular to L1, I need to know the gradient of L1.

I'm gonna use a ratio table to help me.

So to get from 7 to 11, the change in x is plus 4, from 6 to 1 is -5, so our gradient is -5/4.

A perpendicular line then will have gradient 4/5.

Now, I don't need to do any extra work here 'cause the coordinate I've been given is actually the y-intercept.

If it wasn't, you just substitute it into y = mx + c like you did previously, but thankfully I can just write my equation as y = 4/5x + 8.

I'd like you to try exactly the same thing for this one.

So x is increased by 7, y is increased by 21, so our gradient is 3.

The gradient of a perpendicular line then is -1/3.

Don't forget that step.

Luckily the coordinate I was given is the y-intercept.

So that means our line must be y = -1/3x + 1.

And that is the basics of finding the equations of perpendicular lines.

We're just gonna have a little play around now with when the gradients are not rational numbers just to practise some of our surd skills.

So how can we find the gradient of a perpendicular line to y = root 50x? Well, let's do this one together.

The gradient of this line is the square root of 50.

What we need now is the negative reciprocal.

So root 50 X -1 over root 50 is -1.

Any line of the form y = -1 over root 50x + c will be perpendicular.

It's quite a mouthful to say, but it's exactly what we've been doing with our previous questions.

Aah, but Izzy says, "That is not in the simplest form." What does Izzy mean? Can you remember anything from your surd skills? Right.

There's actually two issues with this.

To start with, a surd is in its simplest form when the radicand is an integer with no perfect square factors greater than 1.

So the radicand is the value within the radical symbol, so in this case, 50, so that should be an integer which does not have square number as a factor.

Well, that's not true because 50 has a factor of 25, and 25 is a square number.

So we can write root 50 as root 25 X 2 'cause 50 is 25 X 2, which is the same as root 25 X root 2.

Root 25, of course, is 5, so this can be written as 5 root 2.

So we could write this as y = -1/5 root 2x + c.

But Sofia's got a point.

Her notes on surds say a simplified fraction should not contain a surd in the denominator.

So what we can do is something called rationalising, and it's the process of removing radicals from an expression because we're gonna rationalise the denominator.

That means making sure the denominator does not include a surd.

How we do this with this fraction is we'd multiply the numerator and the denominator by root 2.

By multiplying the numerator and denominator by the same thing, we are getting an equivalent fraction, and you'll see why we choose root 2.

So 1 X by root 2 is root 2, 5 root 2 X by root 2 is the same as 5 X by root 2 squared, when you multiply a value by itself, you square it, or root 2 squared is 2.

So you get root 2 over 5 X 2, or root 2 over 10.

You may want to take some time to practise your surd skills if you want to attempt the trickiest questions in this upcoming task.

So we could write this as y = negative root 2 over 10x + c.

Now, most calculators will write irrational numbers in their simplest form, so we can use them to check.

So if I type into my calculator -1 over root 50, you'll see that it simplifies it for me, and it tells me it's negative root 2 over 10.

It's a good way of checking, particularly at the early stages of developing these skills.

We could go one step further and check that this is a negative reciprocal to our original gradient.

So our original gradient was root 50, we multiply that by negative root 2 over 10.

You can see it's -1, and we know that perpendicular lines have gradients with a product of -1.

So which of these would be perpendicular to the equation y = negative root 8x + 4 root 2? See if you can find them all.

Well done.

If you said that 1 over root 8x would be perpendicular, negative root 8 X by 1 over root 8 gives you -1.

But root 8 is the same as root 4 X 2, which gives us 2 root 2.

If we rationalise the denominator, that gives us root 2/4.

If we multiply the numerator and the denominator by root 2, that gives us root 2/4.

Let's have a practise then.

Most of these questions are not involving those surd skills.

They come into the trickier questions towards the end.

So have a go at these three.

When you're happy with your answers, come back for the next bit.

Don't forget, you can check your answers by substituting back in.

You might want to allow yourself your calculator to support you with that.

Well done.

Another couple for you to have a go at.

Think carefully about your gradients.

You're gonna need to do some rearranging first.

Off you go.

And finally, exactly the same thing, but we've got some irrational gradients this time.

See if you can work out the equations of those lines, even better if you can put them in their simplest form.

Do as much as you can and then come back for the answers.

Well done.

So the first one you get y = -4/3x - 4.

Second one, we're gonna have y = 1/5x - 1.

For 3, we need to find the gradient of L1.

So the gradient of L1 is -4/5, so perpendicular gradient will be 5/4, and then we can substitute our coordinate 12, 4.

Our final answer, y = 5/4x - 11.

Then for 4, we're gonna need to rearrange first to find our gradient.

So our gradient is gonna be 4, for L2, so our perpendicular gradient is -1/4.

Now, we're told that it intersects this line on the y-axis, so we do need to know the y-intercept of L2.

That's fine.

We worked it out before.

The y-intercept is -2, so that means the y-intercept of my new line is -2.

So I get y = -1/4x - 2, but we were asked to write it in a different form.

So we need to multiply everything by 4.

So 4y = -x - 8, and then rearrange so it's equal to 0.

So you get x + 4y + 8 = 0, or you could have had -x - 4y - 8 = 0.

And question 5.

0.

15 can be written as 15/100, which is 3/20.

So a perpendicular line will have gradient -20/3.

And then we just need to do some substitution.

And it was up to you how you wanted to leave your answer.

You could have had it in the form y = -20/3x -40/3, or if you prefer having integer coefficients, you could write that as 20x + 3y + 40 = 0.

It's entirely up to you.

If you want to spend some more time checking your working, pause the video now.

And finally, so our gradient for this first one, we divide everything by 2.

That gives us 3 pi over 4.

Our negative reciprocal then is -4 over 3 pi.

Because it passes through the origin, we have a constant of 0.

y = -4/3 pi x.

For 7A, a perpendicular gradient would be 4 over root 45.

That can be simplified.

We're gonna do that in part B.

So one thing we can do is we can find a square factor of 45.

45 has a factor of 9.

So you can write that as 4 over root 9 X 5, or 4/3 root 5.

Of course, we can rationalise the denominator as well.

If you multiply the numerator and denominator by root 5 you get 4 root 5/15.

And finally, the line passes through the point 60, 20, what is its equation? This was quite tough, so don't worry if you didn't quite get to this final bit.

The fact that you've got some working out so far is showing you've got these skills.

So substitute r coordinate.

So 20 = 4 root 5/15 X 60 + c.

Well, 60 divided by 15 is 4, so that's 4 root 5 X 4 + c, or 16 root 5.

So we've now got 20 = 16 root 5 + c.

So c must be 20 subtract 16 root 5.

We cannot write that in a simpler way, so our final answer is y = 4 root 5/15x + 20 - 16 root 5.

I know that last one was tough.

What we're doing here is we're practising some of our other skills, and we're seeing how we can connect them together.

The more practise you do, the better you get as with everything.

So you should be really, really proud of yourself however far you got through that today.

Let's have a look at what we've learned today then.

We've seen that if two equations are in the form y = mx + c, then you can identify if they're perpendicular from the equations.

We know that the product of the gradients for two perpendicular lines is -1.

What that means is if two gradients are negative reciprocals, then their lines are perpendicular.

We know as well that if equations of lines are not written in the form y = mx + c, they can be rearranged so they are, unless they're vertical lines.

But we know that vertical lines and horizontal lines are perpendicular, so we can use that to help us with those.

Once we know the gradients for perpendicular lines, we can use these to solve linear graph problems involving perpendicular lines.

And we got really stuck in with those at the end of today's lesson.

Thank you for joining us today and for all your hard work, and I look forward to seeing you again.