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Hello there.

Welcome and thank you for joining us for today's lesson.

My name is Ms. Davis, 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 today's very exciting lesson on parallel and perpendicular lines on coordinate axes.

For today's lesson, you are going to need a compass and you're gonna need a ruler and a pencil.

You might wanna make sure you've got some paper, maybe even square paper to hand to help you.

We're gonna look at some problems involving parallel and perpendicular lines on a coordinate axes.

If you're not sure about the definitions of perpendicular and parallel, have a read of those now.

So we're gonna get started straight away with constructing some perpendiculars.

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

So this graph shows a line segment.

It's a line segment, because it has a start and an end.

This segment joins the points with coordinates 0, 3 and 4, 11.

M is exactly halfway along the line.

It has coordinates 2, 7.

How can we find an equation of the perpendicular bisector of this line segment? Pause the video and just have a think.

What is a perpendicular bisector? How we're gonna find the equation do you think? Right, so a perpendicular bisector cuts it exactly in half and at a right angle.

Perpendicular means lines that meet at a right angle.

A bisector cuts something in half.

So we're looking for an equation of the line perpendicular to that line segment which passes through its midpoint.

If it needs to cut it in half, it needs to go through its midpoint.

So let's start by finding the gradient of PQ.

The gradient of PQ is 2, so a perpendicular line will have gradient negative a half.

That means an equation of a perpendicular line will be in the form y equals negative a half x plus c.

But of course we said it must bisect the line segment, so it must pass through the middle of PQ, and that has coordinates 2, 7.

So our line must go through the point with coordinates 2, 7.

So we can substitute them for x and y in our equation.

Seven equals negative a half, multiply by 2 plus c, so c must be 8.

We can write our equation then as y equals negative a half x plus 8.

We could sketch that and see if it looks right.

That's gonna be our y intercept, sketch negative a half as a gradient.

And that looks about right.

So how else could we find the perpendicular bisector of this line segment? We can draw it using our construction skills.

Sam remembers this.

We can use a pair of compasses and a ruler, so make sure you have those ready.

And we're gonna do a bit of a practise.

So let's remind ourselves how to construct a perpendicular bisector.

Doesn't matter if you've forgotten, we're gonna do it together.

So remember, we're trying to construct a line perpendicular to our line segment, which goes through the midpoint.

So I'm gonna construct a perpendicular bisector of a 5.

5 centimetre line.

You're gonna have go doing the same with a 4.

2 centimetre line, and we're gonna do them at the same time.

So for my one, I'm gonna draw a line which is 5.

5 centimetres long.

I'm gonna draw it anywhere on my page for now, just to practise.

Can you do the same on your page? Draw a line which is 4.

2 centimetres long.

Very good.

Now I'm gonna get my compass.

And you need to make sure that the width of your compass is over half the length of AB.

So stretch your compass, put the needle of your compass on one end of the line, and then stretch it so the pencil is over halfway.

Then we're gonna draw two arcs, one each side of AB.

There's one there and one there.

If you'd rather draw in the whole semicircle, that's absolutely fine, but sometimes it can get confusing where all your lines are.

Right.

Have a go at doing that on your one.

Good.

You should have two arcs at the moment.

Now we're gonna move the needle of the compass to the other end of our line segment, and you need to make sure you don't change the width of your compass.

It needs to stay the same as it was when we drew our arcs previously.

And again, we're gonna draw an arc at the top and an arc at the bottom.

And it's important that our arcs intersect with our previous arcs.

See if you can do the same on yours.

Hopefully your diagram looks like this.

So now drawing a line segment between the two intersections creates a perpendicular bisector.

Make sure you're using a ruler and that you line those up as accurately as you can.

If you've done it correct, that should be perpendicular.

Try it yourself.

Right.

So hopefully that is perpendicular.

You could use a protractor or you could use the corner of a piece of paper, anything with a right angle to check that it's roughly correct, and we should have split our line exactly in half.

So for my one, that means half should be 2.

75.

Try it yourself.

Your one, half should be 2.

1.

Measure it and see if it is.

There we go.

If you'd like to practise that further, all you need to do is draw yourself some line segments on your page and practise constructing the perpendicular bisectors.

Right.

Let's apply that now to our line segments.

So just like before, put the needle of our compass on one of the ends.

Make sure the compass is over halfway.

Draw some arcs, swap the needle of your compass to the other side and draw some arcs.

Where they intersect, we're gonna join up with a line.

Now, of course, because this is a line rather than a line segment, it should go across the entire axis.

So our line then has gradient negative a half, and y intercept 0, 8.

So it has the same equation that we worked out algebraically before.

Which method did you prefer, working out algebraically or constructing it with a compass and then finding the gradient and the y intercept? What do you think? Let's see what some of our pupils thought.

Sam said, "I found when I did the constructions, my line was not perfectly perpendicular.

I think algebraically is better." Now, that is true.

It's hard to be completely accurate when you are doing things by hand.

Laura says, "But it was nearly perfect.

We gave you a rough idea of where the gradient and the y intercept would be." That's absolutely true, Laura.

Even if you can't get a perfect perpendicular bisector to doing it by hand, you're gonna be close enough to the gradient and the y intercept to get an idea of where you should be.

Andeep says, "I find it hard to remember that perpendicular gradients are negative reciprocals.

Constructing the line helps me see that." Yeah, it's very easy to forget that perpendicular gradients have to have the opposite sign to each other and have to be reciprocals of each other.

So by constructing the line, it reminds you of that fact.

And Aisha says, "Finding the equation algebraically is more precise, but I like that I can see this works and check with my construction skills." Pause the video if you wanna have a think now about what is your favourite method.

Okay, so we can use our construction skills to draw perpendiculars, which go through any point on the axis.

So here's a line segment, joining the points P and Q, and I've got A, which is -3, 7.

So let's see if we can construct this perpendicular then that goes through A.

So start by drawing a circle or an arc with centre A that needs to intercept PQ at two points.

I'm just gonna draw an arc, because the whole circle is gonna go off my axes and is gonna get in the way.

So I'm drawing a arc and it's crossing my line segment at two clear points.

So you might wanna mark these points with a little cross to know what we're working with.

It doesn't matter that it's hovering in the middle of a grid square, that's perfectly acceptable.

In fact, that's likely to happen.

Right, now I can place my compass needle on one of the intersection points.

You might need to change the width so it's over halfway.

And then we're gonna do exactly the same as what we did with our perpendicular bisectors.

Draw our arcs from one side, move the compass needle to the other side, and draw our arcs again, and then draw in the perpendicular.

If we've done it correctly, our line should go through A, and it should go through the two intersection points.

Now let's find the equation of the line.

Pause the video and give this a go.

Now if we haven't done this exactly, it's gonna be quite tough.

It looks to have roughly a gradient of -2, but it's not exactly -2, not the way I've drawn it anyway.

In fact, this is gonna have a gradient of -5 over 3.

If you look at our y intercept, we've got a y intercept of 2.

So our equation will be y equals -5 over 3 x, plus 2.

Let's check this by looking at our gradient triangles.

We can see that they are rotations of each other and the gradient of PQ is 3 over 5.

Our gradients then are negative reciprocals.

We can also check that the point -3, 7 is on this line.

Yes it is.

Okay, a bit of a recap then on some construction skills.

Here is step one of an attempt to construct a perpendicular of AB through E.

So I want a line perpendicular to AB, but it has to go through point E.

I've started by drawing a circle with centre E.

And this is the next step.

So the pair of compasses has been set ready for the second step.

Why will this not work? What do you think? It is the fact that the compass width is too short.

When constructing the arcs, the width of the compass should be over half the diameter of the circle, otherwise they won't intersect.

Right, which of these circles will help in the construction of the perpendicular to QP through point Z? What do you think? Well done if you said c, a does not intersect QP at all and b only intersects it at one point.

We need two intersection points, so that we can create our arcs.

So which of these is the equation of the perpendicular to the line segment PQ through the point 2, 4? Well done if you said, it's y equals x plus 2.

We definitely need a positive gradient to be perpendicular to PQ, and we need the equation to work for the coordinate 2, 4, y equal x plus 2 is the only one of those that works for that coordinate.

Time to have a go then.

I'd like you to read these questions carefully.

Sometimes you are being asked to work out the equation of the perpendicular line algebraically and then draw the perpendicular.

Sometimes you're being asked to construct the perpendiculars first to work out the equation of the line and then do it algebraically.

So read those questions carefully and give this a go.

And there we have question 2.

And now have a go at question 3.

For question 4, I'd like you to construct the perpendicular for PQ using your pair of compasses first.

Your perpendicular needs to go through point A.

And then I'd like you to find the equation of your constructed line being careful to look at the axis so you get the right gradient.

Question c is gonna require a little bit of thinking.

See if you can explain what is happening with c.

Off you go.

Let's have a look.

So the equation should be y equals 3x plus 7.

Hopefully you manage to construct a perpendicular bisector and indeed it is y equals 3x plus 7.

We have a gradient of 3 and a y intercept of 0, 7.

For this one, we're gonna construct the perpendicular bisector first, and then we need to find the equation of this.

If you've done it perfectly, you'll get a gradient of -3 over 2 and a y intercept of 12.

If you're slightly off, that is to be expected sometimes when we're drawing things by hand.

The midpoint of PQ is 4, 6.

So we're gonna show that that is on our bisector.

So 6 equals -3 over 2, times 4 plus 12.

And that is true.

So the gradient of PQ is two thirds, therefore it must be perpendicular to our line, because the gradients have a product of -1.

And question 3, if we work out the equation of the line, the gradient is 4, the perpendicular gradient is negative a quarter, and it needs to go through 4, 4.

So it's y equals negative a quarter x plus 5.

Now let's have a go at constructing.

Does your line have the same equation as your answer to a? If you've done it perfectly, it will do.

It will have gradient of negative a quarter and y intercept of 5.

If it doesn't, it's probably a little bit of human error with you drawing your perpendiculars.

And finally, if you construct the line, you should have an equation, something like y equals 25 over 2 x, minus 25.

Now P is a segment of the line with equation y equals -2x plus 10.

Our equation had gradient 25 over 2.

That's not a negative reciprocal of -2.

The problem being the scales on the axes were not in the ratio of one to one.

The correct equation for a perpendicular to y equals -2x plus 10 through the point 4, 25 would be y equals a half x plus 23.

If I drew this using technology, you can see that they do not look perpendicular at all, and that is, because I've skewed the scale.

The Y-axis goes up by five.

The X-axis only goes up in steps of one.

If I change the scale, so that they both go up in steps of five, so it's a ratio of one to one, you'll see that they do now look perpendicular.

It's just something to look out for when you're working out perpendicular lines on axes that have been skewed.

Right.

Now we're gonna play around with some shape problems. So we can use our parallel and perpendicular line skills to identify shapes drawn on axes.

So how can we tell if this shape is a parallelogram? Well, parallelograms have opposite sides that are parallel.

So all we need to do is calculate the gradients of opposite sides, and see if they are the same.

What could we do to tell if this shape is a rectangle? Well, all angles are right angles in a rectangle, so we could calculate the gradients of adjacent sides and see if they're perpendicular.

So let's try that.

Parallel lines have the same gradient.

So let's calculate the gradient of AB.

So our gradient is 2.

Now let's try the opposite side, which is CD.

Again, our gradient is 2.

The gradients are equal, so the sides are parallel.

We can do exactly the same for the other pair of sides.

The gradient of BC is going to be negative a half, and the gradient of AD is also negative a half.

So this quadrilateral is a parallelogram as it has two pairs of parallel sides.

Why do we not need to do any more work to show that this could be classified as a rectangle? What do you think? We've already calculated the gradients of all 4 sides.

So all we need to do is show that adjacent sides are perpendicular.

Well, let's look at those gradients again.

AB was 2, BC was negative a half.

They multiply to -1.

So they are perpendicular.

That means angle ABC is 90 degrees.

Now we've already shown that AB and CD are parallel, so we also know that the other angles are 90 degrees.

Equally, you could return to the gradients and show that BC is perpendicular to CD, and CD is perpendicular to AD, and so on.

Now this shape is actually a square.

How could we prove this? Any ideas? Pause the video.

Have a think.

Right, so a square is a rectangle with all 4 sides the same length.

So you could use Pythagoras' theorem to work out the length of all four sides and show that they are the same.

Absolutely nothing wrong with that.

However, the diagonals of a square intersect at right angles.

So we can show that the diagonals are perpendicular by calculating the gradients.

So this is a useful fact to know.

The diagonals of a rhombus, including squares, are perpendicular to each other.

They also bisect each other.

So for any rhombus, the diagonals are perpendicular and they bisect each other.

So we can use this fact to show when a parallelogram is actually a rhombus or when a rectangle is actually a square.

So let's try it.

Let's calculate the gradient of the diagonal BD, that has gradient -3.

And let's calculate the gradient at the diagonal AC, and that's a third.

The diagonals have gradients which are negative reciprocals, therefore they're perpendicular.

If we hadn't done any other work, we'd have to go one step further and check that they bisected each other.

However, we've already shown that this is a rectangle.

So the fact that the diagonals meet at right angle, mean it must be a square.

So how can we show that this quadrilateral is a parallelogram? What do you think? Perfect.

We need to show the opposite sides are parallel.

Let's give it a go.

So I don't need to draw the lines to work out the gradient.

I can just use the coordinates.

What we can actually do is find the change in Y and divide it by the change in X.

So we look at how much the Y values have increased by.

9 subtract 3 is 6.

And then look at what the X values have increased by.

7 subtract 3 is 4.

So our gradient is gonna be 6 divided by 4, change in Y divided by the change in X.

Now that's a nice quick way to calculate gradient, but you've gotta be really careful with negative values getting everything the right way round.

So by all means, use the ratio table to support you.

So let's look at the gradient of DC.

So the change in Y is 6, the change in X is 4.

So we have a gradient of 3 over 2.

Therefore they are parallel.

Right, your turn to have a go.

What is the gradient of BC? It's -2 over 3, 9 subtract 7 is 2, 7 subtract 10 is -3, 2 divided by -3 is -2 over 3.

Why are BC and AD parallel? Of course, it's 'cause their gradients are the same.

So what more do we now need to demonstrate to show this is a rectangle? Right.

We've already shown the opposite sides are parallel.

To show that it's therefore a rectangle, we need to show that adjacent sides are perpendicular.

Okay, well they're the 4 gradients we have calculated.

So 3 over 2, multiplied by -2 over 3 is -1.

Therefore AB and BC are perpendicular.

BC and CD are perpendicular.

CD and AD are perpendicular.

Hence this must be a rectangle.

So what can we do to show this is not a square? We could show them that the diagonals are not perpendicular.

Let's have a look.

The gradient of BD is 8.

The gradient of AC is four sevenths.

The gradients of the diagonals are not negative reciprocals, so they do not intersect at right angles.

This is therefore a rectangle as previously shown, but not a square.

Okay? Aisha has drawn the diagonals onto this shape.

The diagonals of this shape are perpendicular to each other, but this is not a rhombus.

What shape has Aisha drawn? Why is it Aisha's statement correct? Right.

Aisha has drawn a kite.

This is something to be aware of.

The diagonals do meet at a right angle, but this is not a rhombus, and that's because there was that extra bit.

For a rhombus, the diagonals meet at a right angle and they bisect each other.

So the diagonals are essentially perpendicular bisectors.

If the opposite sides were parallel, then the diagonals would bisect each other.

So we're absolutely fine to show the opposite sides are parallel and the diagonals meet at a right angle in order for something to be a rhombus.

But if we haven't shown that the sides are parallel, we would also need to show that the perpendiculars bisect each other.

There's quite a few interesting things about shapes coming out here.

You might wanna just spend some time sketching some shapes and having a look at those diagonals to help you.

Okay? Time for you to have a go.

I'd like you to prove the shape is a parallelogram and then prove it's not a rectangle.

By prove, you need to make sure you're showing all your steps clearly and that you are really explaining what it is you are doing.

Off you go, Right.

Is shape ABDC a rhombus, and how do you know? And then I'd like you to move point B to point E, which has coordinates of 2, 4, and tell me what shape you have now made.

Give that one a go.

Over to you now.

Can you prove this shape is a square? Off you go.

And finally, Andeep thinks this shape is a rhombus.

I'd like you to show that Andeep is incorrect.

Give it a go.

Let's have a look.

So to prove it's a parallelogram, we need to prove it has two pairs of parallel sides.

So calculate the gradients in any way you like.

You have gradients of four sevenths and gradients of -8.

So then showing that this is not a rectangle becomes easy, 'cause four sevenths multiplied by -8 is not -1.

So adjacent size are not perpendicular.

Okay, is it a rhombus? How do you know? Well, the quickest way, being we've already proven it's a parallelogram, is to check the gradients of the diagonals.

The diagonals are perpendicular.

So this is a rhombus, and that's because we've already proved it's a parallelogram.

If we hadn't already proved it's a parallelogram, we would need to show that the diagonals also bisect each other.

To do that, you need to calculate the midpoints of the diagonals and show it's the point where they intersect, which requires a bit more work, and being we've already proven that this is a parallelogram, it's not necessary.

If point B was moved to 2, 4, well, I've drawn the diagonals on to help me with this, 'cause you can see that 2, 4 is still on the same diagonal.

So that means this will be a kite.

The diagonals will still intersect each other at right angles, but they'll no longer bisect each other.

So it is not a rhombus anymore.

It is now a kite.

So to prove it is a square, you can do this in many ways.

I think the quickest way is to show that it's a rectangle and then to show that the diagonals are perpendicular.

Now you can show that all the sides are the same length using Pythagoras, and then show that adjacent sides are perpendicular.

That would work as well.

We're gonna do it this way.

So those are my 4 gradients.

I've shown the opposite sides are parallel and adjacent sides are perpendicular, as the gradients are negative reciprocals.

Then to show it is a square, I can calculate the gradients of the diagonals and show that they are perpendicular.

You might wanna pause and read through that in full.

And finally, there are many ways to show that it is not a rhombus.

The quickest way is to look at the gradients of XW and UV and show that they are not parallel.

So the gradients of opposite sides are not the same, therefore not parallel.

This cannot be rhombus, 'cause it's not a parallelogram.

Well done, there was loads of concepts that we're bringing together there in that lesson.

We've talked a lot about parallel gradients and perpendicular gradients.

We know how to find the equations of perpendicular lines, and we've also played around with different shapes and how to show what a shape on a coordinate axes actually is using its key features.

I know there were some tough bits in that lesson, but I hope you had fun exploring some of that and that you've learned something today.

I look forward to seeing you again.