WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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In today's lesson, we are going to talk about writing linear functions.
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And in particular, we are going to discuss two forms of these functions.
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The first form is the slope-intercept form; and this is a very useful form of the equation, because it can help you to graph a linear equation.
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The slope-intercept form of a line is y = mx + b, where m is the slope, and b is the y-intercept.
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Recall that the y-intercept is the point at which the graph of the equation (the line) intersects the y-axis.
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For example, if we look at an equation in this form, and it is given as y = -2x + 1, then the slope is -2, and the y-intercept, or b, is 1.
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This information alone allows me to graph the line.
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Before, we talked about graphing the line by finding a couple of points.
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And we, in particular, used the intercept method, where we found the x-intercept and the y-intercept, and graphed that.
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This time, I am going to use a slightly different method.
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So, here I have the y-intercepts: this is the y-coordinate where the graph of the line is going to cross the y-axis.
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That is going to be at y = 1; x will be 0, and y is 1.
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Now, I also have the slope; the slope is the change in y, over the change in x.
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And this is written as -2; but you can think of it in your mind as -2/1.
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For every 2 that y is decreased, x is increased by 1.
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And I am already thinking of what I expect this graph to look like.
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Recall that, if the slope is negative, the line is going to decrease going from left to right--it is going to go this way.
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If the slope is positive, the line is going to increase going from left to right.
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I am starting right here, and I am going to decrease y by 2--1, 2--for every increase in x by 1.
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So now, I have another point: 1, 2, and then increase x by 1.
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And I now have plenty to go ahead and graph this with.
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So, you can see how this form of the equation is very helpful in getting information about the line and actually graphing the line.
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The second form of these linear equations that we are going to look at today is point-slope form.
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The point-slope form of the equation of a line is y - y₁ = m (x - x₁).
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And you will see that this is related to slope and the slope formula.
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So, since the slope is (y₂ - y₁)/(x₂ - x₁), you can look and see that that is pretty familiar.
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Now, previously we talked about having two points: (x₁,y₁), and the other (x₂,y₂).
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Well, when we are working with the point-slope form, we have one point (x₁,y₁),
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and then the other point could be anywhere on the line; it is just another point (x,y) that isn't specified.
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So, look at how you could manipulate this to be similar to this,
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if I said, "OK, the slope is some point on the line, minus a given point, over some point on the line, minus the given point."
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Now, I am going to multiply both sides of this by (x - x₁)--both sides of the equation.
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These cancel out; and that is going to give me (x - x₁) times m equals (y - y₁).
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A little bit of rearranging: I am going to put the y portions on the left side of the equation, and the x and the slope on the right side of the equation.
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And you see, that gives me the point-slope form.
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And this is useful, because imagine if I am given some facts about a line, and I am told that the slope equals 4, and that this line passes through a certain point.
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And I am told that it passes through the point (-2,6).
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Knowing this, I can write the equation for this line in point-slope form.
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So, point-slope form: y - y₁ = the slope times (x - x₁).
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So, y is any other point on this line.
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y₁ is 6; slope is 4; x is another x point on this line that goes along with this y coordinate, minus x₁, which is -2.
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Simplifying this: a negative and a negative is a positive.
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What is helpful about this point-slope form is: with a little bit of work, I can put this into the slope-intercept form.
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Recall that the slope-intercept form (this is the point-slope form of the equation) is y = mx + b.
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I am going to multiply this out; this is 4x + 8; and I want to isolate y by adding 6 to both sides.
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So now, I have it in slope-intercept form; this is a very useful form of the equation.
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And again, this can allow us to graph the equation.
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Parallel and perpendicular lines: first, recall that parallel lines have the same slope.
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So, if I have two parallel lines, line 1 and line 2, and the first one has a slope of m₁, and the second one has a slope of m₂, those are equal.
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Perpendicular lines: the product of the slope of two perpendicular lines is equal to -1.
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The slope-intercept form and point-slope form can be used to solve problems involving parallel and perpendicular lines.
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And the reason is: if I am told that a line is parallel to another line, I have the slope.
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With a little bit more information, I can write the equation in slope-intercept form.
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Or I may be given the equation in slope-intercept form and told that a line is parallel to that line.
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The same with perpendicular lines: by having these forms of the equation, which involve slope,
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and knowing the relationship between two lines and their slopes, I can write the equation for the second line.
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I can graph the lines, and it is all about knowing the relationships between these two lines.
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For example, if I am told that the graph of a line is parallel to the line described by the equation y = 1/2x - 4,
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and I am also told that the line I am looking for passes through a point (4,-3),
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I can graph the line I am looking for; I can also write an equation for the line I am looking for.
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So, the graph of a line is parallel to the line y = 1/2x -4, and the line I am looking for passes through a certain point.
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Well, since parallel lines have the same slope, and I am looking at this, and it is in slope-intercept form, which is y = mx + b, I now have the slope.
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So, the slope equals 1/2; since these lines are parallel, I also have the slope of the line I am looking for, and it is 1/2.
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The line I am looking for passes through (4,-3); so that is (4,-3), and I am going to plot that out: (4,-3).
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And I have the slope, so I know that when I increase y by 1, I increase x by 2; therefore, I can plot the line.
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OK, and this is as expected, because it has a positive slope, so it is increasing as it goes to the right.
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The same holds true for perpendicular lines: for example, if I am told that the graph of a line is
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(this stands for perpendicular) perpendicular to the graph of the line defined by the equation y = 1/4x + 6,
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and the line passes through some point (say (1,2)), I can graph this line.
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And the reason I can graph it is that I know that this slope is 1/4, and I recall that the two slopes are related by this formula.
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So, if I am given a point on a line, as well as the knowledge and relationship between that line and a parallel or perpendicular line,
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I have the point; I can find the slope; I can graph the line.
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OK, first example: Find the equation in slope-intercept form of the line with the slope 2/3 and passing through (2,-4).
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Slope-intercept form is y = mx + b; and I am given slope, so I am given m = 2/3; so let's start from there: y = 2/3 x + b.
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Well, in order to write this out, I also need to find b; and b is unknown.
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However, I am given an x value and a y value; and since I have m, x, and y, I can solve for b.
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So, substituting in -4 for y, and 2 for x, now I can solve for b.
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So, -4 = 2 times 2...that is 4, so that is 4/3, plus b.
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Subtract 4/3 from both sides...equals b...and you could really think of this as -1 and 1/3; it might be easier to look at it that way.
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This is -5 and 1/3...equals b.
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OK, going back to the beginning here: y = mx + b, so y equals...m is given as 2/3; x; and then b is -5 and 1/3.
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Using the facts that I was given, I am able to write this in slope-intercept form.
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I was given the slope, and I was given a point on a line.
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The point on the line, and the slope: by plugging those into this equation, I could find the y-intercept; and therefore, I could write this out in slope-intercept form.
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Example 2: find the equation in slope-intercept form of the line passing through these two points.
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Slope-intercept form is y = mx + b, so I need to have the slope, and I need to have the y-intercept, in order to write this.
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The slope is the change in y, over the change in x; and I am given two points, so I can find the change in y over the change in x.
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I am going to call this (x₁,y₁), and this (x₂,y₂).
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OK, so m = y₂ (which is -3) minus -7, over -6 minus -2.
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-3...and that negative and negative becomes a positive, so plus 7, minus 6--a negative and a negative--that is plus 2.
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-3 + 7 is 4; -6 + 2 is -4; I have the slope now--the slope is -1.
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My next thing is to find b, which is the y-intercept; so, I have y = mx (-1x; we usually just write this as -x,
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but we are writing it out right now as -1x) + b.
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I need to solve for b, and I can do that, because I have some x and y values I can substitute in.
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You could choose either set of coordinates, either ordered pair.
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I am going to go ahead and choose this first one: y is -7; it equals (instead of writing -1, I am just going to write) -x (which is -2) + b.
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Solving for b: -7 equals...a negative and a negative is a positive, so that is 2 + b.
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Subtracting two from both sides, b equals -9.
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In order to write this equation, I needed to have my slope, which I do.
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And so, I have that m equals -1, and b equals -9.
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So now, I can go ahead and write this as y = -x (that is mx, where m is -1) + b (and that is -9): y = -x - 9.
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This is the equation for this line, written in slope-intercept form.
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Example 3: Find the equation in slope-intercept form of the line passing through the point (-2,-3), and parallel to the graph of y = 3x - 7.
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Slope-intercept form is y = mx + b: the first thing I need to do is to find the slope.
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I am not directly given the slope; however, I am told that this line is parallel to the line described by this equation.
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Parallel lines have the same slope, so if I know this slope, I know the slope for the line I am looking for an equation of.
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Well, this is in slope-intercept form; therefore, y = mx + b, so I have the slope.
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The slope of this line is 3, and the slope of the parallel line (which is my line) is also 3.
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So now, I have y = 3x + b; and in order to fully have this in slope-intercept form, I need to have b.
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Well, I have a point on this line, (-2,-3); so I am going to substitute in; I am going to let x equal -2 and y equal -3, and then solve for b.
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OK, so y is -3; x is -2; and I have the slope: so 3 times -2 plus b; that gives me -3; 3 times -2 is -6; plus b.
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Adding 6 to both sides, b equals 3.
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I have the slope; I have the y-intercept; I can write this out in slope-intercept form: y = (slope is 3) 3x...and b is 3.
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OK, so this one was a little bit more complicated than before, because they didn't directly give us the slope or two points on the line.
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But what they did give us is the fact that this line is parallel to the line described by this equation.
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Knowing that means that I have this slope (which is 3), and since my line is parallel, it is the same slope.
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Once I have the slope, I just take the point on that line, substitute that in for x and y, and solve for b.
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So, to write it in this form, I am essentially given m; I can figure out b; and this is the equation in slope-intercept form.
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Find the equation in slope-intercept form of the line passing through (2,-1) and perpendicular to the graph of 2x - 3y = 6.
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In order to write this in slope-intercept form, y = mx + b, I need to have the slope of this line.
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I am not given the slope directly; however, I am told that it is perpendicular to the graph of this line.
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Recall that the slopes of perpendicular lines are related by this equation.
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So, if you have the slope of two lines that are perpendicular, and you take their product, it is equal to -1.
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So, let's let the slope of this line equal m₁; and m₂ is the slope of my line, the line I am looking for.
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Let's go ahead and figure out m₁: what I need to do is write this equation in slope-intercept form,
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and that will give me the slope of this line, which in turn will give me the slope of the line I am looking for.
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First, I am going to subtract 2x from both sides; then, I am going to divide both sides by -3.
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That is going to give me y = -2x/-3 + 6/-3.
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Simplify: the negatives cancel out, and I will get 2/3x - 2.
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OK, since this is in slope-intercept form, this is m; this is the slope; so m₁ equals 2/3.
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I now want to find m₂, which is the slope of the perpendicular line, or the line that I am looking for.
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m₁ times m₂ equals -1; and I am given m₁--I am given that m₁ is 2/3.
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So, it is 2/3 times m₂ equals -1.
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If I multiply both sides by 3/2, I can isolate m₂, and it is -1 times 3/2.
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Therefore, the slope of the line that I am looking for is -3/2.
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So now, I have the slope: I have that, for my line, y = -3/2x + b.
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I have slope; I need the y-intercept; well, I am also given a point on this line, which means I have an x and y value to substitute here.
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So, y is -1; x is 2; the 2's cancel out, and that is going to give me -3 plus b.
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I am going to add 3 to both sides, which is going to give me b = 2.
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So now, I have the slope; I have the y-intercept; so I can write the equation for this line in slope-intercept form: slope is -3/2; b is 2.
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Again, we approached this by realizing that the slope of the line we are looking for is related to the perpendicular line
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by the equation m₁ times m₂ (the product of the slopes of the perpendicular lines) = -1.
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So then, I went about looking for the slope of this line; rewriting it in slope-intercept form gave me y = 2/3x - 2.
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So, I know that this slope is 2/3; once I have this slope, I can find my slope: m₁ times m₂ equals -1.
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So, that is 2/3 times m₂ equals -1; m₂ equals -3/2.
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I go back to this form y = mx + b, now knowing the slope of my line.
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Substitute in x and y values and the slope to solve for b; b equals 2.
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I have b; I have the slope; that allows me to write this equation in slope-intercept form.
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That concludes this lesson of Educator.com.