WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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In today's lesson, we will be discussing slope.
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Recall the definition of slope: the slope, m, of the line passing through a point (x₁,y₁),
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and another point (x₂,y₂), is given by:
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the slope equals the change in y, over the change in x (the change in the y coordinates, over the change in the x coordinates).
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Therefore, if you have any two points on a line, you can use that to find the slope.
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For example, imagine that you are given the graph of a line that looks like this, and you are asked to find the slope.
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Well, I have a point right here at x is 0, so my (x₁,y₁) point is going to be x is 0, y is 2.
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I have a second point over here, and I am going to call this (x₂,y₂); and here, x is -2, y is 0.
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So, I am using the intercepts; but I could have used any other point on this line.
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OK, an important point is that I could have assigned either one of these as...
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I could have said this is (x₁,y₁), and this is (x₂,y₂).
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It doesn't matter, as long as you are consistent (I can't say (x₁,y₂), (x₂,y₁)).
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As long as I am consistent, and I follow that consistency when I am subtracting, I am fine.
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So, I arbitrarily assigned this as (x₁,y₁), and this point as (x₂,y₂).
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So, I found two points on the line; now, my y₂ value is 0, and I am going to subtract my y₁ value (which is 2) from that.
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The x₂ value is -2; my x₁ value is 0, so minus 0; so here, the slope equals -2 over -2, or 1.
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So, I could have used any two points on this line, and found the difference between those x and y coordinates--the change in y over the change in x.
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So, what this is telling me is that...we say m = 1, but what we really mean is m = 1/1; think about it that way, and say that.
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For every increase in y by 1, I am going to increase x by 1; increase y by 1, increase x by 1.
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So, I may be given a line and asked to find the slope; or I may directly be given a set of points and asked to find the slope.
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So, if I were given two points on the line, but not given the line itself, or a graph, or anything, I could easily find the slope.
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This is telling us the vertical change over the horizontal change.
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OK, some cases to keep in mind when you are dealing with slope: a horizontal line has a slope of 0.
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And let's think about why this is so: consider a horizontal line up here at, say, 4; OK, so this is 4.
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And if I look at any point--for example, right here--here, x is 1; y is 4; and then maybe I look over here at this point; here x is 3; y is still 4.
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Now, recall that the slope formula is the change in y over the change in x.
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So, I have my two points; and I am going to call this (x₁,y₁), (x₂,y₂).
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Again, I could have done it the other way around.
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All right, so I have y₂ (that is 4), minus y₁;
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and then I have my x₂ (which is 3), minus x₁ (which is 1).
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It actually doesn't matter what I have down here, because, since I have a horizontal line, the y-values everywhere here are going to equal 4.
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So, 4 - 4 is going to be 0; therefore, the slope is 0.
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So, a horizontal line has a slope of 0, because the y-values are the same at every point, so the difference in y is going to be 0.
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OK, vertical lines are also a special case: and they have an undefined slope.
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For example, let's look at a vertical line right here at x = -3.
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OK, if I look right here at (I could pick any point)...I am going to pick this point, x is -3; y is 1.
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Then, I will pick another point right here; let's say x is -3; y is -2.
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So, I have my (x₁,y₁), and my (x₂,y₂).
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Again, slope is change in y over change in x; change in y is y₂ (that is -2), minus 1, over change in x (-3 - -3).
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This is going to give me -3 over...-3 minus -3 is -3 plus 3; that is over 0.
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And this is not allowed; this is undefined.
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We say that a vertical line has an undefined slope because, since the change in x is 0,
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because x is the same at every point, what you end up with is a denominator that is 0, and that is undefined; that is not allowed.
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So, for a horizontal line, the slope is 0; for a vertical line, slope is undefined.
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A couple of other things to keep in mind regarding slope: one is that in a line that rises to the right (such as this one), m is a positive number.
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So, it is increasing to the right; the values are increasing.
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If you have a line that falls to the right, such as this one, here the slope is positive; in this case, the slope is negative.
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Rises to the right--slope is positive; falls to the right--slope is negative.
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A horizontal line has a slope of 0; a vertical line has an undefined slope.
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Parallel lines: two lines are **parallel** if and only if they are vertical or have the same slope.
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Let's first check out the case where the lines are not vertical, but they are still parallel.
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If I have two lines that are parallel, it is going to look like this, for example.
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These are going to have the same slope; so if this is my line 1 and line 2, m₁ will equal m₂--the slopes will be equal.
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And that is because they are changing at the same rate--these two lines are changing at the same rate, so they never intersect.
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Now, we set vertical lines aside; we put them separately; we don't say they have the same slope.
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And remember that the reason is because a vertical line has an undefined slope.
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So, I can't say these two have the same slope, because their slope is undefined.
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However, we recognize that they are still parallel; so we say that two lines are parallel
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if they are vertical (that is a separate case), or if they have the same slope; but both of these represent parallel lines.
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Perpendicular lines: two non-vertical lines are **perpendicular** if and only if the product of their slopes is -1.
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So, "perpendicular" would tell me that these two lines intersect at right angles.
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And if this is my line 1 and line 2, and then I have a slope m₁ and m₂, what this is stating
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is that the product of m₁ and m₂ is -1.
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Let's look at an example: let's say I have an m₁ that is equal to 4, and I say that it is perpendicular to a line L₂.
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So, this is line 1, and this is line 2--the perpendicular line.
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I can find the slope of the perpendicular line, because I know that m₁ times m₂
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is -1 (since these are perpendicular lines), and I am given m₁.
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So, 4 times m₂ equals -1; therefore, the perpendicular line would have a slope of -1/4.
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So, the perpendicular line has -1/4 as the slope, and that is the negative reciprocal of m₁.
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Knowing the relationship between parallel lines and the slopes of parallel lines and the slopes of perpendicular lines
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can be helpful in answering problems and graphing lines; and we will see that right now with the examples.
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OK, in this first example, find the slope of the line passing through the points (-9,-7) and (-6,-3).
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Well, the definition of slope is the change in y over the change in x.
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And I could choose to assign this either way; but I am just going to go ahead
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and say this is (x₁,y₁), and this is going to be my (x₂,y₂).
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And it doesn't matter which way you assign, as long as you are consistent.
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So, I could have called this (x₂,y₂), and it would have been fine; we would have gotten the same answer.
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OK, y₂ is -3, and y₁ is -7, so minus -7; over x₂ (which is -6), minus x₁ (which is -9).
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Simplifying this: this is -3; a negative and a negative gives me a positive; and that gives me 4/3.
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So, the slope of a line passing through these points is 4/3, just using my slope formula.
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OK, Example 2: Graph the line passing through the point (-2,-1) with a slope of -2/3.
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Well, I am given the slope, and I am given a point on the line.
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So, let's start out with the point that I am given, which is (-2,-1); that is right here.
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Remember that the slope is the change in y over the change in x.
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So, if the change in y is -2, then I am starting out here at -1; I am going to go to -2, -3.
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For every 2 that y is decreased in value, x is going to increase by 3: 1, 2, 3; so, I am going to end up with this point right here.
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You could have also looked at this as this, and I could say that for every 2 y is increased, x is decreased by 3: 1, 2, 3, right here.
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The same thing--you could look at it either way.
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OK, so now, I am able to graph this line because I had a starting point right here, and I have the slope.
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So, I know, starting from here, how much the x is going to change and how much the y is going to change.
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So then, I simply connect these lines to give my graph.
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And notice that this line is decreasing to the right: and that fits with what I have, which is a negative slope.
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So, simply plot the point you are given; and then use the change in y over the change in x to find additional points on the line.
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OK, graph the line passing through (1,-3) and parallel to the graph of this equation.
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Well, I have my point; x is 1; y is -3; so, I have my starting point.
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I need my slope; and I am not directly given the slope; however, I am given a parallel line.
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And recall that parallel lines have the same slope.
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That means that, if I find the slope of this line, I am going to have the slope of the line that I am looking for.
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In order to find the slope, I need to be able to find the change in y over the change in x; so I need two points on this line.
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So, the slope is (y₂-y₁)/(x₂-x₁).
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Thinking of an easy way to find a couple points on this line, I am going to use the intercept method.
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And I could find these points and graph, or I could just find these points and plug them into my slope equation.
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First, I am going to let x equal 0 to find the y-intercept.
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OK, so this gives me -3y = 6, or y = -2.
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When x is 0, y is -2; when y is 0, let's go ahead and plug that in; 2x - 3(0) = 6, so this is 2x - 0 = 6; 2x = 6.
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Divide both sides by two; that gives me x = 3.
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OK, so when x is 0, y is -2; when x is 3, y is 0.
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Now, I can find the slope of the sign, because this can be my x₁;
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this can be my y₁; this can be my x₂; this can be my y₂; so let me find the slope.
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This is y₂, which is 0, minus y₁, which is -2.
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This is x₂, which is 3, minus 0, which is x₁.
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This is going to give me 0 minus -2 (0 plus 2), over 3 minus 0; so the slope is 2/3.
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Therefore, the slope of this line is 2/3, and since parallel lines have the same slope, the slope of the line I am looking for is also 2/3.
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Now, I was given this point, (1,-3); now I know that the vertical change is going to be an increase in y by 2;
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and there is going to be a change in x--an increase by 3--1, 2, 3; so I have another point.
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Now that I have this second point, I can go ahead and graph this line.
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And this is the line I was asked to graph, so I have completed what I have been asked to do.
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But just to look at the idea that this is indeed a parallel line, I am going to go ahead and plot this other line, as well.
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This is the line we were asked to find; this other line--when x is 0, y is -2; when x is 3, y is 0.
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And you can plot that out and see that, visually, this does look like a parallel line.
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And these two are changing at the same rate: the slope of both of these is 2/3.
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I was able to graph the line passing through this point and parallel to the graph of this line
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by finding the slope of this line, and then realizing that my parallel line is going to have the same slope, which is 2/3.
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Example 4: Graph the line passing through the point (-2,1) and perpendicular to the graph of this equation: 3x + 4y = 12.
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OK, I am given a point; and let's see, let's call this -2, -4, -6, -2, -4, -6, 2, 4, 6, 8; OK.
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-2 would be right here; and then 1 would be right about there; so that is my point.
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But I also need the slope: recall that the slopes of perpendicular lines--the product of those slopes is equal to -1.
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So, the product of perpendicular lines' m₁ and m₂ (I have two perpendicular lines;
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the slope of one is m₁, and the other is m₂)--their product is -1.
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So, I am going to call this first line line 1, and its slope is going to be m₁; I am going to find m₁.
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My line I am going to call line 2; and its slope is going to be m₂.
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I need to find the slope of this line, recalling that slope is the change in y over the change in x.
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So, in order to find this change, I need to find two points on this line.
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I am going to go for something easy, so I am going to let x equal 0, and I am going to plug that in here.
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I am just going to find the intercepts as my two points; but I could have found any two points on that line to get the slope.
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This is going to give me y = 3; when x is 0, y is 3.
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For my second point, I am going to let y equal 0; that is 3x + 4(0) = 12; so, if 3x = 12, x = 4.
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So now, I have two points; two points means I can find the slope.
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I am going to say that this is my (x₁,y₁), (x₂,y₂), so I can keep track of everything.
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y₂ is 0; y₁ is 3; so 0 - 3, over x₂ (which is 4), minus 0.
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OK, I have found m₁: I have found the slope of this line.
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And I know that the line I am looking for, which is going to have a slope m₂, is perpendicular to this line.
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So, I am going to go right here and say that m₁...I know that is equal to -3/4; and I am looking for m₂.
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m₁ times m₂ equals -1; so now, all I have to do is substitute this in.
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-3/4 times m₂ equals -1; I am going to multiply both sides by -4/3 to move this over to the right.
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And that is going to give me -1, times -4/3, or m₂ = 4/3.
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OK, now I found this slope, and I need to graph the line.
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I have my starting point here at (-2,1); and the change in y is going to be 4, so that is going to be 1, 2, 3, 4,
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for an increase in x of 3; so that is 1, 2, 3, right there.
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Increase in y by 4: 1, 2, 3, 4; increase x by 3: 1, 2, 3--so right about there; OK.
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I can go ahead and plot this line out.
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Reviewing what we did: we were given a point that the line passes through, and we needed the slope in order to graph it.
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We were told that this line is perpendicular to the line described by this equation.
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Therefore, we found the slope of this line; and we did that by finding two points in the line (the x- and y-intercepts) and plugging that into the slope formula.
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Once I found the slope of this perpendicular line, I wanted to find the slope of my line.
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And I did that by recalling that the product of the slopes of perpendicular lines equals -1.
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So, -3/4 times the slope of the perpendicular line is -1.
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I figured out that the line I am looking for has a slope of 4/3, and then I just took my point that I was given and increased y by 4, and then x by 3.
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That concludes this lecture about slope on Educator.com.