Lesson Video: Slopes of Parallel and Perpendicular Lines

Video Transcript

In this video, we’ll be looking at and working with parallel and perpendicular lines. We’ll go through the definitions of the terms parallel and perpendicular, and then we’ll look at their equations and go over some questions. Two lines are said to be parallel if they’re in the same plane, and no matter how far you extend them in each direction, they’re the same distance apart. Another popular definition also says that they must never intersect. And the effect of that definition is that if you have two lines that are in fact the same line, one on top of each other, they’re not parallel because they intersect in an infinite number of different places. So although they’re always the same distance apart, zero, we don’t count them as being parallel because they do intersect.

Now here’s an example where we’ve got two different lines. They’re both on the 𝑥𝑦-coordinate plane and they’re parallel. Wherever you measure the distance between them, you’ll always get the same answer. Two lines are said to be perpendicular if a right angle is formed at their point of intersection, like it is here. So these two lines are both on the 𝑥𝑦-coordinate plane and they intersect at the point two, zero. And the angle between them there is 90 degrees. In the rest of this video, we’re gonna be working only with lines in the 𝑥𝑦-coordinate plane. And we’ll be looking at the equations of these lines, picking out aspects that tell us if they’re parallel or they’re perpendicular.

And equations of straight lines are really an important part of this topic, so let’s just do a quick recap of 𝑦 equals 𝑚𝑥 plus 𝑏. So the general form of an equation of a straight line is 𝑦 equals 𝑚𝑥 plus 𝑏 or maybe 𝑦 equals 𝑚𝑥 plus 𝑐, depending on where you live. The 𝑚-value, the multiple of 𝑥, the coefficient of 𝑥, tells you the slope of the line. And the value of 𝑏 tells you where it cuts the 𝑦-axis. Now the slope, or the gradient, of the line is defined as being the amount that the 𝑦-coordinate changes by, when we increase the 𝑥-coordinate by one.

Now, I’ve drawn two points 𝑎 and 𝑏 on my line and I’ve increased the 𝑥-coordinate by one to get from 𝑎 to 𝑏. So the difference in their 𝑥-coordinates is positive one. And when I do that, the difference in the 𝑦-coordinates from 𝑎 to 𝑏 will be an increase of 𝑚. So whatever value this is, okay, that is the amount that the 𝑦-coordinate increases by, every time I increase my 𝑥-coordinate up by one, moving along the line. So if that line had been for example 𝑦 equals 0.5 𝑥 plus three, the multiple of 𝑥, the slope, the 𝑚-value is 0.5. This means that every time I increase my 𝑥-coordinate by one, my 𝑦-coordinate will increase by 0.5.

So if we arrange the equation of the line in this format 𝑦 equals 𝑚𝑥 plus 𝑏, then it’s a really simple matter to read off the slope of that line. Just look at the coefficient of 𝑥, the multiple of 𝑥; that is your slope. So here we have a question.

Which of the following straight lines have the same slope or gradient? And then we’ve got five lines. A is 𝑦 equals three 𝑥 minus seven. B is 𝑦 equals minus a half 𝑥 plus three. C is 𝑦 equals minus three 𝑥 plus seven. D is 𝑦 equals minus a half 𝑥 plus five. And E is 𝑦 equals three 𝑥 plus nine. Now, all of those equations are already in the 𝑦 equals 𝑚𝑥 plus 𝑏 format for us, so what we need to do is look at the 𝑥-coefficients. And if they’re the same number with the same sign, then they’ll have the same slope or gradient.

So B and D have the same slope, or gradient, of negative a half. And A and E have the same slope of three. C has got a slope of negative three, so it’s a different sign to A and E. So that is not the same slope or gradient. Now when two lines have the same slope or gradient, we call them parallel. So here, A is parallel to E and B is parallel to D.

So now we can recognise parallel lines just by looking at their equations, so long as they’re in the 𝑦 equals 𝑚𝑥 plus 𝑏 format. So thinking about the slope of perpendicular lines, if two lines are perpendicular, then the product of their slopes, or gradients, is negative one.

Well, why would that be? Let’s take a look. So I’ve got two lines here, 𝐿 one, which I’ve given the equation 𝑦 equals 𝑚𝑥 plus 𝑎, and 𝐿 two, which I’ve given the equation 𝑦 equals 𝑛𝑥 plus 𝑏. Let’s take that point of intersection and then a point on each of those lines which has an 𝑥-coordinate which is one greater than that 𝑥-coordinate at the point of intersection. For line one, the difference in the 𝑦-coordinates between those two points, this one and this one, is going to be 𝑚. And for line two, the difference in those coordinates is going to be 𝑛. Well in fact, it’s going to be negative 𝑛. So line two is a downhill line, which means it’s going to have a negative gradient.

Now I’m interested at the moment in the distance here. And that is a positive distance. So I need to take the negative of that negative gradient to work out what that actual distance is going to be. Now if I label those points A, B, and C, we know that triangle ABC is a right-angled triangle because the two lines are perpendicular. So the angle ABC is a right angle. And in right-angled triangles, we can use the Pythagorean theorem to say that the square of the hypotenuse is equal to the sum of the squares of the other sides. So length AC squared is equal to length AB squared plus length BC squared. And we know that length AC is 𝑚 plus negative 𝑛. So AC squared is 𝑚 plus negative 𝑛 all squared. Now if we look back at our diagram, we can see that we’ve got two more right-angled triangles. We’ve got this one here and this one here.

And again, we can use the Pythagorean theorem to work out the length of the hypotenuse of each of those. We know that the top triangle has a height of ℎ and a width of- sorry, a height of 𝑚 and a width of one. And the bottom triangle has a height of negative 𝑛 and a width of one. So those lengths AB and BC are the square root of 𝑚 squared plus one squared and the square root of negative 𝑛 all squared plus one squared. And now we can start to simplify this. 𝑚 plus negative 𝑛 is just 𝑚 take away 𝑛. So the left-hand side becomes 𝑚 minus 𝑛 all squared. The square root of 𝑚 squared plus one squared all squared is just 𝑚 squared plus one. And negative 𝑛 all squared is just 𝑛 squared. So the square root of negative 𝑛 all squared plus one squared all squared is just 𝑛 squared plus one.

Now 𝑚 minus 𝑛 all squared means 𝑚 minus 𝑛 times 𝑚 minus 𝑛 and just tidying up the right-hand side, I’ve got 𝑚 squared plus 𝑛 squared plus two. Now multiplying each term in the first bracket by each term in the second bracket on the left-hand side, I get 𝑚 squared minus two 𝑚𝑛 plus 𝑛 squared. So I can subtract 𝑚 squared from both sides and I can subtract 𝑛 squared from both sides, which eliminates 𝑚 squared from both sides and eliminates 𝑛 squared from both sides. So I’ve got minus two 𝑚𝑛 is equal to two. Now if I divide both sides by negative two, I get 𝑚 times 𝑛 is equal to two over minus two, which is equal to negative one. Now remember, 𝑚 was the slope of my first line and 𝑛 was the slope of my second line. So 𝑚𝑛 is the product of the two slopes of the lines. So when those lines were perpendicular, it doesn’t matter what the actual slopes were, we knew that when we multiply them together, we’ll always get this answer of negative one.

Now if you got a bit lost along the way during that explanation, don’t worry. That bit’s not important. This is what you’ve got to remember. If two lines are perpendicular, then the product of their slopes, or gradients, is negative one. So if we call 𝑚 the slope of the first line and 𝑛 the slope of the second line, 𝑚 times 𝑛 is equal to negative one. Or if I divide both sides by 𝑛, I get 𝑚 is equal to minus one over 𝑛. Or if I divide both sides by 𝑚, I get 𝑛 is equal to minus one over 𝑚. In other words, each slope is the negative reciprocal of the other. This means that if I know one of the slopes, I can find the perpendicular slope by just changing the sign and flipping that number.

So for example, if 𝑚 was five, 𝑛 would just be negative one over five, the opposite sign and then flip that number. If 𝑚 was minus three, I take the opposite sign to make it positive and flip that number, three over one becomes one over three. And if 𝑚 was equal to two-thirds, 𝑛 will be negative three over two. So knowing this rule means that if you know the gradient, or the slope, of a particular line, it’s very easy to work out what the gradient, or the slope, of a perpendicular line to that one would be.

So to sum up the basic facts, two lines are parallel if they have the same slope but different 𝑦-intercept. For example, 𝑦 equals seven 𝑥 minus five and 𝑦 equals seven 𝑥 plus two. They both have a slope of seven and their 𝑦-intercepts are different, minus five and positive two, so they’re parallel. And two lines are perpendicular if the product of their slopes is equal to one. For example, 𝑦 equals three 𝑥 minus one and 𝑦 equals minus a third 𝑥 plus nine. The slope of the first is three and the slope of the second is negative a third. So the product of those slopes is three times negative a third. But three is the same as three over one, so to make this a fraction calculation, I’ve got three times negative one over one times three. Well that’s negative three over three, which is equal to negative one. So it meets the criteria, so the lines are perpendicular.

And one more example, 𝑦 equals negative two-sevenths 𝑥 plus eight and 𝑦 equals seven over two 𝑥 plus eight. The slopes are negative two over seven and seven over two. They’re the negative reciprocal of each other. And if I multiply the slopes together, negative two over seven times seven over two is equal to negative fourteen over fourteen which is equal to negative one. So those two lines are perpendicular. And with perpendicular lines, it doesn’t matter that those intercepts were both at positive eight. The slopes are different, so they are different lines. They are definitely perpendicular. Let’s have a look at some typical questions then.

Two lines, A and B, have slopes three-quarters and minus four over three, respectively. Are they parallel, perpendicular, or neither?

Well, the slopes aren’t equal, so they’re definitely not parallel. Now if we multiply those slopes together, we get three-quarters times negative four over three, which is negative twelve over twelve, which is negative one. So it looks like those two sl- lines would be perpendicular.

Next, which of the following lines are parallel to each other? And then we’ve got five equations. A) 𝑦 equals eight 𝑥 minus five. B) Two 𝑦 is equal to eight 𝑥 plus three. C) Eight 𝑥 minus 𝑦 plus two equals zero. D) A half 𝑦 minus four 𝑥 equals twelve. And E) 𝑦 equals minus an eighth 𝑥 plus seven.

Well with A and E, they’re already in the 𝑦 equals 𝑚𝑥 plus 𝑏 format, so it’s easy enough to read off what the slope is. But for B, C, and D, we’re gonna have to need to do a bit of rearranging to get them in the right format to be able to read off their slopes. For equation B, I’m gonna have to divide both sides of that equation by two. So the left-hand side, half of two 𝑦 is just 𝑦. And then dividing each term on the right-hand side by two, half of eight 𝑥 is four 𝑥 and half of three is three over two. For equation C, I can just add 𝑦 to both sides which will eliminate it from the left-hand side, and give me just 𝑦 on the right-hand side. So that gives me eight 𝑥 plus two equals 𝑦. Now I’m just gonna write this the other way around 𝑦 equals eight 𝑥 plus two, because that’s the format that we’re familiar with.

Now D needs a little bit more work. I’ve got the 𝑦 and the 𝑥 term on the left-hand side and then just the number on the right. So first of all, I’m gonna add four 𝑥 to both sides, which gives me a half 𝑦 is equal to, well twelve plus four 𝑥 or four 𝑥 plus twelve. And then doubling each side of that equation to leave me with just 𝑦. Two lots of half 𝑦 are 𝑦, two lots of four 𝑥 are eight 𝑥, and two lots of twelve are twenty-four. Now we’ve got our equations in the right format. It’s a pretty straightforward matter of finding the slopes, so we can see which lines are parallel to each other. In A, the slope is eight. In B, the slope is four. In C, the slope is eight. In D, the slope is also eight. And in E, the slope is negative an eighth. So A, C, and D, all have a slope of eight. So the answer is A, C, and D are parallel.

Now we’ve got to write the equation of a line which is parallel to 𝑦 equals eight 𝑥 minus four.

So it’s got to be parallel, which means it’s got to have the same slope of eight. So it’s gotta start off with 𝑦 equals eight 𝑥. And then we can add anything we like cause wherever that line cuts through the 𝑦-axis, it doesn’t matter. It’s gonna be parallel to this line. The only thing that you shouldn’t use is eight 𝑥 minus four. Don’t make it exactly the same line because most people would say that they’re not parallel because it’s the same line. So you can write anything you like here eight 𝑥 plus a thousand. Here we go, that’s a line which is parallel to 𝑦 equals eight 𝑥 minus four.

Next, we’ve got to write the equation of a line which is perpendicular to 𝑦 equals three 𝑥 minus two.

Well, the slope is three, so the slope of that perpendicular line must be the negative reciprocal, so that’s minus one over three. So our equation is gonna start off 𝑦 equals minus one over three 𝑥. And we can add anything we like. We could just leave it as minus one over three 𝑥, or we could add any number we like to make a perpendicular line.

Now, we’ve got to write an equation for the straight line that is parallel to 𝑦 equals a half 𝑥 plus five and passes through the point six, 10. Well, we know that the slope of the line 𝑦 equals a half 𝑥 plus five has a slope of a half. So if our line is gonna be parallel to that, it must also have a slope of a half. But of course, that line could cut the 𝑦-axis anywhere, so we could move that line up or down. What we’re told is that it must pass through the point six, 10. And that means that when 𝑥 is six, 𝑦 needs to be 10.

Now if we use the general form of our equation 𝑦 equals 𝑚𝑥 plus 𝑏, we know the slope 𝑚 is equal to a half. Now, we’ve got to find out the value of 𝑏. But we know a particular coordinate pair that sits on the line, when 𝑥 equals six, 𝑦 equals 10. So replacing 𝑥 and 𝑦 with six and 10, we’ve got 10 is equal to a half times six plus 𝑏. So 10 is equal to three plus 𝑏. And then subtracting three from both sides, gives me seven is equal to 𝑏. Now, we know the value of 𝑏; we can finish off our equation. 𝑦 is equal to a half 𝑥 plus seven.

Lastly, write an equation for the straight line that is perpendicular to 𝑦 equals three-quarters 𝑥 minus four and passes through the point four, 11.

So we’re trying to find a gradient, or a slope, which is perpendicular to three-quarters. So the slope of our perpendicular line is gonna be negative four over three, the negative reciprocal, remember. Three-quarters times negative four over three is equal to negative one. So that means it’s a perpendicular line. So we’ve got the slope of the line and we also know it’s gonna go through the point four, 11. So when 𝑥 is equal to four, then 𝑦 is equal to 11. So using the format of the equation 𝑦 equals 𝑚𝑥 plus 𝑏, we know that 𝑚 is minus four over three. And we know that when 𝑥 is four, then 𝑦 is 11. So we can use all that information to work out the value of 𝑏.

So 11 is equal to negative four-thirds times four plus 𝑏. Well, four is the same as four over one. So that becomes a fraction calculation, minus four-thirds times four over one. So 11 is equal to negative sixteen-thirds plus 𝑏. So if I add sixteen-thirds to both sides, I’ve got 11 plus sixteen-thirds is equal to 𝑏. And that is equal to 49 over three. So I can put that back into our original equation and get my answer, 𝑦 equals negative four-thirds 𝑥 plus 49 over three.