Video: Finding the Equations of Perpendicular Lines

Use the fact that the slopes of perpendicular lines have a product of negative one to find the equations of perpendicular lines. Understand the term negative reciprocal.

15:57

Video Transcript

In this video, we’re going to see the relationship that exists between the equations of perpendicular lines. And we’ll see how to find the equation of a line that is perpendicular to another given line.

So, first, let’s look at the relationship between the equations of perpendicular lines. In the diagram here, I have two lines, 𝑦 equals π‘š one π‘₯ plus 𝑐 one and 𝑦 equals π‘š two π‘₯ plus 𝑐 two. Those are the equations of those two lines in slope-intercept form. Now, two lines are perpendicular if they meet at right angles. In the case of perpendicular lines on a coordinate grid, there’s a special property between the slopes of the two lines.

And it’s this, the product of the two slopes, so that’s π‘š one and π‘š two, is equal to negative one. Another way that this could be written is that π‘š one is equal to negative one divided by π‘š two, and π‘š two is equal to negative one divided by π‘š one. π‘š one and π‘š two are referred to as the negative reciprocals of one another. The reciprocal of a value is just one divided by that value and a negative reciprocal is negative one divided by that value as we have here.

So, for example, the negative reciprocal of two is negative one over two, negative a half. The negative reciprocal of negative three is negative one divided by negative three, which is positive a third. In the case of a fraction, the negative reciprocal of two-fifths is negative one divided by two-fifths, which simplifies to negative five over two. As when we divide by a fraction, we invert that fraction and multiply.

This also gives us a handy key fact that we’ll need in working with the slopes of perpendicular lines. And it’s this, the negative reciprocal of a fraction π‘Ž over 𝑏 is negative 𝑏 over π‘Ž. So, we invert the fraction, and we change the sign. If it was positive to start off with, then the negative reciprocal will be negative. And if it was negative to start off with, then the negative reciprocal will be a positive value. So, now, we’ll see how to use this relationship that exists between the slopes of perpendicular lines to help us in finding their equations.

Find, in slope-intercept form, the equation of the line perpendicular to 𝑦 equals two π‘₯ minus four that passes through the point A three, negative three.

So, we’ve been asked to find the equation of this line in slope-intercept form, which means we’re looking for it in the form 𝑦 equals π‘šπ‘₯ plus 𝑐. Reading the question, we can also see that we’re told this line is perpendicular to the line 𝑦 equals two π‘₯ minus four. We’re also told the coordinates of a point that this line passes through. So, we need to find the values of π‘š and 𝑐 for this line. Remember, π‘š represents the slope and 𝑐 represents the 𝑦-intercept.

Let’s start off by calculating the slope. And in order to do this, we need to remember the key fact about the slopes of perpendicular lines. That key fact was this, that π‘š one multiplied by π‘š two is equal to negative one. The product of the slopes is negative one. So, looking at the equation of the line we already know, we can see straightaway that the slope of this line is equal to two. We can see that because it’s already in slope-intercept form.

So, to work out the slope of the line perpendicular to this, we need to find the negative reciprocal of two. So, that’s negative one divided by two, or negative a half. So, we know the slope of the line. And therefore, we can substitute it into our equation for the line. Now, we need to work out the value of 𝑐. And in order to do this, we need to use the coordinates of the point that we know that lies on the line, point A.

As this point lies on the line, it tells us that within this equation, when π‘₯ is equal to three, 𝑦 is equal to negative three. And therefore, I can substitute this pair of values for π‘₯ and 𝑦 to give me an equation that I can solve in order to work out the value of 𝑐. So, if I substitute both of the values, I now have negative three is equal to negative a half multiplied by three plus 𝑐. This simplifies to negative three equals negative three over two plus 𝑐. And now to solve this equation for 𝑐, I need to add three over two to both sides.

At the same time, I’ve expressed that negative three, on the left-hand side of the equation, as negative six over two in order to make the addition more straightforward. This tells me then that 𝑐 is equal to negative three over two. Now, I have the value of 𝑐. The final step is to substitute it back into the equation of my line. So, we have that the equation of this line in the requested slope-intercept form is 𝑦 is equal to negative a half π‘₯ minus three over two.

Write, in the form 𝑦 equals π‘šπ‘₯ plus 𝑐, the equation of the line that is perpendicular to the line negative five π‘₯ plus two 𝑦 equals negative six and that intercepts the π‘₯-axis at 20.

So, we’re asked to find the equation of this line in slope-intercept form, 𝑦 equals π‘šπ‘₯ plus 𝑐. We’re also given two pieces of information about this line. Firstly, that it’s perpendicular to the line negative five π‘₯ plus two 𝑦 equals negative six. And secondly, that it intercepts the π‘₯-axis at 20. This is all the information we need in order to find the equation of this line.

So, we need to work out both the slope and the 𝑦-intercept of this line. Let’s begin by looking at the slope. We’re told that it’s perpendicular to the given line. And in order to work out the slope, we need to remember a key fact about the slopes of perpendicular lines, which is that the product of the slopes is negative one. They’re negative reciprocals of each other. So, if we can determine the slope of the first line, we can then use this relationship in order to calculate the slope of the second line.

In order to find the slope of this first line, I’m gonna rearrange it into slope-intercept form, 𝑦 equals π‘šπ‘₯ plus 𝑐. I’m going to begin by adding five π‘₯ to both sides of the equation. This gives me two 𝑦 is equal to five π‘₯ minus six. Secondly, I’m gonna divide both sides of the equation by two. So, now I have 𝑦 is equal to five over two π‘₯ minus three. Now, this line is in slope-intercept form, and so I can see that its slope is equal to five over two.

In order to find the slope of the perpendicular line, I need to find the negative reciprocal of this value, which in the case of a fraction means inverting the fraction and changing the sign. Therefore, the slope of the line I’m interested in, the perpendicular line, is negative two-fifths. So, we have the start of our equation for this line, it’s 𝑦 equals negative two-fifths π‘₯ plus 𝑐. We still need to work out the value of 𝑐.

So, let’s look at the other piece of information we’re given in the question, which is that the line intercepts the π‘₯-axis at 20. This means that the line passes through the point with coordinates 20, zero. Or phrased slightly differently, when π‘₯ is equal to 20, 𝑦 is equal to zero. This means I have a pair of values for π‘₯ and 𝑦 that I can substitute into the equation of my line in order to calculate the value of 𝑐.

So, substituting zero for 𝑦 and 20 for π‘₯, I now have zero is equal to negative two-fifths multiplied by 20 plus 𝑐. This gives me zero is equal to negative eight plus 𝑐. And if I add eight to both sides of the equation, I have that 𝑐 is equal to eight. So, I’ve worked out the values of both π‘š and 𝑐. The final step then is to substitute these values into the equation of the line. So, we have then that the equation of this line in the requested form is 𝑦 equals negative two-fifths π‘₯ plus eight.

If the straight line passing through the two points two, eight and three, three is perpendicular to the straight line whose equation is three π‘₯ plus π‘˜π‘¦ plus eight equals zero, find the value of π‘˜.

So, within this question, we are told about two straight lines. And the key fact is that they’re perpendicular to each other. This means that the product of their slopes is equal to negative one. So, the approach that I’m going to take for this question is to calculate the slope of the line passing through the points two, eight and three, three first of all. I’ll then use the fact that it’s perpendicular to the second straight line in order to calculate the slope of this straight line. I’ll then compare that with the equation of the second straight line in slope-intercept form in order to find the value of π‘˜.

So, first of all, the slope of the line passing through the points two, eight and three, three. Remember, the slope of a line can be calculated from two points, using change in 𝑦 over change in π‘₯, or 𝑦 two minus 𝑦 one over π‘₯ two minus π‘₯ one. Using the values in this question, this gives me eight minus three over two minus three. This is five divided by negative one, which is negative five. So, we know the slope of this first line, which I’m going to refer to as π‘š one.

Now, using the key fact about the slopes of perpendicular lines, I can now calculate the slope of the second line, π‘š two. So, π‘š two is going to be the negative reciprocal of negative five. It’s negative one divided by negative five which just simplifies to one-fifth. So, now we know the slope of this second line. And in order to work out the value of π‘˜, I now need to rearrange this line into slope-intercept form.

Slope-intercept form, remember, is 𝑦 is equal to π‘šπ‘₯ plus 𝑐. So, the first step is I’m going to subtract both eight and three π‘₯ from both sides of this equation. And in doing so, I have that π‘˜π‘¦ is equal to negative three π‘₯ minus eight. Next, I’m gonna divide both sides of this equation by π‘˜. So, now I have 𝑦 is equal to negative three over π‘˜π‘₯ minus eight over π‘˜.

Now, if I compare this version of the equation of the line with the slope-intercept form, I can see that the slope of the line is equal to negative three over π‘˜. But remember, I’ve also calculated the numeric value of the slope of the line. I know that it’s equal to one-fifth. So, this is what I’m going to do. I’m gonna set these two values equal to each other to give me an equation that I can now solve in order to find the value of π‘˜.

So, I have one-fifth is equal to negative three over π‘˜. Now, to solve this equation, I want to multiply by π‘˜ because it’s in the denominator of a fraction. At the same time, I’m also going to multiply by five. In doing this, I have that π‘˜ is equal to negative three multiplied by five. So, this gives me a value for π‘˜. π‘˜ is equal to negative 15.

Which of the following lines is perpendicular to the line 19π‘₯ minus three 𝑦 equals five?

So, we’re given the equations of five other lines, and we’re asked to determine which of them are perpendicular to this given line. So, in order to answer this question, we need to remember a key fact about the slopes of perpendicular lines, which is that the product of their slopes is equal to negative one. So, our approach then is going to be to calculate the slope of the line that we’re given, and then calculate the slope of each of the other lines and see for which of the lines this relationship holds. I’m going to find the slope of each line by rearranging it into slope-intercept form, 𝑦 equals π‘šπ‘₯ plus 𝑐.

So, let’s begin with the main line in this question, 19π‘₯ minus three 𝑦 is equal to five. My first step is going to be to subtract 19π‘₯ from both sides of the equation. This gives me negative three 𝑦 is equal to negative 19π‘₯ plus five. Next, in order to get this to the 𝑦 equals π‘šπ‘₯ plus 𝑐 form, I need to divide both sides of the equation by negative three. Now, the negative in the denominator of that first fraction will cancel with the negative in the numerator, leaving me with 𝑦 is equal to 19 over three π‘₯ minus five over three. So, comparing this with the slope-intercept form, I can see that the slope of this first line is 19 over three.

So, for any line to be perpendicular to this, its slope will have to be the negative reciprocal of this value. For a fraction, that means I invert the fraction and change the sign. So, lines that I’m looking for will have a slope of negative three over 19. So, what we need to do is algebraically rearrange the equation of each of these lines and see which, if any of them, have a slope of negative three over 19.

Now, I’ve just been through an example of how to do this for one line. So, I’m not going to go through all five. I’m going to give you the rearranged versions and leave it as an exercise for you to rearrange them yourself and persuade yourself that you do, in fact, get these same rearranged equations.

So, here are the rearranged equations of those five lines. And if you’re not sure how any of those came about, then go back and just replay the rearrangement I did at the start of this example. So, remember, I’m looking for a line which has a slope of negative three over 19. The first line has a slope of two over 19. The second has a slope of 19 over three. The third line, this one, does have a slope of negative three over 19. So, this line is perpendicular to the required line.

Looking at the final two, they have slopes of negative 19 over three and positive three over 19. So, neither of these are perpendicular to the required line. So, my answer to the problem then is that only this third line with equation two minus 19𝑦 equals three π‘₯ is perpendicular to the given line.

In summary, then, the key fact about perpendicular lines is that their slopes are the negative reciprocals of each other, which means that they multiply to negative one, π‘š one π‘š two is equal to negative one. Using this fact in conjunction with our other methods for finding the equation of a straight line enables us to find the equation of perpendicular lines.

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