Lesson video

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 perpendicular linear graphs.

By the end of the lesson, you'll be able to prove the relationship between the gradients of perpendicular lines.

We're gonna do a lot of exploring, a lot of playing around, and see if we can figure things out for ourselves today.

With that in mind, there's a few pieces of equipment you might need.

So you might want to make sure that you've got a ruler and a protractor to hand.

You might also want to make sure you've got some squared paper that you could draw some things on and play around with as we go through.

The two lines are perpendicular if they meet at right angles, and this is what we're exploring today.

As part of our proof, we're also gonna use this phrase if and only if.

So IFF is an abbreviation of if and only if.

This is an implication that goes both ways.

For example, a number is positive if and only if it is greater than zero.

What that means is we can draw out two statements.

A number is positive if it is greater than zero, but also if a number is greater than zero, it is positive.

So to start with, we're gonna have go at drawing some perpendicular lines.

So Jacob has plotted these two lines.

One is horizontal and one is vertical so they make a perfect cross.

What do you think Jacob means by this and how could he improve his statement? Okay, well any two straight lines that are not parallel will intersect somewhere on a grid.

So the fact that they cross is not anything particularly special.

What he's probably referring to is the fact that they meet at a right angle.

A better way to say that would've been, "They intersect at a right angle," or, "They intersect at 90 degrees." Or he could say, "They are perpendicular." And then we understand a lot better what point he's trying to make.

So two lines are perpendicular if they meet at right angles.

What do you think about these lines? Are they perpendicular? And how could we show that? How could we check? Well, Alex says we can use a protractor.

So let's have a quick recap of how to use a protractor to check these are perpendicular or not.

So we want to put our protractor on our axes.

What we want to do is we want the bottom centre of the protractor to be where the two lines intersect.

So in this case, it's on the coordinate 0,2.

We want to make sure that this bottom centre of our protractor is on that point.

Now what we need to do is rotate our protractor so that one of the zero degrees marks, you'll see that there's one on both sides, we just need one of those zero degrees marks to be on one of the lines.

So you'll see that we've perfectly lined up that purple line, that's the one with positive gradient, so that it sits on the zero degree mark on the right-hand side of my protractor.

We can see then this is not 90 degrees if you want to be more exact.

If you read round from the zero degree marker, we are looking at the inside numbers, we can see that this angle is 98 degrees.

So these lines are not perpendicular.

Let's have a look at these two lines.

So this time I put my protractor where the two lines intersect.

And I've lined up the green line, which has got the positive gradient this time, with the zero degree on the right-hand side.

And again I can read round and I can see clearly they intersect at 90 degrees.

Jacob reckons you can do this without a protractor.

You can use the grid lines.

Let's have a look.

So we can use a gradient triangle to show that we can go two right, one up, two right, one up, everywhere along this line.

So it has a gradient of 1/2.

Equally, if you look at the other line, we can go two down, one right, two down, one right.

You might wanna pause this video for a moment and see if you can draw some perpendicular lines on your squared paper by counting the squares.

Don't worry if you don't understand exactly how this works at the moment.

We are gonna look at this in more detail in a moment.

You might wanna just have a play around for a bit and see if you can see a connection.

So we can use the gradient triangle to help us draw perpendicular lines.

So I've drawn a line and I want a perpendicular line so I can draw in a gradient triangle.

So we go three right, two up, and that takes me to the next integer coordinate.

Now, if we rotate that gradient triangle 90 degrees, we can see our perpendicular line will have a slope, which can be described as down three, right two.

So what we've done is rotated that gradient triangle 90 degrees.

It doesn't matter where we put that gradient triangle.

For now, I'm keeping it on the coordinate 0,2.

So that our centre of rotation is 0,2.

It doesn't need to be.

That gradient triangle can actually be moved so it is anywhere on the axis.

'Cause all we want is a perpendicular line, doesn't matter where it intersects our other line.

For now though, let's keep them intersecting at 0,2.

And now we can draw our line in.

Again, you might want to pause the video and have a go at drawing this on your own square paper.

So Lucas says, "These lines should be perpendicular.

One line slopes one square right, three squares up, and the other slopes one square left, three squares up." We can see they're not perpendicular.

What mistake has Lucas made? So these lines are reflections of each other.

If you look at his gradient triangles, they are reflections of each other, not a rotation.

So perpendicular to the purple line, which is the one with positive gradient.

If we rotate that gradient triangle through 90 degrees, we can now see that our line will slope one square down, three squares right.

Then we can draw in our perpendicular line.

How else could Lucas have shown these lines were not perpendicular? So he could have used a protractor.

Just like we did on the previous slides, we could use our protractor to see whether they intersect at 90 degrees.

If he didn't have a protractor, he could use something like a piece of paper, anything with a right angle, as a rough guide.

So time to have a play around with this idea of perpendicular lines.

For each pair of lines, I'd like you to decide whether they are perpendicular or not.

You could use a protractor.

You could count the squares and look at a gradient triangle and think about rotating your gradient triangle.

You might start to see some other patterns emerging as well.

However you wish to do it, decide whether they're perpendicular or not.

Even better if you could convince somebody else that you are definitely right.

Give that a go.

I'd like you to have a go at drawing a line perpendicular to the ones on the grid.

For question three, I'd like you to draw a perpendicular line, but it has to go through the point I've marked with a cross.

When you're happy playing around with those, come back.

And finally, we're gonna bring some of our graph skills in here.

So the line L1 with equation y = 1/2x + 3 is drawn on the axis.

I'd like you to draw a line which is perpendicular and intersects at the marked point.

So that's 0,3 for question four.

Then can you tell me the equation of your line? Think about your gradient and your y-intercept.

Can you write your equation? For question five, the line with equation y = -3/2x + 3 has been drawn on the axis.

I'd like you to draw the line which is perpendicular and passes through the marked point.

So again, count the squares, use your gradient triangle to find your perpendicular line, and then make sure it goes through that point.

When you are happy you've got the right line, can you tell me the equation of your line? And when you're ready, come back for the answers.

Okay, let's see if we got these.

The two lines in A are perpendicular.

B is not.

C, the two lines are not perpendicular.

D, the two lines are not perpendicular.

They look close, but if you count squares, the one with positive gradient goes along five and up two, but the one with negative gradient only goes down four and right two.

The two lines in E are perpendicular.

And the two lines in F are perpendicular.

For A, you could have any line that is parallel to the one that I've drawn.

What you need to check is that your line goes right one, down one, right one, down one, and so on.

It doesn't matter where it intersects the other line, as long as it goes right one, down one.

For B, you are looking for a line that goes right one, down two.

And again it can intersect the other line wherever you want it to.

So as long as it's parallel to mine, it'll be perpendicular to the original line.

For question three, you probably wanted to start on the point that was marked, and then a perpendicular line will go right one, up one, right one, up one.

For B, again, we want to rotate that gradient triangle.

So we need a line that goes right two, up one, right two, up one.

So you'll want to start at the point that's been marked, go right two, up one, mark a new point, and so on, and then make sure your line carries on with that gradient.

For C, the original slope of the first line is right one, up three.

So we rotate that we're going down one, right three, and so on.

So start on the coordinate we've got, down one, right three, down one, right three.

And then the last one was pretty hard to do.

So well done if you got this one.

So you need to find two points on the grid so that we can draw our gradient triangle.

If you do that, you can see that the slope of the line goes right four, up three.

So we need to rotate that.

So we're looking for a gradient, which goes down four, right three.

And again you're gonna want to start on the point that's marked.

So start on the point that's marked, then we do the opposite.

So left three, up four, left three, up four.

You'll see that the two lines do not intersect on a point where two grid lines meet this time.

For question four, you want to check that your line looks like mine.

The equation of the line is y = -2x + 3.

For question five, again, check your line looks like mine.

A key point is that intersects the y-axis at 0,-2.

That means our equation is y = 2/3x - 2.

Well done if you got those last two.

So now we're gonna talk in depth about the gradients of perpendicular lines.

So we know that the gradients of parallel lines are the same, but there's also a relationship between the gradients of perpendicular lines.

Lucas says, "When lines are perpendicular, one has a positive gradient and the other a negative gradient." Alex says, "So two lines with positive gradients cannot be perpendicular." What do you think about these statements? So Lucas is correct if we ignore vertical lines.

Vertical lines have an undefined gradient.

So if we ignore those lines, then if we had two perpendicular lines, one would have to have a positive gradient and one would have to have a negative gradient.

And that means Alex is perfectly correct.

It's not possible for two lines with positive gradients or two lines with negative gradients to be perpendicular.

Jacob says, "So would the lines with gradients 5 and -5 be perpendicular?" Well, we're gonna draw y = 5x and y = -5x to check.

We're not interested in the y-intercept at the moment.

We're just gonna have a look at the gradients.

Okay, those are definitely not perpendicular.

So we can't just multiply a gradient by -1 to get a perpendicular gradient.

There's something else happening here.

So let's look at some examples we've already come across and see if we can spot a pattern.

So that was one of our previous questions.

We've got a gradient of -2 and a gradient of 1/2, and they're perpendicular.

We've got a gradient of 3 and a gradient of -1/3, and they're perpendicular.

Here we had a gradient of 2/3 and one of -3/2, and they were perpendicular.

And here we have a gradient of 3/4 and -4/3, and they were perpendicular.

Can you describe the pattern? Is it something that will always work? How does that link to what you've seen from gradient so far? Pause the video and have a think.

So Jacob says, "The absolute values are reciprocals of each other." You might have said something similar.

Or you might have said, "If you ignore the negative, the values are reciprocals of each other." 2 and 1/2 are reciprocals, 3 and 1/3 are reciprocals, 2/3 and 3/2 are reciprocals.

However, we know that one has to be positive and one has to be negative.

Lucas told us that earlier.

So that's why Jacob has says the absolute values are reciprocals of each other.

So we've already determined that one gradient has to be positive and the other negative.

If you think about how we drew the gradient triangles, so, for example, if one slope was along one, up two, the other would be along two, up one in the negative direction, and that is how you end up with this reciprocal.

In order to get the gradient, you essentially divide them the other way around.

And, of course, one has to be positive and one has to be negative.

So what we can say is that the gradients of perpendicular lines are negative reciprocals of each other.

If reciprocals are values that multiply to one, then what do you think negative reciprocals are gonna multiply to? Have a think.

Right, negative reciprocals are values that multiply to -1.

So if two numbers are negative reciprocals, then their product is -1.

So for example, 3 and -1/3 are negative reciprocals because 3 X -1/3 is -1.

3/5 and -5/3 are negative reciprocals of each other 'cause 3/5 X -5/3 is -1.

Let's summarise what we've seen so far and make sure we're using that mathematical language.

So we have seen that two lines are perpendicular only if the product of their gradients is -1.

But also the products of the gradients of two lines is -1 only if they are perpendicular.

There is this direct link between two lines being perpendicular and the fact that their gradients multiplied to -1.

Now we're gonna return to this idea of if and only if.

So some relationships in mathematics only hold true in one direction and some work in both directions.

We can use the phrase if and only if to describe relationships with work in both directions, and this is one of those cases.

So we are now gonna prove that two lines are perpendicular if and only if the product of their gradients is -1.

So far, we've played around with a few different gradients.

We've started to spot a pattern between gradients.

We've made this statement that the gradients of two perpendicular lines multiply to -1.

And we've tried to justify it by using our gradient triangle and how we rotate our gradient triangle and how that impacts our gradient.

However, it's a really useful skill to be able to prove something definitely works for all cases, and that's what we're gonna do now.

Now this proof does have some complicated elements to it, so I really recommend that you take your time watching it through, see if you can get your head around what we're doing.

By all means, draw some diagrams. But don't panic if you don't understand every element in detail at this stage.

But by proving that this always works, it's then something that we can use without having to prove it every single time.

Two lines are perpendicular if and only if the product of their gradient is -1.

We're gonna start with if two lines of perpendicular, the product of their gradient is -1.

So we're gonna assume that two non-vertical lines called L1 and L2 are perpendicular.

So we're starting with the assumption that they're perpendicular, and we're gonna prove that means their gradients have to multiply to -1.

We're gonna ignore vertical lines at the moment 'cause their gradients are undefined.

Okay, so L1 is gonna have equation y = m1x + c1.

If this was the equation of our line, it would have gradient m1.

The other line then will say y = m2x + c2.

So this would have gradient m2.

Because we're focusing on the gradients, the constants are not going to matter for this proof.

Okay.

Well, we know that these two lines are perpendicular.

That's our starting point.

That means there's gonna be a point where these intersect at 90 degrees.

We're gonna call that point P.

I'm just gonna sketch a quick diagram to see what we've got.

So you've got L1, we've got L2, we've got point P where they intersect.

For the purposes of this proof, I've gone with L1 as the positive gradient and L2 as the negative gradient.

You might wanna do the same so your proof lines up with mine.

Okay, well we can define another point Q, which is also on L1, but which is a step of one in the positive x direction.

So if point P was on the line, if we go right one and then back up to my line, there's gonna be a new coordinate Q.

Now that coordinate Q is gonna be one away horizontally, and then the vertical distance is gonna be the gradient, isn't it, 'cause that's what gradient is.

It's the change in y when x increases by one.

So there's a new point Q there and the diagram is gonna help you here.

Now we can define a third point R, which is on L2 with the same idea.

We're gonna take a step of one in the positive x direction and then we're gonna draw a line down to reach our original line, and there's our point R.

Now the important thing about R is that m2 must be a negative number 'cause that gradient is gonna be negative.

If those two lines meet to a right angle, we can see clearly that that's gonna have to be negative.

That's gonna be important in a future step that m2 is a negative number.

Okay, so there's the diagram of what we've got so far.

What we're gonna use is Pythagoras's theorem.

So if we use Pythagoras' theorem, the length of PQ, because we've got a right angle gradient triangle, haven't we? So the length of PQ is gonna be 1 squared + m1 squared and square rooted.

PR is gonna be 1 squared + m2 squared and square rooted.

Doesn't matter that m2 is negative.

'Cause when you square a negative number, you get a positive number.

Now this is where you've gotta pay attention to your negative numbers.

The length QR is m1 - m2.

Say that m1 was 1 and m2 was -2, the length of that line would be 1 - -2, or 3, okay? So we're subtracting m1 and m2, not adding, because m2 is a negative value.

If you want to pause the video and just see if you can work out why that is the case, do so now.

Just try it with a couple of numbers.

Okay, but remember we started this by saying that these two lines were perpendicular.

Because these two lines meet at a right angle, PQR is a right angle triangle, that means Pythagoras' theorem will hold for the side length, PQ, PR, and QR.

So that means squaring PR, squaring PR, and adding them together will get you the square of QR.

So let's use Pythagoras's theorem then on this triangle.

You might wanna make sure you've got a picture of the triangle to hand as we do this bit.

So we're gonna square that value we had for PQ, and we're gonna square the value we had for PR, add them together, and that's gonna give us the square of QR, which was our vertical line.

Now if you think about what we have for that first section, if you square root something and then square it, you're gonna get back to what you started with.

So 1 squared + m1 squared, square root and square gives you 1 squared + m1 squared.

The same happens when we square the length of PR.

We get the square root of 1 squared + m2 squared, all squared, which gives us 1 squared + m2 squared.

Just pause and make sure you're happy with that so far and then we'll look at the right-hand side.

Now the right hand side is the product of two binomials.

It's m1 - m2 X m1 - m2.

If we multiply every term by every term, that gives us m1 squared - m1m2, minus the product of m1 and m2 again, + m2 all squared.

Now we can simplify.

So 1 squared is 1.

So I don't need to keep writing that as 1 squared.

So the left-hand side is m1 squared + m2 squared + 2, and the right-hand side is m1 squared + m2 squared - 2 lots of the product of m1 and m2.

So 2m1m2 Now we've got m1 squared + m2 squared on both sides of my equation.

So I can simplify this by subtracting those from both sides.

So that leaves me with 2 = -2 lots of m1m2.

If I divide by 2, I get 1 = -1 lots of m1m2.

If I multiply those by -1, I can see that m1m2, which is m1 X m2, = -1.

Hence, with a little bit of effort, we have shown that if L1 and L2 are perpendicular, the products of their gradients, m1 and m2, is -1.

That last statement 12 is saying that multiplying m1 and m2 gives us -1.

Now that was a lot of hard work there.

It's really good that we are able to formally prove that that always works 'cause then we can always use it.

Now we said that this implication goes both ways.

So that means if the product of the gradients of two lines is -1, they are perpendicular.

So we can prove this going in the other direction.

So we can use a similar diagram.

We've done most of the hard work.

The only thing that's different is that we don't know that the lines are perpendicular.

That's what we're trying to prove.

So we can't start by saying that they're perpendicular and therefore they meet at a right angle.

Instead, we can start with the fact that their gradients multiply to -1.

So we've got our same diagram, but this time, we're gonna assume that m1m2 = -1, but we don't know that they're perpendicular.

As before, we can still use Pythagoras' to get the length of PQ and PR because our gradient triangles are right angle triangles.

Just as before, the length of QR is m1 -m2, where m2 is a negative number.

Right, we don't know that this is a right angle triangle, so we need to show that it's a right angle triangle.

If Pythagoras' theorem holds, then this must be a right angle triangle.

So we're going to use the side length.

And if Pythagoras' theorem does work for the side lengths, then we know that this is a right angle triangle.

Pythagoras' theorem is another implication that goes both ways.

If the square of the two smaller sides sums to the square of the hypotenuse, then it's a right angle triangle.

But also if it's a right angle triangle, the squares of the two smaller sides sums to the square of the hypotenuse.

It works for all right angle triangles and it only works for right angle triangles.

So we're gonna show that PQ squared = QR squared - PR squared.

That's how we're gonna do this.

So we're gonna do QR squared - PR squared.

And if we end up with PQ squared, that must be a right angle.

Okay, so QR squared - PR squared, just like before, m1 - m2 all squared is the product of two binomials.

And by squaring something that we've already square rooted, we get back to what we started with, which is 1 squared + m2 squared.

Now because we're subtracting, we need to make sure we're subtracting the whole of 1 squared + m2 all squared, which is why I've put that in brackets.

Little bit simplifying then, I've got m1 squared + m2 squared - 2m1m2, like we did previously.

Then I'm subtracting 1 and subtracting m2 squared.

Pause a video if you need to catch up with what we've done there.

Then for eight, I've just simplified because I've got m2 squared and -m2 squared, and they are a zero pair.

When I add them, I get zero.

So that means I'm left with m1 squared - 2m1m2 - 1.

Now, remember, we started off by assuming that m1m2 is -1, and we haven't used that yet.

We can use that here.

So we can substitute that into our expression.

So we've got m1 squared - 2 lots of -1 - 1.

So that's m1 squared + 1.

Well, hang on a minute.

That is PQ squared.

That is the length of PQ all squared.

So that means Pythagoras's theorem holds.

Well, if Pythagoras' theorem holds, it must be a right angle triangle an L1 and L2 are perpendicular.

Hence, we have now shown that if the product of the gradients of two lines is -1, they are perpendicular lines.

Do not worry if there's bits of that that were a little bit confusing.

As we said with the other proof, doesn't matter if you can't replicate that perfectly at the moment, but the fact that we've shown you that it always works means you can now use this fact.

Okay, quick check of the important bits.

What is the product of two values which are negative reciprocals of each other? What do you think? Well done.

Negative reciprocals multiply to -1.

Which of these is the correct expansion of m1 - m2 all squared? What do you think? Well, I don't know if you notice it's that bottom one, <v ->m2 X -m2 + m2 squared.

</v> Okay, time to have a practise with these things we've talked about today.

So for question one, I'd like you to calculate the gradient of both lines and then tell me in which questions are the lines perpendicular.

Give that a go.

Well done.

For question three, we're having a look at that first proof that we looked at together.

The if two lines of perpendicular, the product of their gradient is -1.

I want you to see if you can put the elements to this proof in order so that it works as a succinct proof.

There's a diagram in the top right-hand corner to help you.

Give that a go.

Well done.

So the gradient of L1 is 2 and the gradient of L2 is -1/2.

They are perpendicular.

Their gradients are negative reciprocals 2 X -1/2 is -1.

The gradient of L3 is 2/3.

The gradient of L4 is -3/2.

They are negative reciprocals of each other.

2/3 X -3/2 is -1.

For C, the gradient of L5 is four.

The gradient of L6 is -1/5.

They're not perpendicular.

They're not negative reciprocals.

4 X -1/5 is not -1.

Let's look at this proof then.

So we've got F, then C, then B, then K, then J.

Pause the video if you want to make sure you've got those right so far.

Then D, then A, then E, then H, then G, and then we need our summary statement, therefore, if L1 and L2 are perpendicular, the product of their gradient is -1.

Pause the video again if you need to to check the second half of that proof.

Fantastic work.

Especially with that last bit, guys.

There was lots of algebra skills being pulled together there.

So today we've worked with this idea that two lines are perpendicular if they meet at right angles.

We've seen that we can check whether they're perpendicular using a protractor, or by using a gradient triangle rotated 90 degrees.

Then we explored how the products of the gradients of two perpendicular lines is -1.

And then we proved that using Pythagoras' theorem, we played around with this proof that two lines of perpendicular if and only if the products of their gradients is -1.

There are some super exciting maths there today, guys, and you've really pushed the limits of the things that we can do with our algebra skills and you are showing some really high-level mathematical reasoning there.

So well done.

Thank you for joining us today, and I look forward to seeing you again.