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 July 13th, 2011, 04:03 PM #1 Newbie   Joined: Jun 2011 Posts: 8 Thanks: 0 Converting a linear graph to a linear equation? Ok so I'm really furious, I lost a good chunk of my mark on a test I just wrote, all because one of the problems asked for something I had never done before. It wanted me to convert a linear graph into a linear equation in standard form. I have the slope-intercept form down pat, can write a graph to go along with it or vice versa no problem. Standard form is different though, I can write a graph out of a standard form equation, but I can't do it the other way around. An example of slope-intercept form being: "y = 2/3 times X + 4" While an example of Standard Form is: 3x + 2y = 6 Writing it as slope-intercept is logical because you just multiply X by the slope and then add the y-intercept. Standard form is different, no matter how hard I examine a graph, I can't figure out a way to write a linear equation in the form of AX + BY = C.
 July 13th, 2011, 04:32 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 464 Math Focus: Calculus/ODEs Re: Converting a linear graph to a linear equation? Suppose you are given 2 points on a line: $$$x_1,y_1$$$ and $$$x_2,y_2$$$ and you are instructed to find the equation of the line written in standard form. One way to do this is to use the point-slope equation: $y-y_1=m$$x-x_1$$$ Compute the slope m: $m=\frac{y_2-y_1}{x_2-x_1}$ and we have: $y-y_1=\frac{y_2-y_1}{x_2-x_1}$$x-x_1$$$ Multiply through by $x_2-x_1$ $y$$x_2-x_1$$-y_1$$x_2-x_1$$=$$y_2-y_1$$$$x-x_1$$=$$y_2-y_1$$x-$$y_2-y_1$$x_1$ Now, rearrange into standard form: $$$y_1-y_2$$x+$$x_2-x_1$$y=y_1$$x_2-x_1$$-$$y_2-y_1$$x_1$ $$$y_1-y_2$$x+$$x_2-x_1$$y=x_2y_1-x_1y_1-x_1y_2+x_1y_1$ $$$y_1-y_2$$x+$$x_2-x_1$$y=x_2y_1-x_1y_2$ Thus we see: $A=y_1-y_2$ $B=x_2-x_1$ $C=x_2y_1-x_1y_2$ You can of course change the signs of all 3 values if needed to avoid most of them being negative.
 July 13th, 2011, 05:19 PM #3 Newbie   Joined: Jun 2011 Posts: 8 Thanks: 0 Re: Converting a linear graph to a linear equation? Thanks, that clears everything up nicely. I'm still furious that the test had that question though, I really hope the teacher has the good sense to not dock me any marks, 10% off my total mark is pretty steep for something they failed to cover or even mention before.
 July 13th, 2011, 05:56 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 464 Math Focus: Calculus/ODEs Re: Converting a linear graph to a linear equation? Let's look at the definition of slope: Given two points $$$x_1,y_1$$$ and $$$x_2,y_2$$$ $m=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}$ Now, suppose we let the second point instead be an arbitrary point on the line (x,y), we then have: $m=\frac{y-y_1}{x-x_1}$ Rearranging, we find the point-slope formula for a line: $y-y_1=m$$x-x_1$$$ Now, suppose the first point is $$$x_1,y_1$$=(0,b)$ this gives us: $y-b=m$$x-0$$$ Solving for y, we obtain, the slope-intercept formula: $y=mx+b$ Now, suppose we know the x-intercept (a,0) and the y-intercept (0,b). We compute the slope: $m=\frac{b-0}{0-a}=-\frac{b}{a}$ Using the slope-intercept formula, we find the line is given by: $y=-\frac{b}{a}x+b$ $\frac{b}{a}x+y=b$ Divide through by b: $\frac{x}{a}+\frac{y}{b}=1$ This is the two-intercept formula for a line. If we multiply through by ab, we have: $bx+ay=ab$ This is the line in standard form in terms of its intercepts.

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# how to turn equations to a linear graph equation

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