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Contents

- 1 Coordinates
- 2 Coordinates
- 3 Equation for Coordinates
- 4 Equation for Coordinates
- 5 Vertical Lines
- 6 Vertical Lines
- 7 Vertical and Horizontal Lines
- 8 Slope Intercept Form
- 9 Slope Intercept Form
- 10 Point Slope Form
- 11 Point Slope Form
- 12 More Point Slope Form
- 13 Point Slope Practice 1
- 14 Point Slope Practice 1
- 15 Point Slope Practice 2
- 16 Parallel Lines
- 17 Parallel Lines
- 18 Many Parallel Lines
- 19 Perpendicular Lines Equations
- 20 Perpendicular Lines Equations
- 21 Perpendicular Lines
- 22 Slopes of Perpendicular lines
- 23 Parallel or Perpendicular 1
- 24 Parallel or Perpendicular 1
- 25 Parallel or Perpendicular 2
- 26 Parallel or Perpendicular 2
- 27 Equation Analysis
- 28 Slope of a Parallel Line
- 29 Slope of a Parallel Line
- 30 Equation of a Parallel Line
- 31 Equation of a Parallel Line
- 32 Slope of a Perpendicular Line
- 33 Slope of a Perpendicular Line
- 34 Equation of a Perpendicular Line
- 35 Equation of a Perpendicular Line
- 36 Practice 1
- 37 Practice 1
- 38 Practice 2
- 39 Practice 2
- 40 Practice 3
- 41 Practice 3
- 42 Practice 4
- 43 Practice 4
- 44 Practice 5
- 45 Practice 5
- 46 Practice 6
- 47 Practice 6

We're going to continue look in our graphs, but this time, we're going to be writing equations and join connections between those equations and the graphs. Lets start with a horizontal line. Here's a flat horizontal line and here are some points. I want you to tell me the coordinates of each of these points. You can put your numbers here.

Well for A, we travel four units to the left and then tjjeree units down, so negative 4, negative 3. For B, we don't move horizontally, we just move down 3, so 0, negative 3. For C, we move to the right 3 units and down 3 units, so positive 3 for the right, negative 3 for down. And finally for D, 5 units right, understanding of points.

I think I see a pattern, maybe you see it too. Think about what is the same for each of these points, is the x value the same or is the y value the same and what number does it equal? That should help you write the equation. Also I want you to put the slope of the equation in this box, you can enter the number or if you think it is undefined type the letter u.

Well, when I look at the points, I note that the Y values are all the same. Y is always equal to negative 3, it doesn't matter what x is. So the equation for this horizontal line is y equals negative 3. And remember, we have a flat line here so the slope must be zero. It has lots of run but no rise. This quiz was pretty tough. I know I haven't taught you how to write the equation of a line, so don't worry if you didn't get it right on the first try.

Here are some other points that form a vertical line. I want you to use the same sort of approach we used last time. Try and label these points and then figure out what's the equation for the line. Then I also want you to find the slope. You can put the number in this box or if you think the slope is undefined, write the letter u. Good luck.

If I label the points A, B and C, I can write their coordinates. Notice that the x-coordinate is always the same, it's always 2. It doesn't matter the value of y. We just know that the x value is 2. For every point on this line, we're 2 units to the right of the origin. For the slope, it's rise over run, and we know that vertical lines have tons of rise but zero run. Uh-oh. Zero in the denominator? That means it's undefined, so u. I hope these patterns are helping you figure out the equations.

Vertical and horizontal lines are somewhat tricky. They go against our intuition. When we think of vertical lines, we usually think about the variable y. But the equation is actually x equals. It's because the x-coordinate for every point on this line was 2. When we usually think about the horizontal direction, we think about the variable x. But horizontal lines are always in the form, y equals. All the points on this line have a y-coordinate of negative 3, that's why it's y equals. When math goes against our intuition, it's helpful to stop and think, what's true for all the points? What is this line really saying? The coordinates of each point can be a lot of information, and they help us find the equations of the lines.

Let's continue writing some equations, but this time, let's try generating equations from words instead of a graph. I want you to write the equation above line that has a slope of negative 2 and that passes through the point 0 negative you'll be able to figure out. You can write your equation here.

I know the slope is negative two, so this is my m. This point is actually the y intercept. The x coordinate is zero, and then we move down four units on the y-axis, so the b is negative four. So I make the subsitution for m and b, y equals negative 2x plus negative 4. When I simplify this, I can just write y equals negative 2x minus 4. This is the equation of my line.

What if instead I didn't give you slope and the Y intercept, but I just gave you by the difference of the x values. We can calculate the slope to be negative 3 halfs. And, again, we don't have the y intercept so, we need a different approach to actually write our equation. We're going to use something called point-slope form. We can have one point be x y since both variables appear in linear equations. At least, equations that are not horizontal or vertical lines. If we multiply both sides by the denominator x minus x 1 we can reduce this to make 1. This is our point-slope form. We call it point slope form since we have, well, a point and the slope. We know these two points define a line so let's use the same process to find the equation of our line. I know the slope is negative one of these points. So what do you think? Which point is best to use for x1 y1? Should we use this point, or should we use this point. Which one do you think would be easier to substitute in here?

I think this point is easier. I know the x and y coordinate are both positive, so it's going to be easier to plug in, to this equation. I won't have to deal with two negative signs. So for x1, I'm going to put 1. And for y1, I'm going to put 2. Keep in my mind, it doesn't matter which point you use. You'll still wind up with the same equation. I'm going to multiply both sides by x minus 1. X minus 1 divided by x minus 1 makes 1. So, I have x minus 1 times negative 3 halves. I move the negative 3 halves in front of the x minus 1 because we usually list constants outside of parentheses first. Im going to multiply both sides of the equation by 2. I only need to multiply the negative 3 halves by 2, because this number will be distributed to both terms. However, on this side, I need to multiply 2 to each of these terms. 2 times negative 3 halves is negative 3. And by distributing this two, I get 2y and negative 4. Continuing with distribution I have negative 3x plus 3 and 2y minus negative 4 on the right. I'm going to try and solve this equation for y, I'm going to get it in slope intercept form. Adding 4 to both sides, I get 2y equals negative 3x plus 7. I added 4, because remember, I'm trying to get the y alone. I never used to seen y on the left side of the equation but this is okay. We can rewrite our equation in the end with the y on the other side and this on the right. Finally, we divide each term by 2 and we get y is equal to negative 3 halves x plus 7 halves. This is our equation in slope-intercept form. Notice I've just written the y on the left side, and this expression on the right side. Here's my equation. There's also a quick check here. We want to make sure our slope is negative 3 halves. Is that what we found before? We can check and see the slopes are both negative 3 halves. So, yeah, we have at least have this part of the problem right.

The key thing to remember is point slope form. If we understand the equation for slope then we know point slope form. We can fix one of the points to be x and y and then we just plug in another point once we know the slope. This can lead us to the equation.

Find the equation of a line with a slope of 1 4th that passes through the point 6, negative 1. I want you to write your answer in standard form. Remember, ax plus by equals c is standard form. A, b and c are just numbers.

For this problem, we start with our slope formula. And, we already know the slope, it's 1 4th. We also want to fix one of these points to be x and y, and then this other point will be x 1 y 1. So I have the slope is 1 4th, I fix the point x,y, and then this last point is 6 negative 1. So 6 for the x, negative 1 for the y. After substituting the values, here's the equation I get. You can think about cross multiplying here, or just multiplying by the least common denominator, which would be 4 times x minus 6. The 4 is reduced to 1, and the x minus 6, and the x minus 6 also reduced to 1. 1 times x minus 6, is well, just x minus 6. And then, I have 4 times the quantity y plus 1. So I need to distribute this four to both terms. So 4 y plus 4. I want to write this equation in standard form, like this. So I need to move this constant to the right and the 4 y to the left. I'm going to add 6 to both sides to get a constant on the right so I get x equals 4 y plus 10. And now I subtract 4 y from each side. So I get x minus 4 y equals 10. I just list these terms here because they are not like. If you got x minus 4 y equals 10, amazing work. I know this isn't the easiest problem and we haven't even covered how to do this part here. So, if you got it, excellent work.

Try this one out. Write the equation of a line that passes through the point 3, -1 and -2, 5. This time write your answer in slope intercept form, y equals mx plus b.

Parallel lines never intersect, and they share a unique property. What are the slopes for these two parallel lines?

Well, both the equations are in y equals mx plus b, or slope-intercept form. So, this slope is negative 2 3rds and this slope is negative 2 3rds.

It turns out any parallel lines will have the same slope. I can draw more of these lines on this graph and all of these lines have a slope of negative 2 move to another point in the line, we travel two units down and three units right. It's the same slope for both lines. Pretty cool.

We've seen parallel lines. Now, let's check out perpendicular lines. Here are 2 lines that are perpendicular. Before we talk about perpendicular lines, I want to see if you can find these equations. Use the graph to write the equations in slope intercept form. For this line, write the answer here. And for this line, write it here.

For this blue graph, I know the y intercept is at 1. So my y intercept, or my b is positive 1. To find the slope, we just move from this point to another point in the line. I'm going to move to this point. So I move up 4, and then right 3. These are both positive directions. So I have 4 3rd. For the red line, I start at the y intercept, which is at negative 3. To find the slope, I can go to another point. I see another point there. So I travel down 3 units, and write 4 units. Here, this direction is down. So it's negative 3, and the 4 is positive, since I moved to the right. So, negative 3 4th. Congrats if you got both of those right.

You might be wondering what perpendicular means. Well, it means that the lines cross at a right angle. We've seen this before with complementary angles, they formed a 90 degree angle together. With complementary angles, we had two angles that added up to 90 degrees. In this case, we have two lines that create four 90 degree angles. That's what it means to have perpendicular lines.

If we take a closer look at the slopes, we can see that these slopes are actually reciprocals of one another and opposite in sign. The slopes of perpendicular lines are always opposite reciprocals of each other. The reciprocal of 4 3rds is 3 4ths, and then we make it negative, we change the sign. Now that we know a little bit more about parallel and perpendicular lines, lets put this knowledge to use.

For this skills check, I want you to determine if each pair of lines is perpendicular, parallel, or neither. So, if you think these 2 lines are perpendicular, you would write a in this box. If you think they're parallel, write b. And if you think they're neither, write c. Give these problems some thought, and make sure you think about what you're looking for. What does it mean to be perpendicular? What does it mean for lines to be parallel? Good luck.

Remember, for perpendicular lines, they're negative reciporicals of one another. So, we're looking for slopes that are y over x, and negative x over y. While the slope of these two lines are opposite in sign, they're not reciprocals. I know the reciprocal of 2 is 1 half. So these lines can't be perpendicular. Parallel lines have the same slope. These aren't the same, so they're not parallel. It must be neither. For the second pair of lines, we know that they must be parallel. The slope is 3 in both cases, so b. Here, I do have opposite reciprocals of one another. The reciprocal of 2 5ths is 5 halves and the slopes are opposite in sign. I don't know what these lines look like, but I know that they cross at a 90 degree angle because they're perpendicular. This last one is really tricky. We need to remember that there is a 1 as the co-efficient in front of the variables. The reciprocal of 1 or 1 over 1 is still lines are perpendicular. If you got some of these wrong, be careful that you weren't looking at the b or the y intercept. The y intercept doesn't determine whether or not lines are perpendicular or parallel. It's only the slope, the m.

Here are three more sets of equations. We know these equations really represent lines, so tell me, is each pair perpendicular, parallel, or neither? Put the letter of the correct choice here, and as a hint, notice that all of these equations are in standard form. We can find the slope in the standard form using negative A over B. So what do you think for each of these?

For this first equation, I have an A of negative 2, and a B of positive 3, so my slope is 2 3rd. For the second equation, the A is positive 3 and the B is positive 2. So I get a slope of negative 3 halves. The slops are opposite in sign, and reciprocals of one another, so they must be perpendicular. For the second pair of lines I know that the first slope is 3 4th and the second slope is 4 3rd.. These slopes are reciprocals of one another, but they're not opposite in sign so they can be perpendicular and are not the same so they can be parallel so they must be neither these lines will just intersect somewhere. For the last two equations, we can find that their slopes are both negative 2. So they're the same. They must be parallel, b. Through the last two skills checks, I hope you've gained a better understanding of what it means for lines to be perpendicular and parallel.

What's really interesting about the last set of equations is that you can tell they are parallel just by looking at them. Remember, in standard form, we get the slope by finding negative A over B. So, we only need to look at the left side of the equation. Since these two equations have the same left side, that means, their slopes have to be equal. Equations with equal slopes are either the same line or they're parallel lines. We know these two lines can't be the same, because the right-hand side is different, so these lines must be parallel.

Lets continue using our knowledge of parallel and perpendicular lines to solve this problem. I want you to find the equation of a line that passes through the point 1,2 and thats a parallel to this line. Before we find that equation, lets break it down into smaller steps. Whats the slope of this line, slope of our line be? Put the slopes in these boxes.

We know for a standard form the slope is negative A over B, so our slope is negative 3 halves. If we want a line that's parallel to this line, well, our slopes have to be the same, so this also has to be negative 3 halves. Way to go if you got those right.

So, we know the slope of our new line must be negative 3 halves and we have a point, so that means we can use point slope form to find the equation. Now that we know the slope and the point, we can use our point slope form to figure out the equation of the line. So I want you to figure this out. What's the equation of our line in standard from? Write your answer here.

With point slope form, we have one point that's fixed and then we'll plug in the coordinates into this other point. 1 for the x and 2 for the y. Notice that I'm always using parentheses when I plug in. That's because I don't want to make a mistake with my negative signs. Since both these numbers are positive, I can drop the parenthesis. And on the other side of the equation, we have our slope, negative 3 halves. Now, I'm going to cross-multiply. Then, I carry out the distribution. Negative 3 times x is negative 3 x, and negative 3 times negative this line up here and now I'm going to solve for standard form. I want the constants on the right, and the x and y on the left. If you got negative 3 x minus 2 y equals negative 7, excellent work. Notice how every single term is negative. Well, we could just multiply every term by negative 1 and make the entire equation and all it's terms positive. So, an alternative answer would be just been multiplied by a negative 1.

This time we're going to find the equation of the line that's perpendicular to this line. It still passes through the same point 1, 2. We knew the slope of this line was negative 3 halves. So, what was the slope of our line be if it's going to be perpendicular to this line? Write your answer here.

I know perpendicular lines have slopes that are opposite in sine. And they're reciporical of one another. This is negative 3 halves. So I know my slope must be positive. And it must be the reciprocal of 3 halves. So it must be 2 3rds. If you put positive 2 3rds, nice work.

So now that we have the slope, and a point on our line, I want you to find the equation of our perpendicular line. Write the equation in standard form here. Good luck.

We start with our point-slope form. Our slope is 2 3rds, x 1 is 1, and y 1 is distribute, and I get 2x minus 2, equals 3y minus 6. I've written this equation again, up here, and now I put it in standard form. I add 2 to both sides, to get equal to negative 4. Here's our perpendicular line. That's some serious algebra, so if you got it right, excellent work.

Well you have learned a lot about lines. We've learned about slope. We've learned about equations of lines. We've learned about horizontal and vertical lines. Let's do some practice problems, shall we? For this first practice problem, I want you to find the equation of the line. The first line is a vertical line that goes through the point negative 5, 6. You can put your answer here. For the second one, I want you to find the equation of a horizontal line that goes through the point negative 5, 6. And you can put your answer in this box. If you're not sure how to solve these problems, feel free to go back and review the videos on horizontal and vertical lines before you answer.

The equation of the vertical line that goes through the point negative 5 6 is x equals negative five. What do we know about vertical lines? We know that vertical lines look like this. We know that vertical lines have undefined slope, and we also know that the equation of a vertical line is always going to be x equals a constant. Since this vertical line goes through the point Negative 5, negative 5. So, the equation of the vertical line through the point negative 5, line that goes to the point negative 5, 6 is Y equals 6. What do we know about horizontal lines? We know they go left to right, or right to left. We know that the slope is 0, and we know their equation looks like y equals a constant. Now if we know that this horizontal line goes to the point negative 5,6, if I plot the point negative 5,6 on this horizontal line what I notice is no matter what value x is, y is always going to be 6. So I get the equation y equals 6. If you got these answers right, great work.

Okay, let's try to find a couple more equations of lines. For these two problems we're given slightly different information. For the first problem, I want to find the equation of a line that goes through the point 3 negative 4, that has the slope of 0. You can put your answer in this box. For the second problem, I want you to find the equation of a line that goes through the point 3 negative

Did you get y equals negative 4 for the first answer, and x equals 3 for the second answer? If so, way to go! Let's see how we got those answers. Similar to the last practice problems that we did, for the first problem we know that a slope of 0 means a horizontal line, and a horizontal line looks like this. And we know the equation is going to be y equals sum constant. If I put a point on the line, 3 negative 4. I know that, no matter what x equals, y is always going to equal negative 4. Which is why we get the equation, y equals negative 4. For the second problem, the key here is the undefined slope. We know that undefined slope means its a vertical line. And we know a vertical line looks like that, and I also know the equation of a vertical line is x equals a constant. When I plot the point on my vertical line, 3 negative 4, I can see that no matter what the value of y is, the x value always stays at 3. So, the equation is x equals negative 3.

Now that you've had some practice finding equations of horizontal and vertical lines, let's try some other kinds of equations. For this problem I want you to find the equation of the line that goes through the points negative 2, negative 3 and 2, 7. I want you to write your answer in this box in standard form. Remember standard form is ax plus by equals c. Now remember you're going to have to find the slope between these two points before you can find the equation of the line. Just try your best.

If you got 5x minus 2y equals negative 4, you did a great job. Let's see how we got that answer. Remember, the first thing we have to do is find the slope between these two points. We know that the equation for slope is y2 minus y1 over x2 minus x1. Now, to make sure I put the values in the right places, I always label one of the my points, x1, y1 and the other point, x2, y2. That way, I make sure I plug them into the right places. So when I find the slope, y2 minus y1 is, y2 is 7, minus y1 is negative 3 over x2 which is 2 minus x1 which is negative 2. 7 minus negative 3 is 10. 2 minus negative 2 is 4 and 10 over 4 can be reduced to 5 halves. So we know that the slope of our line. So we know the slope of our line is 5 halves. Now that we know the slope, we can find the equation of the line. In order to find the equation of the line, we use the equation m equals y minus y1 over x minus x1. Now I plug in what I know. The slope is 5 halves, and that's going to equal y minus y1, y1 is negative 3 over x minus x1 which is x minus negative 2. Let's go ahead and write these again. 5 halves equals y minus negative 3 is y plus 3 and x minus a negative 2 is x plus I have 5x, plus 5 times 2 is 10, equals 2y plus 2 times 3 is 6. Since I want this equation in standard form, I'd need the x and y terms on the left side and the constant on the right side. So, I can subtract 10 from both sides to move the constant over to the right. And I have 5x equals 2y minus 4. 6 plus negative final answer of 5x minus 2y equals negative 4.

This next practice problem should be a little bit easier than the last one. We want the equation of the line through the point 0,2 and this time the slope is given to us and the slope is 3 4ths. Once again, write your answer in standard form and in this box.

Did you get 3 x minus 4 y equals negative 8? If so, that is awesome. Now what makes this problem easier than the last one? Well first, the slope is given, so we don't have to calculate the slope. The other is if we look at the point 0, 2. The point 0, 2 is actually the y intercept because x is 0 and y is 2. Now if we know the slope and we know the y intercept, we can very easily use a slope intercept form and write y equals mx plus b which equals 3 4th times x plus 2 which is the y intercept. Since we want to write the equation in standard form, I'm going to go ahead and multiply the equation through by 4. Giving me 4y equals 3x plus 8. And if I want to write this in standard form. I need to move the 3x over to the other side. Which would give me negative 3x plus 4 y equals they're actually the same. If I were to take this equation and multiply the whole equation through by negative 1, I'm going to get negative 1 times negative times positive 8 gives me negative 8. So as you can see this equation negative Our convention typically use a math is that we do not have a negative coefficient to our x variable.

This next practice problem, we're going to go back to finding the equation of a line through two points. But in this case I want you to write the answer, in slope intercept form here.

The answer here is y equals negative 3 halves x plus 7 halves. Now remember, just because you see fractions, doesn't mean it's wrong. Let's see how we got that answer. Again, this is a two-part question. First, we have to find the slope. And then, we have to find the equation. So, first, let's find the slope. I know the equation of slope is y2 minus y1 over x2 minus x1. I'm going to go ahead and label my points, this is x1, y1 and this is x2, y2. Again, this helps me make sure I put the numbers in the right places. So, the slope y2 is negative negative 1 minus 2 is negative 3 over 3 minus 1 is 2. So, we know the slope is negative 3 halves, which is why you see negative 3 halves x in the equation. But, how did we get 7 halves? There's a couple different ways to do this. We also know that the slope is equal to y minus y1 over x minus x1. So, our slope was negative 3 halves and that's going to equal y minus y1, y minus 2 over x minus x 1, which is 1. And again, I can cross multiply. When I cross multiply, I have negative 3 times x minus 1 equals 2 times y minus 2. Now, I have to distribute. After I distribute, I have negative 3x. Negative 3 times negative 1 is positive 3 equals 2y. 2 times negative 2 is negative 4. To put this in slop- intercept form, to put this equation in slope-intercept form, the first thing I'm going to do is subtract 3 from both sides, and I get negative 3x equals 2y minus 7. If I just go ahead and add 4 to both sides, I get negative 3x plus 7 equals 2y. And I can still divide everything through by 2. And I get y equals negative 3 halves x plus 7 halves. And we know that negative 3 halves x plus 7 halves equals y, is the same as y equals negative 3 halves x plus 7 halves. Now, we're going to try a couple of problems with parallel and perpendicular lines.

We've had practice finding the equations of horizontal and vertical lines, as well as finding the equations of lines given two points or given a point and a slope. This last problem, we're going to look at parallel and perpendicular lines. So, this last problem, we want to find the equation of a line. The first line goes through the point 2,4 and is parallel to the line 3x plus y equals 7. For the first problem, find the equation of the line that goes through the point find the equation of the line that goes through the point 2,4 and is perpendicular to the line 3x plus y equals 7. You can put your answers here and put them in slope-intercept form.

These last two practice problems bring together a lot of what we learned in unit then find the slope of a line that is parallel to that line. So let's find the slope of the line 3x plus y equals 7. In order to find the equation of a line that's parallel to this line. I first have to find the slope of this line. If I simply subtract 3x from both sides I get that y equals negative 3x plus 7, which means the slope of this given line is negative 3. I know that parallel line if I want to find the slope of a line that's parallel to the given line. I know that parallel lines have the same slope. So, what I write is m parallel equals the original slope, which equals negative 3. This is just a shorthand notation to indicate parallel. So, the slope of the parallel line is equal to the slope of the original line which is equal to negative 3. To find the equation of the line, I want to label my points x1 and y1. And I can use the formula m equals y minus y1 over x minus x1. Since I know my slope is negative 3, I can write negative 3 over 1. Any whole number can be written as a fraction with 1 as the denominator. And that's going to equal y minus y1 is 4 over x minus x1 is 2. When I cross multiply, I have negative 3 times x minus 2 equals 1 times y minus negative 3 times negative 2 is positive 6 equals y minus 4. Since I want the answer in slope intercept form. If I just add 4 to both sides, I get negative 3x plus 10 equals y. And we know that's the same as y equals negative 3x plus 10. And, one quick check, we know that the slope of the parallel line is the same as the slope of the original line. And, since we put the answer in slope intercept form I can verify that it has the same slope. Now what about the perpendicular line? Since we want to find the equation of the perpendicular line. To 3x plus y equals 7. Once again, I start with the original slope which is negative 3, but I want the perpendicular slope. So the perpendicular slope is the negative reciplical of the original slope, which is going to be negative 1 over the original slope was negative 3. So a negative time a negative is a positive, so the slope of the perpendicular line is positive 1 3rd. Now that I know the slope, I can use my equation m, y minus y1, over x minus x1, which gives us 1 again. When I multiply x minus 2 times 1, I'm just left with x minus 2 and I have 3 times y and 3 times negative 4 is negative 12. Since I want the equation in slope intercept form I need y equals, so I'm going to add 12 to both sides and I get x plus 10 equals 3y and when I divide everything through by 3, I get y equals x over 3 or 1 3rd x plus 10 over 3. And that's the same as y equals 1 3rd x plus 10 3rd. And again, I can verify my slope is supposed to be 1 3rd, because it's the negative reciprocal of negative 3. And since we put it in slope intercept form, the slope is 1 3rd. If you got both these answers right, I would say you've learned a lot about lines.