Video: GCSE Mathematics Foundation Tier Pack 1 β€’ Paper 1 β€’ Question 26

GCSE Mathematics Foundation Tier Pack 1 β€’ Paper 1 β€’ Question 26

03:46

Video Transcript

A line 𝐿 one has equation 𝑦 equals two π‘₯ minus five, and a line 𝐿 two has equation two 𝑦 plus four π‘₯ equals negative seven. Determine whether the two lines are parallel. You must show all of your working out and explain your answer clearly.

Now two lines are parallel if they have the same gradient. So what we’re gonna have to do in this question is look at the gradient of each of the lines and compare them.

So first, let’s have a think about the general form of the equation of a straight line. 𝑦 is equal to π‘šπ‘₯ plus 𝑐. So π‘š is a number that tells us about the gradient of the line, and 𝑐 is another number which tells us the 𝑦-intercept, where it cuts the 𝑦-axis. So we need to rearrange these two equations into the 𝑦 equals π‘šπ‘₯ plus 𝑐 format and then compare the gradients.

Now in the first case for line one, 𝑦 is equal to two π‘₯ minus five, we’ve already got the equation in the right format. In this case, π‘š is equal to two, the gradient of the line is equal to two, and 𝑐 is equal to negative five. Let’s just make a note of that in our working out.

Now you don’t need to do this, but just by way of explanation if 𝑐 is equal to negative five, it cuts the 𝑦-axis at negative five. Let’s say that’s about here. And then the gradient is two. So every time I increase my π‘₯-coordinate by one, my 𝑦-coordinate will go up by positive two. So we’ll end up with a line looking something like that.

Now let’s think about the gradient for line two. The equation is two 𝑦 plus four π‘₯ is equal to negative seven. We’re gonna need to rearrange this into the 𝑦 equals something times π‘₯ plus another number format. I want to get this 𝑦 term on its own. So in order to do that, I’m gonna have to get rid of this π‘₯ term.

And how do I get rid of plus four π‘₯? Well, I’m gonna need to take away four π‘₯ from the left-hand side. And if I do that to the left-hand side, I also have to do it to the right-hand side. And when I do that, over on the left, I’ve got two 𝑦 plus four π‘₯ minus four π‘₯. Well, if I have four π‘₯ and I take away four π‘₯, I end up with nothing. So I’m left with just two 𝑦. And over on the right-hand side, I’ve got negative seven take away four π‘₯.

Now it doesn’t actually matter whether I start with negative seven and then take away four π‘₯ or if I start with negative four π‘₯ and take away seven. So just because it looks more like the 𝑦 equals something times π‘₯ plus 𝑐 format, I’m actually gonna swap those two round.

But now I’ve got two 𝑦 is equal to negative four π‘₯ take away seven, so I need to halve everything in order to get one 𝑦. Half of two 𝑦 is one 𝑦. Half of negative four π‘₯ is negative two π‘₯. And half of negative seven is negative 3.5. So I’ve got 𝑦 is equal to negative two π‘₯ minus 3.5. Then comparing that to the format 𝑦 equals π‘šπ‘₯ plus 𝑐, we can see that π‘š is equal to negative two, so the gradient is negative two, and 𝑐 is equal to negative 3.5.

Now we don’t actually need to do this, but in this video, let’s take a look at it anyway. If we were to sketch that graph, we can see that this line cuts the 𝑦-axis at negative 3.5. And as I move along the line, every time I increase my π‘₯-coordinate by one, my 𝑦-coordinate goes down by two.

So this line is going down at this angle. So 𝐿 one is an uphill line; it’s got a positive gradient. And 𝐿 two is a downhill line; it’s got a negative gradient. Clearly, they’re not parallel. So we can write out our answer. 𝐿 one has a gradient of positive two. 𝐿 two has a gradient of negative two. These are different. The lines are not parallel.

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