WEBVTT
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In this lesson, we will learn how to find the equation of a line in standard and point-slope forms given two points, a slope and a point, or a graph.
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So what the objectives are is to recognize the point-slope form of a straight line and be able to write this given a slope and a point on the line.
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We’re gonna determine equations that represent straight lines, identify the graph of a line given a point-slope form, and then identify the intercepts of a straight-line graph.
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But before we get on and have a look at how we’re gonna do this with examples, what we’re gonna look at is our forms of a straight-line equation.
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So what we’re gonna have a look at here is three ways that we can write the equation of a straight line.
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However, we’re only gonna really concentrate on two of them for this lesson.
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So the first general form of the equation of a straight line is the slope intercept form, which is 𝑦 equals 𝑚𝑥 plus 𝑏, where 𝑚 is the slope, but also known as the gradient, and 𝑏 is the 𝑦-intercept.
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Again, you might sometimes see this written as 𝐶.
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Okay, so now let’s look at the next form.
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Well, the next form is known as the standard form.
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And this is 𝐴𝑥 plus 𝐵𝑦 equals 𝐶.
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And then, finally, what we have is the point-slope form, which is 𝑦 minus 𝑦 sub one equals 𝑚 multiplied by 𝑥 minus 𝑥 sub one, where once again 𝑚 is the slope or gradient and 𝑥 sub one, 𝑦 sub one is a point on the straight line.
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It is worth noting here that you might sometimes see 𝑥 sub one, 𝑦 sub one written as 𝑥 sub 𝐴, 𝑦 sub 𝐵.
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It doesn’t matter what the letter is.
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It’s just whatever’s used to represent the coordinates of that point on the straight line.
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So here we’ve talked about three forms of the equation of a straight line.
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However, in this lesson, we’re just going to be concentrating on the second two: the standard form and the point-slope form.
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So now what we can do is look at some questions.
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And the first question we’re gonna have a look at is a question that involves the point-slope form.
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Find, in point-slope form, the equation of the graph with slope four that passes through the point two, negative three.
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Well, the first thing we’re going to do is remind ourselves what the point-slope form is.
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And it is 𝑦 minus 𝑦 sub one equals 𝑚 multiplied by 𝑥 minus 𝑥 sub one.
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And this is where 𝑚 is the slope and 𝑥 sub one, 𝑦 sub one is the coordinates of a point on the line.
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So therefore, first of all, we know from our problem that 𝑚, the slope, is gonna be equal to four.
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Our 𝑥 sub one, 𝑦 sub one, so our point on the line, is two, negative three.
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So when we substitute in our values, we’re gonna get 𝑦 minus negative three is equal to four multiplied by 𝑥 minus two.
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Well, the key here is that we want the answer in point-slope form.
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Well, we’re almost there.
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We just want to simplify the left-hand side.
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So on the left-hand side, we have 𝑦 minus negative three.
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Well, if you subtract a negative, it’s the same as adding a positive.
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So we get 𝑦 plus three.
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So therefore, we can say that, in point-slope form, the equation of the line with slope four that passes through the point two, negative three is 𝑦 plus three equals four multiplied by 𝑥 minus two.
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So great, what we’ve done is had a look at an example which gets us to substitute in values into the point-slope form.
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So now what we’re gonna have a look at is a question on the graph.
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And in that question, we’re gonna look at how we can write the equation of the line in the graph using the point-slope form.
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Write the equation represented by the graph shown.
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Give your answer in the form 𝑦 minus 𝑎 equals 𝑚 multiplied by 𝑥 minus 𝑏.
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So in this question, what we’re asked to do is write the answer in the form which is known as the point-slope form, where the slope is represented by 𝑚 and 𝑎, 𝑏 is a point on our straight line.
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It is worth noting as well that even though this form looks slightly different, because you may have seen the form 𝑦 minus 𝑦 sub one equals 𝑚 multiplied by 𝑥 minus 𝑥 sub one, they’re no different apart from the fact that the point on the line is represented by 𝑥 sub one, 𝑦 sub one instead of 𝑎, 𝑏.
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So great, now we know what the form is.
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What we need to do is find the slope of our line.
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Well, as we’ve got a straight line, we know that the slope is gonna be the same all along it.
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So it doesn’t matter which two points we pick on the straight line.
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But a quick tip to remember is to make sure that the point you choose is easy to read, so it lies on the grid lines.
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So we’ve picked our points.
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We got one that we were given at the beginning, which is negative two, six, and then one that we’ve just chosen, which is two, eight.
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Well, to find the slope, what we want to do is find the change in 𝑦 over the change in 𝑥.
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It’s also sometimes spoken about in informal terms as the rise over the run.
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However, if we want to use a formal formula, what we have is 𝑚 is equal to 𝑦 sub two minus 𝑦 sub one over 𝑥 sub two minus 𝑥 sub one, so the change in 𝑦 over the change in 𝑥.
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So to help us to use this, what we’ve done is labeled our points.
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So we’ve got 𝑥 sub one, 𝑦 sub one and 𝑥 sub two, 𝑦 sub two.
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So when we substitute in our values, we’re going to get 𝑚 is equal to eight minus six over two minus negative two.
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So what we’re gonna get is a slope 𝑚 of two over four.
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So then if we divide both the numerator and denominator by two, because we can cancel it out, what we’re gonna have is 𝑚 is equal to a half.
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So our slope is equal to a half.
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So next, what we have is a point.
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And we were given a point at the beginning.
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And this is at negative two, six.
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So we can say the point 𝑎, 𝑏 is equal to negative two, six.
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So we now have all the parts we need to to substitute back in to our point-slope form.
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So what we’re gonna get is 𝑦 minus six is equal to a half multiplied by 𝑥 minus negative two.
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So therefore, we can say that the equation represented by the graph shown is 𝑦 minus six equals a half multiplied by 𝑥 plus two.
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So great, what we’ve done now is had a look at a couple of questions with a point-slope form written in a couple of different formats.
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So now what we’re gonna do is have a look at one of our other learning objectives.
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And that is to look at some equations and decide which ones may represent a straight line.
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Which of the following equations represents a straight line?
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(A) 𝑦 equals root 𝑥 plus six, (B) 𝑦 plus one over 𝑥 equals negative eight, (C) 𝑥 plus root 𝑦 equals negative five, or (D) negative seven 𝑥 minus two 𝑦 equals negative nine.
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So to enable us to decide which one of these is in fact the correct equation to represent a straight line, what we’re gonna do is look at a couple of the general forms for the equation of a straight line.
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Well, the couple of forms that we’ve got here are the standard form, 𝐴𝑥 plus 𝐵𝑦 equals 𝐶, or we have the point-slope form, which is 𝑦 minus 𝑦 sub one equals 𝑚 multiplied by 𝑥 minus 𝑥 sub one.
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Well, one thing we can notice from both of our forms is the fact that we have here 𝑥 to the power of one and then 𝑦 to the power of one.
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And that’s in the standard form.
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And then in the point-slope form, we also have 𝑦 to the power of one or 𝑥 to the power of one.
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So therefore, with this bit of information and the two forms we have here for the equation of a straight line, let’s have a look at the four equations that we have to see if they in fact represent a straight line.
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Well, if we look at equation (A), what we have is root 𝑥.
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Well, if we write this in exponent form, this is 𝑥 to the power of a half.
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So therefore, this cannot be the correct answer cause this will not represent a straight line.
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Well, in equation (B), what we have is one over 𝑥.
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Well, if we rewrite this in exponent form, this is 𝑥 to the power of negative one.
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So therefore, once again, this cannot be the correct answer.
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And it cannot be a straight line because we don’t have the 𝑥 to the power of one, 𝑦 to the power of one.
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Then, if we take a look at (C), once again, we’ve got a root, and this time it’s root 𝑦.
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So therefore, this is gonna be 𝑦 to the power of a half.
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So therefore, this can’t be the correct equation either.
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But if we check equation (D), what we have here is 𝑥 to the power of one and 𝑦 to the power of one.
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So therefore, this can be the correct equation.
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So this could be the equation of a straight line.
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And what we can also notice is that it takes the standard form for the equation of a straight line, and that is 𝐴𝑥 plus 𝐵𝑦 equals 𝐶.
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So therefore, we can say that the equation which represents the straight line is negative seven 𝑥 minus two 𝑦 equals negative nine.
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And it’s in the standard form 𝐴𝑥 plus 𝐵𝑦 equals 𝐶, where 𝐴 is equal to negative seven, 𝐵 is equal to negative two, and 𝐶 is equal to negative nine.
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Okay, great, so we’ve solved this problem.
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In the next question, what we’re gonna have a look at is a series of graphs.
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And we’re gonna decide which graph represents the equation we’ve been given.
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Which of the following graphs represents the equation 𝑦 minus five equals two-thirds multiplied by 𝑥 minus three?
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And we have five graphs: (A), (B), (C), (D), and (E).
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So what we have is an equation, and what it’s in is the point-slope form.
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So that’s 𝑦 minus 𝑦 sub one equals 𝑚 multiplied by 𝑥 minus 𝑥 sub one, where 𝑚 is the slope and the point that lies on the straight line is 𝑥 sub one, 𝑦 sub one.
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So the first thing we can find from our equation is our 𝑚, so our slope, because our 𝑚 is equal to two-thirds.
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Well, just reminding ourselves that the slope is equal to the change in 𝑦 divided by the change in 𝑥, so therefore what this means is that for every two units up, the graph goes; it goes three units across.
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And that’s because as we’ve said the top number is the change in 𝑦 and the bottom number is the change in 𝑥.
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So great, we know our slope is two-thirds.
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And what we can also do now is use our equation to find a point on the line.
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So what we can see is that we’ve got a point on the line which must be with the coordinates three, five.
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So we can know that our 𝑥 sub one, 𝑦 sub one is equal to three, five.
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So great, what we have are both bits of information we need to identify which graph is the correct graph.
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So first of all, we’re gonna start with a point on the line.
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Well, if we take a look at graph (A), we can see that the point three, five does lie on the line.
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So therefore, this could possibly be the correct graph.
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Well, if we take a look at graph (B), we can see in fact the point does not lie on the line because three, five, we can see here, is not on the line.
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So therefore, this cannot be the correct graph.
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Taking a look at point (C), we can see that three, five does lie on the line.
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So this could be the correct graph.
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If we take a look at graph (D), we can see that, on graph (D), the point doesn’t lie on the line either cause it’s at the very edge of the axes.
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So this is not the correct graph either.
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So what is worth mentioning at this point is a common mistake that can be made.
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And that’s getting the 𝑥- and 𝑦-coordinates the wrong way round.
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So we can see here this point is in fact five, three.
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Then, if we move on to the final graph, graph (E), we could see that this could possibly be the correct graph because the point does lie on the line cause we have the point three, five on our line.
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So great, what we’ve done now is ruled out two of our graphs, (B) and (D).
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So now what we need to do is take a look at the other bit of information we’ve got, and that’s the slope of our graph, to see if this can help us decide which of the graphs that are left is the correct graph.
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So one thing we know about the shape of the graph is that if we have a positive slope, the line goes up to the right.
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And if we have a negative slope, it goes down to the right.
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Well, if we think about the slope we’ve got, it’s two-thirds, which is positive.
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So therefore, what we’re looking for is the slope that goes up to the right.
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Well, this means we can rule out one more graph.
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We can rule out graph (E) because in fact this is a negative slope.
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So this cannot be the correct graph.
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So great, what we’re left with is two graphs, graph (A) or (C).
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So now, to determine which one of the graphs, (A) or (C), is the correct graph, we can use our slope in a couple of ways.
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First of all, we could go and work out the slope of both of the lines separately.
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However, what we can also do is use what we know this slope means.
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So to use this method, what we’ve done is chosen a point on graph (A).
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And we’ve chosen the point negative three, one.
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It doesn’t matter which point you choose, just one that’s easy to read off the scale.
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So we know from our slope, which is two-thirds, that for every two units up, so the change in 𝑦, our line should go three units across.
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So we can see from graph (A) that if we start on a point on the line, then we get up two units and along three units, we do in fact land at a point again on our straight line.
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So we can say that the slope is two-thirds.
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However, if we take a look at graph (C), and which is a point on our line, if we go up two units and along three units, what this in fact does is takes us beyond our line.
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So therefore, we can say that the slope is not in fact two-thirds.
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But in fact, if we did want to find the slope of our line, we can see that if we go up three units and along two units, so here I’ve picked another couple of points, then we lie back on the line.
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So we could say that the slope of the line would be equal to three over two.
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So what we can say is that the correct graph is graph (A) because this is the graph that represents the equation 𝑦 minus five equals two-thirds multiplied by 𝑥 minus three.
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So great, we’ve looked at a range of different problems now.
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So what we’re gonna have a look at is a final problem.
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So what we’re gonna do with this final problem is find the equation of a straight line given its 𝑥- and 𝑦-intercepts.
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What is the equation of the line with 𝑥-intercept negative three and 𝑦-intercept four?
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So the first thing we’re gonna do in this question is sketch our line.
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So what we’ve got first of all is an 𝑥-intercept of negative three, which means we know that our line crosses the 𝑥-axis at negative three.
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And we have a 𝑦-intercept at four.
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So it crosses our 𝑦-axis at four.
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So now if we join these points, we have our straight line.
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So what we want to do now is find its equation.
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And the form in which we’re gonna find the equation first of all is the point-slope form, which is 𝑦 minus 𝑦 sub one equals 𝑚 multiplied by 𝑥 minus 𝑥 sub one.
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And the reason we’re going to do this is because it’s the easiest way to tackle the problem because what we can do is identify specific points on the line easily and we know the intercepts.
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So therefore, we can easily work out the slope.
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And as you can see from the point-slope form, what we need is the slope, which is 𝑚, and a point on the line 𝑥 sub one, 𝑦 sub one.
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It is also worth noting that you might also see the point-slope form as 𝑦 minus 𝐴 equals 𝑚 multiplied by 𝑥 minus 𝐵, which is exactly the same.
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They’re just giving the coordinates of the point as 𝐴, 𝐵.
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So what we know is that the 𝑥-intercept is negative three and the 𝑦-intercept is four.
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So therefore, we have two points: negative three, zero and zero, four.
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So the first thing we want to do is find the slope of our line.
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And we can do that using a formula.
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And that formula is 𝑚 is equal to 𝑦 sub two minus 𝑦 sub one over 𝑥 sub two minus 𝑥 sub one, so the change in 𝑦 over the change in 𝑥.
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So to help us use this formula, what we’ve done is labeled our points.
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So we’ve got 𝑥 sub one, 𝑦 sub one and 𝑥 sub two, 𝑦 sub two.
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So if we substitute these values in, we get 𝑚 equals four minus zero over zero minus negative three, which is gonna give us 𝑚 is equal to four over three or a slope of four-thirds.
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It’s worth noting that it doesn’t matter which way round we’d labeled our points because it still would’ve given us the same slope.
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So great, we now have our slope, and we also know a point on our line.
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So therefore, we can substitute our information into the point-slope form.
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We could choose any point along our line.
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However, I’m just gonna choose here the first point that we’ve got, which is negative three, zero.
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As we said, any point on our line would work.
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So when we substitute in our 𝑥 sub one, 𝑦 sub one, what we’re gonna get is 𝑦 minus zero is equal to four-thirds multiplied by 𝑥 minus negative three.
00:16:24.840 --> 00:16:28.990
So if we take a look at the right-hand side, what we get is 𝑦 equals four-thirds multiplied by 𝑥 plus three.
00:16:29.360 --> 00:16:32.220
So this is the equation in the point-slope form.
00:16:32.520 --> 00:16:34.500
But we could also write it in standard form.
00:16:35.410 --> 00:16:40.970
So to do this, what we do is multiply the whole thing through by three to give us three 𝑦 equals four multiplied by 𝑥 plus three.
00:16:41.590 --> 00:16:45.520
Then distributing across parentheses gives us three 𝑦 equals four 𝑥 plus 12.
00:16:46.740 --> 00:16:51.930
And then, finally, we can subtract four 𝑥, which gives us three 𝑦 minus four 𝑥 equals 12.
00:16:52.260 --> 00:16:56.170
So therefore, we’ve got the equation in both the point-slope form and standard form.
00:16:57.960 --> 00:17:01.610
Okay, great, we’ve looked at a number of different examples covering all of our objectives.
00:17:01.910 --> 00:17:03.700
So now let’s have a look at the key points of the lesson.
00:17:04.870 --> 00:17:11.390
Well, the first key point is the fact that we have the standard form for the equation of a straight line, which is 𝐴𝑥 plus 𝐵𝑦 equals 𝐶.
00:17:12.290 --> 00:17:20.480
So then we have another form for the equation of a straight line, which is 𝑦 minus 𝑦 sub one equals 𝑚 multiplied by 𝑥 minus 𝑥 sub one, where 𝑚 is the slope.
00:17:20.560 --> 00:17:23.200
And we have a point on the line 𝑥 sub one, 𝑦 sub one.
00:17:23.200 --> 00:17:25.410
And this is called the point-slope form.
00:17:27.160 --> 00:17:36.810
It is worth noting that you might also see this in the form 𝑦 minus 𝑎 equals 𝑚 multiplied by 𝑥 minus 𝑏, where the point is written as 𝑎, 𝑏 rather than 𝑥 sub one, 𝑦 sub one.
00:17:37.700 --> 00:17:45.870
We also looked at how the slope of a straight line remains constant and that if we’ve got a positive slope, then what this does is it slopes up to the right.
00:17:46.100 --> 00:17:49.030
And if we have a negative slope, this slopes down to the right.
00:17:50.430 --> 00:17:59.080
And we also recapped how to calculate the slope of a line, which is the change in 𝑦 over the change in 𝑥, or 𝑦 sub two minus 𝑦 sub one over 𝑥 sub two minus 𝑥 sub one.