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 problem solving lesson with linear graphs.

We're gonna use our knowledge of linear graphs to solve some problems. We're gonna start by talking about midpoints.

So the midpoint of a line segment is halfway between the end points.

So let's start by looking at how we can calculate the midpoint of a line segment.

So a line segment is a part of a line with a fixed start point and a distinct fixed end point.

So this is the line segment AB, and, in this case, it's a section of the line y = 5, but it's got a distinct start point and end point.

So this line segment connects the points with coordinates 2,5 and 6,5.

This means it has a length of 4.

You can see the horizontal distance is 4.

It moves from 2 on the x-axis to 6 on the x-axis.

If I wanted to find the midpoint, I need halfway between the two end points.

So what we need to do is halve the distance between A and B, which would be 2 in this case.

So the midpoint needs to have a distance of 2 from the end of the line segment.

That means the midpoint is gonna have coordinate 4,5.

Check that you're happy with that before we move on.

Okay, so this time, I've taken the grid squares out.

How could we use the coordinates to find the midpoint of the line segment CD? Well, C has an x-coordinate of 2, and D has an x-coordinate of 10.

The y-coordinates are the same 'cause they're a horizontal line.

So we need the middle of 2 and 10.

So they are 8 apart horizontally.

If we halve that, we get 4.

So there's a horizontal distance of 4 between the end points and the midpoint.

Well, if we add 4 on to 2, we get 6.

Or if we take 4 off from 10, we get 6.

So the middle of 2,1 and 10,1 is gonna be 6,1.

Notice that the y-coordinates are all the same because it's on a horizontal line, and the x-coordinate is halfway between the x-coordinates of the end points.

We can do exactly the same for vertical line segments.

So let's say P is the midpoint of the line segment of AB.

See if you can work out what the coordinates are gonna be.

Now you might be able to see this just by looking, but let's have a look at our method.

So the x-coordinates are the same.

So we need halfway between the y-coordinates.

So it has a distance of 6 between 1 and 7.

If we halve that, we get 3.

So P must have coordinates 2,4.

I've added 3 onto the y-coordinate of B.

Or you could have taken 3 off the y-coordinate of A.

You just need to make sure that it's the same distance from point A as it is from point B.

So it's three squares away from point A and three squares away from point B.

Right, which coordinate pair is the midpoint of that line segment? What do you think? Well done if you said 5,7.

We've got a distance of 6.

Half the distance is 3.

Then you can add that onto your x-coordinate.

Okay, which coordinate pair is the midpoint of this line segment? What do you think? So it's gonna have a non-integer value this time.

The distance is 5.

Halving gets us 2.

5.

So it must be 3,4.

5.

Notice that we're looking for a y value which is halfway between 2 and 7.

Our x values are all the same because they're on a vertical line.

So the line segment for which we're finding the midpoint will not always be vertical or horizontal.

A lot of the time we're gonna be looking at diagonal lines.

So let's see if we can work out the midpoint of AB this time.

Well, we can start by looking at the horizontal and vertical distances.

So to get from an x-coordinate of 1 to an x-coordinate of 5, that's a distance of 4.

To get from a y-coordinate of 2 to a y-coordinate of 8, that's a distance of 6.

Now if we want the midpoint, we need to halve those distances.

So I need a horizontal distance of 2 and a vertical distance of 3.

And I can check that's the same either side of the midpoint.

It should be identical both sides of the midpoint.

You can see that we've got identical triangles.

Now all you need to do is add the horizontal distance on to your x-coordinate and the vertical distance on to your y-coordinate.

So we get the coordinate 3,5.

Now that's a lot easier 'cause we've drawn this on a coordinate grid.

I just want you to have a look at the coordinates 1,2 and 5,8 and then the midpoint 3,5.

What do you notice about their values? Can we do this without drawing it on a grid? Right, well, the x-coordinate of the midpoint, which is 3, is halfway between the x-coordinates of the end point.

Halfway between 1 and 5 is 3.

The y-coordinate is halfway between the y-coordinates of the end points.

Halfway between 2 and 8 is 5.

Let's see if you can use that now then.

So which coordinate pair is the midpoint of the line segment AB? What do you think? Well done if you said 4,4.

And our triangles are identical, so we've got that midpoint.

How do we know that 5,2 is not the midpoint of the line segment AB this time? You might have said "The triangles are not identical "therefore P is not in the middle of AB." Lucas is trying to solve this problem.

P is the midpoint of the line AB.

What is the coordinate of B? "That's easy," says Lucas.

The midpoint of those coordinates is 3,3.

What do you think? Right, Lucas is correct.

The midpoint of the known coordinates is 3,3.

But that doesn't answer the question.

Lucas has noticed.

"B is the other end of the line segment." So this time we're told one end and the middle, and we want the other end.

Jun says that we can use identical triangles to find it.

Let's see how that works.

So to get from A to P, we have a horizontal distance of 4 and a vertical distance of 2.

So to get from P to B, we need the same again, horizontal distance of 4 and a vertical distance of 2.

So B has coordinates 9,6.

Time for you to have a go.

I'd like you to find the respective midpoints.

So, for A, you want the midpoint of AB.

B is the midpoint of BC, and C is the midpoint of AC.

Give that a go.

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

So, for 2a, you're given two coordinates and you need to find the midpoints.

The same again for b.

Be careful with your negative values.

You're definitely going to want a sketch to help you out.

And some problems now.

So for both questions, M is the midpoint of AB, and B is the midpoint of XY.

See if you can find the coordinates of the different points you're being asked for.

So engage those problem solving skills.

Draw on your triangles to help you.

Can you work these out? Give it a go.

Well done.

Now we've got a square.

ABCD is a square.

I'd like you to work out the coordinates of point M where M is the middle of the square.

Little bit of a hint at the bottom if you need it.

Question 6, you're using the same idea.

I'd like you to work out the coordinates of the middle of the square and the coordinates of the missing vertex.

Give those a go.

Let's have a look.

So you should have 3,4, 0,0, and -2,1.

Well done if you got all three of those.

See I've drawn a little sketch to help me? I need the middle of the coordinates 1 and 4, which is 2.

5 and the middle of the coordinates 10 and 2, which is 6.

So, 2.

5,6.

For 2b, I need the middle of the coordinates -12 and 4, which is -4, and the middle of the coordinates -9 and -1, which is -5.

So my midpoint has coordinates -4,-5.

So we're gonna do this in two steps.

First, we need the coordinates of B.

Well to get to the midpoint, we have a horizontal distance of 2 and a vertical distance of 3.

So we need the same again to get to B.

So 4 add 2 is 6.

7 add 3 is 10.

So we've got 6,10.

Now we can use that to find Y.

We've got a movement of -3 in the x direction and 5 in the y direction.

So we need the same again.

6 subtract 3 is 3.

10 add 5 is 15.

So you get 3,15 for Y.

Well done.

So the coordinate of X, we're gonna need to do it in two stages.

We're gonna need to find B first.

So we've got a horizontal distance of 5 and a vertical distance of 7.

So, if you do the same again, we get 10,15 for B.

Now between Y and B, we've got a horizontal distance of 8 and a vertical distance of -4.

So if we do the same again, we get 18,11 for X And our square.

So you could pick either diagonal to find the midpoint.

I'm gonna look at D to B.

So the middle of 1 and 7 is 4.

The middle of 2 and 6 is 4.

So our coordinates is 4,4.

You might want to check that works for the other diagonal as well.

For 6, the midpoint, so we're looking at halfway between A and D.

That's 0,3, and then we can use that to find B.

If we look at how to get from C to M, we're subtracting 2 in the x direction and 4 in the y direction.

So then we need to do the same from the midpoint.

So we get negative -2,-1.

Outstanding work with those problems. Let's play around some more with some parallel lines.

Let's have a look then.

This graph shows two different lines.

Aisha thinks they're parallel.

How could we check if Aisha is correct? So parallel lines are always the same non-zero distance apart.

That means they have the same gradient.

So we could calculate the gradient of both lines.

We can see the gradient of both lines is 3.

So they're parallel.

What does this tell you about the equations of the lines? So if you think about what the equations are gonna be, what do we know? Well, we know that if they're written in the form y = mx plus c, the coefficient of x is going to be 3, and that's going to be the same for both lines.

What is different about these two lines? So we've said that they're parallel.

They both have the same gradient.

So what's different? Well done if you said they have different y-intercepts.

So if you think about our equations of our lines.

They both have a gradient of 3, but A has a y-intercept of 5.

So we can write that as y = 3x + 5.

B has a y-intercept of -2.

So we can write that as y = 3x - 2.

Let's have a look at this shape.

Alex has plotted four coordinates, and he thinks they are the vertices of a parallelogram.

How could we check if he is correct? What do you think? Well done if you remembered that parallelograms have to have two pairs of parallel sides.

Therefore we can calculate the gradient of the line segments: AB, BC, CD, and DA, and then show that the opposite sides have the same gradient, so are parallel.

We'll do it together.

There's the line segment AB.

That was gonna make one side of our shape.

The gradient is 2.

Right? Let's look at DC.

Again, we can see that the gradient is 2.

So if we're gonna prove that this is a parallelogram, we need to write a sentence now to say that line segments AB and DC are parallel as they have the same gradient.

Now we need to check the other two sides.

Now let's look at our change in x and our change in y.

So from 3 to 11, that's an increase of 8.

From 6 to 4, that's a decrease of 2.

So we use our ratio table.

Divide both sides by 8.

Negative 2 divided by 8 is -1/4.

You could see that again on your diagram.

If you split that in half, you could see that we go right 4, down 1, right 4, down 1.

Let's do the same with AD.

And, again, we can see we have a gradient of -1/4.

Line segments BC and AD are parallel as they have the same gradient.

Because opposite sides are parallel, this is a parallelogram.

If we didn't have the grid lines, we could still calculate the gradient.

So let's see if these coordinates form a parallelogram.

Well let's look at XW.

Now looking at the x-coordinates, we've gone from -7 to -5.

That's a change of 2.

Then we've gone from 11 to -1.

That's a change of -12.

That means our gradient is gonna be -6.

For every increase of 1 in the x direction, we decrease by 6 in the y direction.

Let's try UV 'cause that's gonna be the opposite side.

From 5 to 6 is an increase of 1.

From 9 to -2, that's a decrease of 11.

Well that's a gradient of -11.

So this is not a parallelogram because opposite sides do not have the same gradient.

We don't need to check the other sides.

We've already shown it is not a parallelogram.

Okay, I would like you to use the ratio table to find the gradient of BC.

Give it a go.

Well done if you said the gradient is -2/3.

The change in x is 3.

The change in y is -2.

We divide both numbers in our ratio table by 3.

We get 1 and -2/3.

So our gradient is -2/3.

Right, why are the line segments BC and AD parallel? What do you think? Well done if you said their gradients are the same? Both of them have a gradient of -2/3 which makes them parallel.

Right, ABCD is a square.

We've marked on the centre, M, similar to one of those problems from the first part of the lesson.

I want to find an equation of a line parallel to a side length but which passes through M.

So we're gonna bring a few of our skills together.

So firstly we need to find the coordinates of M.

Well M is the midpoint of the diagonals, either AC or BD.

So the middle of -2 and 4 is 1, and the middle of 6 and 8 is 7.

So M has to have coordinates 1,7.

And because we've drawn it on an axes, we can actually see that.

Right, we need it to be parallel to one of our sides.

So we need the gradient of one of the sides.

Now there's two different equations we could have here 'cause we could have it parallel to AB and DC, or we could have it parallel to BC and AD.

Let's do parallel to AB and CD.

So both those have a gradient of 2.

So a parallel line will also have a gradient of 2.

So we can then use the gradient, and we can use the fact that we want it to go through 1,7 to count back to the y-intercept.

If we have a gradient of 2, and we have a coordinate of 1,7, we're also gonna have a coordinate of 0,5.

We've just taken off 2 from that y-coordinate.

You can see that that works on our graph, and the equation of our line is y = 2x + 5.

If you don't like that count back method, you could use the known coordinate and substitute it into the equation y = 2x + c We know it has gradient of 2, so we know it must be in the form y = 2x + c.

Substitute the y and the x value, and we get c as 5.

Our equation then is y = 2x + 5.

We got the same by using our count back method.

So what is the gradient of BC? Well done if you said it was -1/2.

So what is the equation of the line parallel to BC passing through M? You've got four options.

Which one's correct? Well done if you said it's y = -1/2 x + 7.

5.

That was a trickier y-intercept to find.

You could have substituted in the coordinate 1/7 and rearranged to get c, or you can look at the graph 'cause we've got the grid lines drawn on.

We can see that it's gonna intercept the y-axis at 7.

5.

Time for you to have a go.

For question 1, can you tell me whether that forms a parallelogram? But I need you to explain your answer.

For 2, I'm telling you that it is a parallelogram, and I'd like you to prove that it is.

Give those a go.

For 3, the line L1 is drawn on the axes.

Point A has coordinates 1,1, and B has coordinates 5,3.

They're shown on that axes for you.

I'd like you to find the equation of the line parallel to L1 but which goes through the midpoint of AB.

So, a few steps.

See what you can work out.

Draw things on the graph to help you.

Can you get that equation? Give it a go.

Question 4, similar to the previous question.

We haven't got the grid lines this time.

So L2 passes through the coordinates 0,1 and 5,6.

And then I want a parallel line to that one, but it needs to go through the midpoint of AB.

So, take your time.

Think about what you can work out and then whether you can use that to get the equation of this line.

Give it a go.

Let's have a look then.

This one is not a parallelogram.

Opposite sides are not parallel.

Let's have a look.

The gradient of AB is -1/3, whereas the gradient of CD is -1/6.

Parallel lines have to have the same gradient, and ours do not.

So let's prove that this is a parallelogram.

So, the gradient of AB.

We've got a change in x of 4 and a change in y of 3.

So a gradient of 3/4.

We can do the same for CD.

So from D to C, we've got an increase of four in the x coordinates and an increase of 3 in the y coordinates, so a gradient of 3/4.

BC, you can see that the x value's increased by 6.

The y value's decreased by 6.

So the gradient is -1.

And we have the same for AD.

Opposite sides have the same gradient so are parallel.

This must be a parallelogram.

So we've got L1.

So we need to know the gradient of L1 to start with.

So the gradient is -3.

So a parallel line will also have a gradient at -3, but it needs to go through the midpoint of AB.

The midpoint of AB is 3,2.

So we need a line that has a gradient of -3 but also goes through 3,2.

Therefore it's gonna have a y-intercept of 11.

We can show that by substituting our coordinate in if we like.

So y = -3x + 11.

Well done if you got that one.

And, finally, let's see how well we did with this.

Doesn't matter if you didn't get to a final answer or you didn't get the correct final answer.

If you've shown your working, you're likely to have had some of the elements correct, and then we'll be able to make some tweaks if we made some little mistakes somewhere.

So, first, we need to work out the gradient of L2.

So from 0 to 5 is an increase of 5.

From 1 to 6 is an increase of 5.

So our gradient is 1.

Now we need to work out what the midpoint of AB is.

Well from 7 to 15 is an increase of 8.

If you halve that, you have an increase of 4.

If we add that onto 7, you get 11.

So the x-coordinate at the midpoint is 11.

And then we also need the middle of 4 and -2.

So that's a distance of 6.

Half it is 3.

Then make sure you're taking that off the y-coordinate, so you get 1.

So 11,1 is the midpoint.

And then, remember, we know we have to have a gradient of 1.

So that's gonna be y = x + something.

So y = 1x + something.

We substitute in our coordinate.

We can see that must be y = x - 10.

And that's true, isn't it? If you look at that coordinate, the y value is the x value minus 10.

And that's our final answer.

Fantastic if you got some or all of that last question correct.

There's lots of skills there that you were bringing together.

Awesome work today, guys.

We have looked at what the midpoint of a line segment is and how to calculate it.

We've used that to solve some interesting problems, and we've talked about our parallel gradients and how we can use that to show that shapes are parallelograms but also to find the equations of lines with specific features.

Well done for all your hard work, and I look forward to seeing you again.