INSTRUCTORS Carleen Eaton Grant Fraser

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Slope

• The slope of a straight line is a measure of how steep the line is. Lines that increase as x increases have positive slopes, those that decrease as x increases have negative slopes.
• Slope is defined to be the change in the y-coordinates of two points divided by the change in the x-coordinates of those points.
• The slope is the same no matter which two points you pick to compute it.
• A vertical line has no slope, a horizontal line has a slope of 0.
• Non-vertical parallel lines have the same slope, non-vertical perpendicular lines have slopes whose product is – 1.

Slope

Find the slope of the line passing through the points ( − 3, − 1) and (1,2)
• Use the slope formula m = [Dy/Dx] = [(y2 − y1)/(x2 − x1)]
• m = [Dy/Dx] = [(y2 − y1)/(x2 − x1)] = [(2 − ( − 1))/(1 − ( − 3))]
• Simplify
m = [(2 − ( − 1))/(1 − ( − 3))] = [(2 + 1)/(1 + 3)] = [3/4]
Find the slope of the line passing through the points ( − 3,3) and ( − 1, − 1)
• Use the slope formula m = [Dy/Dx] = [(y2 − y1)/(x2 − x1)]
• m = [Dy/Dx] = [(y2 − y1)/(x2 − x1)] = [( − 1 − (3))/( − 1 − ( − 3))]
• Simplify
m = [( − 1 − (3))/( − 1 − ( − 3))] = [( − 4)/( − 1 + 3)] = [( − 4)/2] = − 2
Find the slope of the line passing through the points (2,3) and (5,2)
• Use the slope formula m = [Dy/Dx] = [(y2 − y1)/(x2 − x1)]
• m = [Dy/Dx] = [(y2 − y1)/(x2 − x1)] = [(2 − (3))/(5 − (2))]
• Simplify
m = [(2 − (3))/(5 − (2))] = [( − 1)/3] = − [1/3]
Graph the line passing through ( − 3,4) with slope m = [1/2]
• Step 1:Plot the point ( − 3,4)
• Step 2: From ( − 3,4) use the fact that slope = [change in y /change in x] = [rise/run] = [1/2] to plot the next point
• Go Up 1 and Right 2.
• Step 3 : Repeat again. From previous point, use slope to go Up 1 and Right 2.
• Step 4 : Draw the line
Graph the line passing through ( − 2, − 3) with slope m = 3
• Step 1:Plot the point ( − 2, − 3)
• Step 2: From ( − 2, − 3) use the fact that slope = [change in y /change in x] = [rise/run] = [3/1] to plot the next point
• Go Up 3 and Right 1.
• Step 3:Repeat again. From previous point, use slope to go Up 3 and Right 1.
• Step 4:Draw the line
Graph the line passing through (4, − 3) with slope m = − 2
• Step 1:Plot the point (4, − 3)
• Step 2: From (4, − 3) use the fact that slope = [change in y /change in x] = [rise/run] = [( − 2)/1] to plot the next point
• Go Down 2 and Right 1.
• Step 3:Repeat again. From previous point, use slope to go Down 2 and Right 1.
• Step 4:Draw the line
Graph the line passing through (3,4) parallel to the graph of 3x + 4y = 12.
• Recall that parallel lines have the same slope. Therefore, find the slope of the line 3x + 4y = 12
• and use the slope to plot additional points to graph the line.
• Find x - and y - interceps to use as points to find the slope m.
•  X-Intercept Y-Intercept set y=0 set x=0 3x + 4y = 12
•  X-Intercept Y-Intercept set y=0 set x=0 3x + 4y = 12 (4,0) (0,3)
• Use the slope formula to find m
• m = [(y2 − y1)/(x2 − x1)]
• m = [(3 − 0)/(0 − 4)] = − [3/4]
• Now that you know what the slope is, graph the next point starting at (3,4)
• Go Down 3 ,Right 4.
• Graph the line. Notice how it is indeed parallel to the other line.
Graph the line passing through (1,1) parallel to the graph of − 5x + 3y = 15.
• Recall that parallel lines have the same slope. Therefore, find the slope of the line − 5x + 3y = 15
• and use the slope to plot additional points to graph the line.
• Find x - and y - interceps to use as points to find the slope m.
•  X-Intercept Y-Intercept set y=0 set x=0 -5x+3y = 15
•  X-Intercept Y-Intercept set y=0 set x=0 -5x+3y = 15 (-3,0) (0,5)
• Use the slope formula to find m
• m = [(y2 − y1)/(x2 − x1)]
• m = [(5 − 0)/(0 − ( − 3))] = [5/3]
• Now that you know what the slope is, graph the next point starting at (1,1)
• Go Up 5 ,Right 3.
• Graph the line. Notice how it is indeed parallel to the other line.
Graph the line passing through (3,4) and perpendicular to the graph of 3x + 4y = 12.
• Recall that perpendicular lines have the property that the product of their slopes = − 1.
• Therefore, find the slope of the line 3x + 4y = 12
• and take the negative reciprocal to find the slope of the perpendicular line.
• Use the slope to plot additional points to find the graph.
• Find x - and y - interceps to use as points to find the slope m.
•  X-Intercept Y-Intercept set y=0 set x=0 3x + 4y = 12
•  X-Intercept Y-Intercept set y=0 set x=0 3x + 4y = 12 (4,0) (0,3)
• Use the slope formula to find m
• m = [(y2 − y1)/(x2 − x1)]
• m = [(3 − 0)/(0 − 4)] = − [3/4]
• Now that you know what the slope of the first line is, take the negative reciprocal to find the slope of the
• perpendicular line.
• m2 = − [1/(m1)] = − [1/( − [3/4])] = [4/3]
• Use m2, the slope of the perpendicular line to graph the next point starting at (3,4)
• Go Up 4 ,Right 3.
• Graph the line. Notice how it is indeed perpendicular to the other line.
Graph the line passing through (1,1) and perpendicular to the graph of − 5x + 3y = 15.
• Recall that perpendicular lines have the property that the product of their slopes = − 1.
• Therefore, find the slope of the line − 5x + 3y = 15
• and take the negative reciprocal to find the slope of the perpendicular line.
• Use the slope to plot additional points to find the graph.
• Find x - and y - interceps to use as points to find the slope m.
•  X-Intercept Y-Intercept set y=0 set x=0 -5x + 3y = 15
•  X-Intercept Y-Intercept set y=0 set x=0 -5x + 3y = 15 (-3,0) (0,5)
• Use the slope formula to find m
• m = [(y2 − y1)/(x2 − x1)]
• m = [(5 − 0)/(0 − ( − 3))] = [5/3]
• Now that you know what the slope of the first line is, take the negative reciprocal to find the slope of the
• perpendicular line.
• m2 = − [1/(m1)] = − [1/([5/3])] = − [3/5]
• Use m2, the slope of the perpendicular line to graph the next point starting at (1,1).
• Go Down 3, Right 5.
• Graph the line. Notice how it is indeed perpendicular to the other line.

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Slope

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Definition of Slope 0:07
• Change in Y / Change in X
• Example: Slope of Graph
• Interpretation of Slope 3:07
• Horizontal Line (0 Slope)
• Vertical Line (Undefined Slope)
• Rises to Right (Positive Slope)
• Falls to Right (Negative Slope)
• Parallel Lines 7:18
• Example: Not Vertical
• Example: Vertical
• Perpendicular Lines 8:31
• Example: Perpendicular
• Example 1: Slope of Line 10:32
• Example 2: Graph Line 11:45
• Example 3: Parallel to Graph 13:37
• Example 4: Perpendicular to Graph 17:57

Transcription: Slope

Welcome to Educator.com.0000

In today's lesson, we will be discussing slope.0002

Recall the definition of slope: the slope, m, of the line passing through a point (x1,y1),0007

and another point (x2,y2), is given by:0015

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).0020

Therefore, if you have any two points on a line, you can use that to find the slope.0030

For example, imagine that you are given the graph of a line that looks like this, and you are asked to find the slope.0037

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 2.0053

I have a second point over here, and I am going to call this (x2,y2); and here, x is -2, y is 0.0065

So, I am using the intercepts; but I could have used any other point on this line.0076

OK, an important point is that I could have assigned either one of these as...0081

I could have said this is (x1,y1), and this is (x2,y2).0086

It doesn't matter, as long as you are consistent (I can't say (x1,y2), (x2,y1)).0090

As long as I am consistent, and I follow that consistency when I am subtracting, I am fine.0096

So, I arbitrarily assigned this as (x1,y1), and this point as (x2,y2).0103

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.0107

The x2 value is -2; my x1 value is 0, so minus 0; so here, the slope equals -2 over -2, or 1.0120

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.0138

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.0147

For every increase in y by 1, I am going to increase x by 1; increase y by 1, increase x by 1.0156

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.0167

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.0175

This is telling us the vertical change over the horizontal change.0182

OK, some cases to keep in mind when you are dealing with slope: a horizontal line has a slope of 0.0187

And let's think about why this is so: consider a horizontal line up here at, say, 4; OK, so this is 4.0198

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 4.0212

Now, recall that the slope formula is the change in y over the change in x.0229

So, I have my two points; and I am going to call this (x1,y1), (x2,y2).0239

Again, I could have done it the other way around.0246

All right, so I have y2 (that is 4), minus y1;0249

and then I have my x2 (which is 3), minus x1 (which is 1).0255

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 4.0261

So, 4 - 4 is going to be 0; therefore, the slope is 0.0273

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 0.0281

OK, vertical lines are also a special case: and they have an undefined slope.0291

For example, let's look at a vertical line right here at x = -3.0298

OK, if I look right here at (I could pick any point)...I am going to pick this point, x is -3; y is 1.0306

Then, I will pick another point right here; let's say x is -3; y is -2.0315

So, I have my (x1,y1), and my (x2,y2).0322

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).0330

This is going to give me -3 over...-3 minus -3 is -3 plus 3; that is over 0.0349

And this is not allowed; this is undefined.0359

We say that a vertical line has an undefined slope because, since the change in x is 0,0367

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.0372

So, for a horizontal line, the slope is 0; for a vertical line, slope is undefined.0379

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.0393

So, it is increasing to the right; the values are increasing.0407

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.0413

Rises to the right--slope is positive; falls to the right--slope is negative.0428

A horizontal line has a slope of 0; a vertical line has an undefined slope.0433

Parallel lines: two lines are parallel if and only if they are vertical or have the same slope.0439

Let's first check out the case where the lines are not vertical, but they are still parallel.0446

If I have two lines that are parallel, it is going to look like this, for example.0451

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.0459

And that is because they are changing at the same rate--these two lines are changing at the same rate, so they never intersect.0468

Now, we set vertical lines aside; we put them separately; we don't say they have the same slope.0476

And remember that the reason is because a vertical line has an undefined slope.0483

So, I can't say these two have the same slope, because their slope is undefined.0492

However, we recognize that they are still parallel; so we say that two lines are parallel0496

if they are vertical (that is a separate case), or if they have the same slope; but both of these represent parallel lines.0501

Perpendicular lines: two non-vertical lines are perpendicular if and only if the product of their slopes is -1.0512

So, "perpendicular" would tell me that these two lines intersect at right angles.0522

And if this is my line 1 and line 2, and then I have a slope m1 and m2, what this is stating0535

is that the product of m1 and m2 is -1.0544

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 L2.0552

So, this is line 1, and this is line 2--the perpendicular line.0569

I can find the slope of the perpendicular line, because I know that m1 times m20575

is -1 (since these are perpendicular lines), and I am given m1.0584

So, 4 times m2 equals -1; therefore, the perpendicular line would have a slope of -1/4.0591

So, the perpendicular line has -1/4 as the slope, and that is the negative reciprocal of m1.0606

Knowing the relationship between parallel lines and the slopes of parallel lines and the slopes of perpendicular lines0618

can be helpful in answering problems and graphing lines; and we will see that right now with the examples.0624

OK, in this first example, find the slope of the line passing through the points (-9,-7) and (-6,-3).0633

Well, the definition of slope is the change in y over the change in x.0643

And I could choose to assign this either way; but I am just going to go ahead0651

and say this is (x1,y1), and this is going to be my (x2,y2).0654

And it doesn't matter which way you assign, as long as you are consistent.0660

So, I could have called this (x2,y2), and it would have been fine; we would have gotten the same answer.0663

OK, y2 is -3, and y1 is -7, so minus -7; over x2 (which is -6), minus x1 (which is -9).0668

Simplifying this: this is -3; a negative and a negative gives me a positive; and that gives me 4/3.0686

So, the slope of a line passing through these points is 4/3, just using my slope formula.0697

OK, Example 2: Graph the line passing through the point (-2,-1) with a slope of -2/3.0705

Well, I am given the slope, and I am given a point on the line.0717

So, let's start out with the point that I am given, which is (-2,-1); that is right here.0722

Remember that the slope is the change in y over the change in x.0730

So, if the change in y is -2, then I am starting out here at -1; I am going to go to -2, -3.0735

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.0744

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.0756

The same thing--you could look at it either way.0771

OK, so now, I am able to graph this line because I had a starting point right here, and I have the slope.0777

So, I know, starting from here, how much the x is going to change and how much the y is going to change.0783

So then, I simply connect these lines to give my graph.0789

And notice that this line is decreasing to the right: and that fits with what I have, which is a negative slope.0800

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.0806

OK, graph the line passing through (1,-3) and parallel to the graph of this equation.0820

Well, I have my point; x is 1; y is -3; so, I have my starting point.0829

I need my slope; and I am not directly given the slope; however, I am given a parallel line.0835

And recall that parallel lines have the same slope.0843

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.0853

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.0860

So, the slope is (y2-y1)/(x2-x1).0870

Thinking of an easy way to find a couple points on this line, I am going to use the intercept method.0876

And I could find these points and graph, or I could just find these points and plug them into my slope equation.0883

First, I am going to let x equal 0 to find the y-intercept.0890

OK, so this gives me -3y = 6, or y = -2.0901

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 = 6.0913

Divide both sides by two; that gives me x = 3.0933

OK, so when x is 0, y is -2; when x is 3, y is 0.0937

Now, I can find the slope of the sign, because this can be my x1;0946

this can be my y1; this can be my x2; this can be my y2; so let me find the slope.0953

This is y2, which is 0, minus y1, which is -2.0959

This is x2, which is 3, minus 0, which is x1.0967

This is going to give me 0 minus -2 (0 plus 2), over 3 minus 0; so the slope is 2/3.0975

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 2/3.0988

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;1003

and there is going to be a change in x--an increase by 3--1, 2, 3; so I have another point.1011

Now that I have this second point, I can go ahead and graph this line.1019

And this is the line I was asked to graph, so I have completed what I have been asked to do.1027

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.1032

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 0.1038

And you can plot that out and see that, visually, this does look like a parallel line.1047

And these two are changing at the same rate: the slope of both of these is 2/3.1052

I was able to graph the line passing through this point and parallel to the graph of this line1059

by finding the slope of this line, and then realizing that my parallel line is going to have the same slope, which is 2/3.1068

Example 4: Graph the line passing through the point (-2,1) and perpendicular to the graph of this equation: 3x + 4y = 12.1077

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.1090

-2 would be right here; and then 1 would be right about there; so that is my point.1110

But I also need the slope: recall that the slopes of perpendicular lines--the product of those slopes is equal to -1.1116

So, the product of perpendicular lines' m1 and m2 (I have two perpendicular lines;1127

the slope of one is m1, and the other is m2)--their product is -1.1133

So, I am going to call this first line line 1, and its slope is going to be m1; I am going to find m1.1137

My line I am going to call line 2; and its slope is going to be m2.1147

I need to find the slope of this line, recalling that slope is the change in y over the change in x.1153

So, in order to find this change, I need to find two points on this line.1160

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.1167

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.1172

This is going to give me y = 3; when x is 0, y is 3.1185

For my second point, I am going to let y equal 0; that is 3x + 4(0) = 12; so, if 3x = 12, x = 4.1189

So now, I have two points; two points means I can find the slope.1206

I am going to say that this is my (x1,y1), (x2,y2), so I can keep track of everything.1210

y2 is 0; y1 is 3; so 0 - 3, over x2 (which is 4), minus 0.1219

OK, I have found m1: I have found the slope of this line.1233

And I know that the line I am looking for, which is going to have a slope m2, is perpendicular to this line.1239

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 m2.1246

m1 times m2 equals -1; so now, all I have to do is substitute this in.1258

-3/4 times m2 equals -1; I am going to multiply both sides by -4/3 to move this over to the right.1265

And that is going to give me -1, times -4/3, or m2 = 4/3.1277

OK, now I found this slope, and I need to graph the line.1285

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, 4,1291

for an increase in x of 3; so that is 1, 2, 3, right there.1300

Increase in y by 4: 1, 2, 3, 4; increase x by 3: 1, 2, 3--so right about there; OK.1308

I can go ahead and plot this line out.1325

Reviewing what we did: we were given a point that the line passes through, and we needed the slope in order to graph it.1329

We were told that this line is perpendicular to the line described by this equation.1335

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.1340

Once I found the slope of this perpendicular line, I wanted to find the slope of my line.1351

And I did that by recalling that the product of the slopes of perpendicular lines equals -1.1356

So, -3/4 times the slope of the perpendicular line is -1.1363

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 3.1368

That concludes this lecture about slope on Educator.com.1382