### Video Transcript

A line πΏ one has equation π¦ equals two π₯ minus five, and a line πΏ two has equation two π¦ 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 weβre gonna have to do in this question is look at the gradient of each of the lines and compare them.

So first, letβs have a think about the general form of the equation of a straight line. π¦ is equal to ππ₯ plus π. So π is a number that tells us about the gradient of the line, and π is another number which tells us the π¦-intercept, where it cuts the π¦-axis. So we need to rearrange these two equations into the π¦ equals ππ₯ plus π format and then compare the gradients.

Now in the first case for line one, π¦ is equal to two π₯ minus five, weβve already got the equation in the right format. In this case, π is equal to two, the gradient of the line is equal to two, and π is equal to negative five. Letβs just make a note of that in our working out.

Now you donβt need to do this, but just by way of explanation if π is equal to negative five, it cuts the π¦-axis at negative five. Letβs say thatβs about here. And then the gradient is two. So every time I increase my π₯-coordinate by one, my π¦-coordinate will go up by positive two. So weβll end up with a line looking something like that.

Now letβs think about the gradient for line two. The equation is two π¦ plus four π₯ is equal to negative seven. Weβre gonna need to rearrange this into the π¦ equals something times π₯ plus another number format. I want to get this π¦ term on its own. So in order to do that, Iβm gonna have to get rid of this π₯ term.

And how do I get rid of plus four π₯? Well, Iβm 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, Iβve got two π¦ plus four π₯ minus four π₯. Well, if I have four π₯ and I take away four π₯, I end up with nothing. So Iβm left with just two π¦. And over on the right-hand side, Iβve got negative seven take away four π₯.

Now it doesnβt 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 π¦ equals something times π₯ plus π format, Iβm actually gonna swap those two round.

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

Now we donβt actually need to do this, but in this video, letβs take a look at it anyway. If we were to sketch that graph, we can see that this line cuts the π¦-axis at negative 3.5. And as I move along the line, every time I increase my π₯-coordinate by one, my π¦-coordinate goes down by two.

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