Lesson video

Hello there.

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

My name is Miss 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.

So today's lesson is on checking and securing your understanding of finding the equation of the line from two coordinate pairs.

We're gonna look today then at finding the equation of the line from two coordinate pairs as well as from the gradient and one coordinate pair.

So we're gonna investigate a few different types of questions with a few different bits of information we might be given.

We're gonna talk lots today about the gradient and the y-intercept.

So if you need a reminder of what those are, pause the video now.

Perfect.

So, we're gonna split this lesson into two parts.

We're gonna start by using the gradient to help us find the y-intercept.

Then in the second part of the lesson, we're gonna look at another method which we might find slightly more efficient.

So if we know the gradient and the y-intercept of a straight line, we can write the equation in the form y = mx + c.

So, for example, if I want the equation of the line with gradient -5, which passes through the point 0,3, I've been told the gradient, I've been told the y-intercept.

0,3 must be on the y axis.

Therefore the equation of my line is y = -5x + 3.

Now, if we know the gradient and a coordinate pair on the line, which is not the y-intercept, we have to do a little bit more work.

What we can do is we can use the gradient to work out the y-intercept as long as we know one of the coordinates.

So let's try this one.

An equation for the line with gradient 3, which goes through the point 2,8.

I always find it helpful to sketch the graph first.

So I know it's gonna go through the point 2,8 and I know it's gonna have a gradient of 3 and that's what I'm gonna use in a moment.

I'm gonna show you two methods side by side.

You might wanna use one or you might wanna use both together.

So I've also put my coordinate in a table of values.

So I know that when x is 2, y has to be 8.

I want to find out the y-intercept.

So I want to work backwards until I know what y is when x is zero.

And it's the gradient that we're gonna use to find that out.

So because the gradient is 3, if I count back 3, I can see that the coordinate 1,5 will also be on my line.

If I count back 3 again, I can see that the coordinate 0,2 will also be on my line.

Therefore the y-intercept must be 0,2.

I can write my equation then as y = 3x + 2.

We've got a gradient of 3 and I've since worked out that my y-intercept must be 2.

Of course, we can check our equation by substituting our known point so we can double check that 2,8 definitely sits on this line.

So 8 has to equal 3 lots of 2 plus 2.

Well, that is true.

So that must work for our coordinate and we can see that it clearly has a gradient of 3 as well.

So let's try one with a negative gradient.

So we've got a gradient of -4 which goes through the point 3,-2.

So this time when I sketch my axis, I need to make sure I put my coordinate in the right place.

So it's got to be in that fourth quadrant.

And I can use my table if I want to help me.

And this time we're gonna need to add 4.

If we think about how a negative gradient is going to look, if I subtract 1 from my x coordinate, I'm gonna need to add 4 to my y coordinate.

So I can see that 2,2 is gonna be on that line.

And this time we're gonna need to add 4 three times, aren't we, to get from a coordinate of 3 something back to a coordinate of 0 something.

So if I add 4 three times, that's the same as adding 12.

So 0,10 must be my y-intercept.

Just right before we can check that our point 3,-2 satisfies that equation.

And it does.

So we must have the right equation for that line.

Quick check then.

A line has a gradient of 2 and goes through the point 6,10.

Find the y-intercept of this line.

You can use either diagram or both diagrams to help you.

Okay, so we've got a gradient of 2.

So if we take 1 off the x value, we also need to take 2 off the y value.

So 5,8 would also be on that line.

I don't particularly want to count back 2 six times, So I'm gonna think about what that's gonna be an overall movement of.

2 X 6 = 12.

So if I'm counting back 2 six times, that's the same as subtracting 12.

So we have a y-intercept of 0,-2.

So, which of these then is the correct equation of the line? Yeah, of course we're gonna need a gradient of 2.

So it needs to be 2x and a y-intercept of -2, so 2x - 2.

So we can calculate the equation of a line if we know any two coordinate pairs.

So we are gonna write the equation for a line which goes through the point -2,-7 and 3,18.

The first thing we need to do is calculate the gradient.

Now, I'm absolutely gonna sketch these points to help me.

So -2,-7 should be in that third quadrant.

3,18 should be in that first quadrant.

And I can see now that I'm expecting a positive gradient.

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

So from -2 to 3, that's a change of 5.

<v ->7 to 18, that's a change of 25.

</v> So my gradient must be 5.

Now, I know my gradient's 5, I can use either points to help me work out the y-intercept.

I'm gonna pick 3,18 and I'm gonna need to count back 5 three times.

So that's 15 in total.

If I subtract 15 from 18, I get 3.

So that y-intercept must be 0,3.

I reckon my equation then is y = 5x + 3, but I can check.

I'm already pretty sure that it works for 3,18 'cause that was the coordinate I used.

So let's shut the other coordinate, -2,-7.

So -7 equals 5 lots of -2 plus 3.

Yes, it does.

You can always double check that 3,18 works as well if you are not 100% convinced.

So Laura's trying to find the equation which passes through the points 3,-2 and 3,6.

She's drawn a ratio table to help her but she doesn't think that works.

What problem has Laura encountered? What could she do to help her find the equation of this line? What do you think? So the problem is the change in x is 0.

So she's got to a point where she's trying to do 8 divided by 0, but dividing by 0 is undefined so she cannot work out the gradient.

What she should probably do to help her find the equation of this line is draw a sketch and you're gonna see why.

So 3,-2 is there and 3,6 is there.

You might already be miles ahead of me on this one.

That's why calculating our gradient didn't work 'cause we've got a vertical line.

The equation of the line then is x = 3.

And Laura's right.

She should have noticed that the x coordinates were both 3.

Drawing that sketch probably would occlude her in a little bit.

So we're gonna try one together and then you're gonna give one a go.

So let's find the equation of the line passing through -3,2 and 4,9.

We're gonna start by finding our gradient.

So if we look at those x values from -3 to 4, that's an increase of 7.

From 2 to 9, that's an increase of 7 also.

Our gradient then is 1.

I definitely should sketch this out.

I'm just a little bit short on space here.

So I've put this into a table of values instead.

So to count back from the point 4,9, I'm gonna need to take the gradient off four times.

So 4 lots of 1 is 4, so I'm gonna need to subtract 4.

So 0,5 should be my y-intercept.

My equation then is y = x + 5.

And then I'm gonna check that for my coordinates.

And that is true.

Both my coordinates follow the rule y = x + 5.

Time for you to have a go.

Can you find the equation of that line? And check your answer.

Give it a go.

So with our ratio table, when x increases by 2, y increases by 6.

So we have a gradient of 3.

Sketching those points out would've told you that you're expecting a positive gradient, which you should be feeling quite happy now that that's what we've got when we've calculated it using a ratio table.

Now you can count back the gradient on your graph or you can use a broken table.

So I'm gonna need to take off the gradient to 3 five times.

So that is 15.

6 subtract 15 is -9.

So we've got y = 3x - 9.

Then we need to check that works for our two coordinates.

We've already used 5,6.

We're probably fairly confident it works for 5,6.

We can check it anyway.

3 lots of 5 is 15.

Takeaway 9 is 6.

More importantly, we need to check our other coordinate.

So 3 lots of 7 is 21, takeaway 9 is 12.

So it works for that coordinate too.

Great time for a practise.

I'd like you to work out the equations of each of these lines from their given features.

There is a set of axis there to help you if you want to sketch any known points.

Give it a go.

Lovely, so this time I've given you two coordinates and a little sketch to help you.

Can you work out the equations of those lines? Off you go.

Let's see if we can put all those skills together now.

So, can you find the equation of the line which goes through those given points? There's a ratio table to help you and then think about how you're gonna work out your y-intercept.

Well done.

Your final challenge is to see if you can do this all by yourself.

You've got two coordinates.

Can you find the equation of the line that goes through the points with those coordinates? Off you go.

Three final ones to do and then come back when you're ready for the answers.

Let's have a look then.

So, for a, you are actually given the gradient and the y-intercept.

So y = 7x - 5.

B, we need to count back 3 four times.

So we need to subtract 12 from our y coordinate, which gives us y = 3x + 4.

For c, we're given a negative coordinate so you have to be careful here.

To get from -2 to 0, we're adding the gradient twice.

So we need to add 2 to -1, gives us positive 1.

Hopefully by sketching the point, you could see that that was going to work.

So y = x + 1.

And for d, we've got y = 2x - 4.

So for these three, I'd like you to pause the video and check through my working.

You'll see that I've used a ratio table to get the gradient and a broken table of values to count back to my y-intercept.

Have a read through those and then we'll look at the next bit.

Well done.

So for these questions, you were given the two coordinates.

So the first thing you need to do is work out your gradient.

I've used a ratio table.

Then once you've got your gradient, you can use a broken table of values or you can use your graph to count back to your y-intercept.

For question b, it's gonna be easier to write your coordinates the other way around to start with.

To get from 5 to 9, you are adding 4.

To get from 10 to 18, you are adding 8.

Then you can work out your gradient from there.

Read through the rest of those answers now and then we'll look at the other set.

So the first one, we've got y = -3x + 10.

For e, well done if you spotted that the y coordinates are the same in both coordinate pairs.

So that means there's gonna be a horizontal line y = 5.

And then f, got y = -2x + 9.

Then for g, this was similar to the one Laura had difficulties with earlier 'cause this is gonna be a vertical line.

So it's x = -1.

For h, we've got a gradient of 3, and therefore we got y = 3x - 3.

And for i, a gradient of -1.

So y = -x - 3.

So as promised, we're now gonna have a look at a slightly more efficient way maybe of calculating your y-intercept.

Let's see what you think.

So Laura's now trying to find the equation of the line with a gradient -3/5 which passes through the point -32.

She has a good point here.

Using the gradient to count back to the y-intercept is gonna be tough.

The gradient is a fraction, so sketching it accurately and then counting back -3/5, however many times we need to do it, is gonna be trickier.

Also, if we look at the x coordinate of -30, that's a lot of counting that we're gonna need to do to get to zero.

I know we're starting to find some slightly more efficient ways of doing it that's still going to be a little bit of pain.

So let's have a look.

what we can use if we use this knowledge of the form y = mx + c, we already know the gradients so we can half fill in our equation already.

We know our equation must be y = -3/5x + c, just the plus c bit we need to work out.

So what we're gonna do is we're gonna use the coordinate pair to work out the value of c.

We know that -32 is on our line, so it must satisfy our equation.

So we can substitute those values into our equation and work out the missing constant.

So 2 = -3/5 X -30 + c, which is substituted the x and the y coordinate into our y = mx + c.

Little bit of fractional skills.

So if I do -30 divided by 5 and then multiply by -3.

So -30 divided by 5 is -6, multiply by -3 is 18, so 2 = 18 + c.

So c must be -16.

So I know the equation of my line is y = -3/5x - 16.

Or if you wanted to write it in a different form, if we multiply each term by 5, I've got 5y = -3x - 80 or 3x + 5y = -80.

If you want to write those down and just spend some time checking that you know how to rearrange, feel free to do that now.

So let's see if Laura's equation is correct.

Substitute our known point into Laura's equation.

3 lots of -30 plus 5 lots of 2 equals -80.

<v ->90 + 10 = -80.

</v> Yeah, that's true.

So her equation is correct.

So this is gonna be a helpful method when finding the equation of a line between any two points.

So this was similar to the questions we were looking at before.

Definitely gonna sketch my points.

Gives me a really good idea whether I'm looking for a positive or a negative gradient.

So now, I need to work out the change in x and the change in y.

So I can just use the coordinates.

So from -9 to 6 is plus 15.

From -8 to -3 is plus 5.

Now, I can divide both sides by 15, and 5 divided by 15 is 1/3.

So my equation must be y = 1/3x + c.

And my only challenge now is to work out the value of c.

Well, we can use either point to help us.

I'm gonna stick with the positive x value.

So I'm gonna stick with 6,-3.

So substitute in.

<v ->3 must equal 1/3 lots of 6 plus c.

</v> So -3 is 2 + c or -5.

So y = 1/3x - 5.

This does fit with our sketch.

If we think about where we're expecting the y-intercept to be on our sketch, -5 seems reasonable.

We are expecting there to be a positive gradient and a reasonably shallow gradient.

So 1/3 does sound reasonable.

Of course we can substitute the other points to be sure.

So -8 = 1/3 X -9 - 5.

So -8 = -3 - 5, and that is true.

<v ->8 is -3 - 5.

</v> Quick check then.

Aisha is finding the equation of the line passing through the points with coordinates 3,-5 and -2,25.

She thinks it's gonna be y = 6x - 23.

Check using substitution if Aisha is correct.

Give that one a go.

Let's try that first.

Coordinate 3,-5.

And that works.

We get -5 = -5.

We do need to check our second coordinate 'cause we know we need to know two points on the line to get a specific line passing through them.

So try our other coordinate.

Oh, it doesn't work for that second coordinate.

So that is not the correct equation of the line.

So here is her working.

Can you spot where she has gone wrong? Take your time to look through what she's done.

So the mistake was made at the very beginning.

The value she's put into the ratio table are incorrect.

If you rewrite the coordinates the other way around, from -2 to 3 is an increase of 5.

From 25 to -5 is a decrease of 30.

Drawing a sketch would definitely have helped her 'cause she'd be able to see that she should have a negative gradient.

I'd like you to fill in the missing steps of these workings so that Aisha gets the correct equation of the line.

Off you go.

So I've written the coordinates the other way around.

As I said, it would probably help us.

We've got an increase of 5 in the x values, so we should have a decrease of 30 in the y values.

That means our gradient is gonna be -6, so y = -6x + c.

Then we can substitute either coordinate, but I went with 3,-5.

So -5 equals -6 lots of 3 plus c.

So c must be 13.

Our final equation y = -6x + 13.

Of course we're gonna check that as well.

Time for you to put all those skills into practise.

You can use any method that you like.

I'm hoping that you found that method of substituting to find the y-intercept maybe a little bit easier than counting back.

However, when you find a method that you like and it works for you, it's absolutely fine to carry on using that.

So you've got three questions to have a go at.

Once you're happy with your answers for those and you've checked them, come back for the next bit.

Well done.

Three more to do.

Take your time for f.

You've got all the skills that you need to work it out.

Just pay particular attention to some of your fractional calculations.

Off you go.

Now, let's see if we can put this into context.

So a graphic designer uses a linear equation to work out how much to charge for a different length of jobs.

The equation consists of a fixed charge and a price per hour.

So they charge 92 pounds for a two-hour job and 232 pounds for a six-hour job.

Can you write the equation they use for calculating the price of any job, p, with respect to the number of hours, h? Once you've got that, you can use that to work out how much they charge for a five-hour job.

Off you go.

Let's have a look at our answers then.

So for a, when x increases by 5, y increases by 40.

So we have a gradient of 8.

You can use either coordinate.

I've used the one that's slightly smaller value, so 7,41.

So 41 equals 8 lots of 7 plus c.

Little rearranging tells me that c must be -15.

So y = 8x - 15.

For b, I've rewritten the coordinates the other way around to help me.

So from -2 to 2 is an increase of 4.

From 22 to 6 is a decrease of 16.

So an x increases by 4, y decreases by 16.

So that's a gradient of -4.

Again, pick a coordinate to substitute.

I've picked the ones with two positive values.

So 6 = -4 X 2 + c, which gives me c is 14.

Don't forget your final answer, y = -4x + 14.

And then for c, when x increases by 4, y increases by 2.

So the gradient to that line is going to be 1/2.

2 divided by 4 is 2 quarters or 1/2.

I'm gonna use that second point again because they're both positive values.

So 1 = 1/2 X 10 + c.

So c must be -4.

The y = 1/2x - 4.

Let's have a look at the slightly trickier ones.

So this time when x increases by 6, y decreases by 9.

That gives us a fractional gradient of -9/6, 1, -3/2.

Then just substitute in as normal.

Just being careful with my fractions of my negatives.

<v ->3/2 times 2 is -3, so c is 0.

</v> So that's quite helpful.

That gives us an equation as y = -3/2x.

You might wanna check that that works.

Is -3/2 multiply by 8 -12? Yes, it is.

For e, I've got big values here, but I'm doing exactly the same thing.

I'm gonna work from the right hand coordinate to the left hand coordinate.

So from -10 to 20, that's an increase of 30.

From -20 to 16, that's an increase of 36.

So that gives me 36 divided by 30 for my gradients, which is the same as 6/5.

Then I can substitute my known coordinate 6/5 times 20 is 24, so 16 = 24 + c.

So c must be -8.

y = 6/5x - 8.

Of course you could have rearranged that into another form if you wished.

I'm saving myself time and just leaving that in the form y = mx + c.

And then f.

Now, I have decided to stick with the coordinates this way round.

So to get from 12 to 5, you subtract 7.

To get from 1 to 17/3 you add 14/3.

1 is 3/3, isn't it? So you must add 14/3 to get 17/3.

So -7, 14/3.

If you want to, you can multiply 3 by -1, so 7, -14/3.

And then you just need to divide by 7.

Well, -14/3 divided by 7.

That's a nice easy one.

That's -2/3.

Once you get your gradient, substitute as normal.

So y = -2/3x + 9.

If you think I went a little bit fast there, just pause the video and check that through.

See if you can line that up with what you've done and then we'll have a look at the last bit.

And finally, we can essentially create two coordinates here.

We know a two-hour job is 92 pounds and a six-hour job is 232 pounds.

That means that for an extra four hours, it costs an extra 140 pounds.

That means each extra hour would cost an extra 35 pounds.

So I know that I've got 35 pounds per hour, but of course that would get me 70 pounds for a two-hour job and it's not is 92 pounds.

So there must be an extra charge of 22 pounds.

So the price, p = 35h + 22.

And of course you could check that works with both of the pieces of information we were given.

For a five-hour job, we've got 35 X 5 + 22, which is 197.

Well done with that problem solving bit at the end.

So today we've seen then how to calculate the equation of a line from any two pairs of coordinates.

We thought a little bit more about how to calculate our gradient efficiently.

We've talked about how we can find the change in y and divide it by the change in x, which is essentially what we do when we use our ratio tables.

We've also talked about how we can find the y-intercept by counting back using the gradient in one point, but that it might be more efficient to substitute the known point into the equation so far and rearrange to find out the value of c.

Thank you for joining us for today's lesson and I look forward to seeing you again.