Finding the Equation of a LineDate: 01/08/97 at 21:55:55 From: Lauren Subject: Writing Linear Equations in Point-Slope and Standard Forms I am a 9th grade Algebra 1 student and these questions are on slopes. Write the point-slope form of an equation of the line that passes through the given point and has the given slope: (1, 3), m = -2 and (-6, 3), m = -2/3 Write the standard form of an equation of the line that passes through the given point and has the given slope: (2, 13), m = 4 and (8, 2), m = -2/5 Write the point-slope form of an equation of the line that passes through each pair of points: (-5, 2), (4, -1) and (4, -2), (8, -2) Thank you very much for helping me out. Thanks, Lauren Date: 01/09/97 at 19:39:59 From: Doctor Keith Subject: Re: Writing Linear Equations in Point-Slope and Standard Forms Hi, All of your questions revolve around how to get an equation of a line from a point and a slope or two points, so let's start with a quick review of the formulas for a line. We have two basic forms of a line, the point-slope and the standard (the standard is sometimes also called the slope-intercept form). They are: Standard: y = mx+b m is the slope b is the y-intercept (the value of y when x=0) Point-Slope: (y-y1) = m(x-x1) m is the slope (x1,y1) is the given point (some point on the line) When I was learning this, I found it very useful to look at the relation between these two forms of the equation of a line. In fact, the two equations are easily obtained from one another. How, you ask? Well, let's look at the point slope form and see if we can show how you get the standard form from it. Point-Slope: (y-y1) = m(x-x1) y-y1 = mx - mx1 distributive property y = mx - mx1 +y1 additive inverse (y1-y1=0) and adding the same quantity to both sides (y1 in this case) y = mx + (y1-mx1) commutative and associative properties of addition y = mx + b let b = (y1 - mx1) Why bother with this, you might ask (and it is a very good question to ask). Well, it shows that the two forms of the equation of the line are equivalent and it helps you to see the connection. For instance, we now have b = y1 - mx1, which is the formula for calculating the y-intercept of a line from any point on the line and the slope of the line. We can also note that by taking b to the other side of the equation we have y-b = mx, which is a special case of the point-slope form (with the point (0,b)). A final point for why the relation between the two forms is useful is that it helps you to remember them (you can get one from the other if you know the relation). Enough theory. Let's try some problems: 1) You gave me the following point and slope: (1, 3), m = -2. Let's find the line in both forms. The point-slope form is easiest to get, so let's do that first: We have (x1,y1) = (1,3) and m=-2 and the form is (y-y1) = m(x-x1). Thus by direct substitution we have y-3 = -2(x-1) as the solution for the point slope form. Now we want to get the standard form, we can do the algebra again (if we forgot the form of b we found above) or we can use the form we found, namely b = y1-mx1. Let's do both for clarity With algebra we have: y-3 = -2(x-1) y-3 = -2x+2 y = -2x+2+3 y = -2x+5 This is our answer By formula we have: b = (y1-mx1) b = (3-(-2)(1)) b = (3+2) b = 5 By direct substitution we have y = -2x+5, which is the same. The point slope form is obviously very easy to get when you have a point and the slope of the line (which is the whole reason for this form). The standard form takes a little more work but is very useful for drawing and analyzing your line. Hence the need for both. You can use either way you like to go between the two forms. It's all a matter of personal preference. We will do one more quickly and you should be able to answer the rest. Let's get the two forms of the line with point and slope given by: (8, 2), m = -2/5. By direct substitution, we can get the point-slope form: y-2 = (-2/5)(x-8) By algebra we can get the standard form: y-2 = (-2/5)x+16/5 y = (-2/5)x+26/5 By substitution we can also get the standard form: b = y1-mx1 b = 2-(-2/5)8 b = 2+16/5 b = 26/5 So we have y = (-2/5)x+26/5 Now you also had some problems where you were given two points and asked to give the equation of the line. This requires you to calculate the slope of the line from these points and then use the slope and one of the points as we did above to get the equation of a line. So we need a formula for the slope given two points. There are several ways to get this. You can remember that the slope is the rise (change in y) divided by the run (change in x) or you could calculate this form from the point-slope form of a line. If we have points (x1, y1) and (x2, y2), by memory we have: m = (y2-y1)/(x2-x1) By algebra on point-slope form: (y-y1) = m(x-x1) which holds for all (x,y) on the line. In particular, it holds for (x,y) = (x2,y2) (y2-y1) = m(x2-x1) Dividing by x2-x1 (which cannot be zero for distinct points because a vertical line is not a function and a line must be a function) we get: m = (y2-y1)/(x2-x1) Thus we can use our equation for calculating the slope given two points and then pick either point (it does not matter which) and we can get the equation of the line as we did in the first part (we now have a point and the slope). Lets try an example: You gave the following two points: (-5, 2), (4, -1) So what is the slope? m = (y2-y1)/(x2-x1) m = (-1-2)/(4--5) m = (-3)/(9) m = -1/3 Or by algebra on the point-slope equation of a line: (y2-y1) = m(x2-x1) (-1-2) = m(4--5) (-3) = m(9) -1/3 = m Then we can use either {(-5,2), m = -1/3} or {(4,-1), m = -1/3} and solve for the equation of the line in either form as we did before. I have given you a couple of ways of solving the problems, so let's put the options in a little step by step review: Given points (x1,y1) and (x2,y2) find the equation of the line. Option 1 1) get the slope by m = (y2-y1)/(x2-x1). 2) pick either (x1,y1) or (x2,y2) as the point to use with your slope. 3) follow the given point and slope steps below for the form you want Option 2 1) get the slope by algebraic solution of point-slope equation (y2-y1) = m(x2-x1). (Note this already has the substitution (x,y)=(x2,y2).) 2) pick either (x1,y1) or (x2,y2) as the point to use with your slope. 3) follow the given point and slope steps below for the form you want. Given point (x1,y1) and slope m and you want point-slope form: 1) substitute values into the equation (y-y1) = m(x-x1) 2) you're done Given point (x1,y1) and slope m and you want standard form: Option 1 1) get point-slope equation through above steps 2) use algebra to manipulate into the form y = mx+b 3) you're done Option 2 1) calculate the y intercept (b) by b = y1-mx1 2) directly substitute into the formula y = mx+b 3) you're done As you can see, the most important part of working with lines is understanding what the relations between the forms are. Each of the options listed calculate the same thing, some by formula and some by algebraic manipulation, so it really comes down to what you like and feel comfortable with. I would suggest trying both and seeing what you think. If you would like further explanation on some point, or a graph to make this clearer, write back and I would be happy to give you more help. I hope all goes well on the test. -Doctor Keith, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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