Question 396931
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Let's start by using the correct words to describe things.  Not the X-line or Y-line, it is the x-axis and the y-axis.


If the line crosses the y-axis at 1, then the coordinates of that point are (0,1).  If the line crosses the x-axis at 2, then the coordinates of that point are (2,0).


The slope of your line is indeed *[tex \Large -\frac{1}{2}].  Use the slope formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m\ =\ \frac{y_1\ -\ y_2}{x_1\ -\ x_2} ]


where *[tex \Large \left(x_1,y_1\right)] and *[tex \Large \left(x_2,y_2\right)] are the coordinates of two points on your line.  We determined the coordinates of two points by using the intercept values, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m\ =\ \frac{1\ -\ 0}{0\ -\ 2}\ =\ -\frac{1}{2}]


I'm not sure where you got the idea that *[tex \Large \Delta x\ =\ 4] and *[tex \Large \Delta y\ =\ -2] unless there is another point or two defined on your line that you didn't tell us about.  Using the coordinate differences on the two intercepts (reading left to right), *[tex \Large \Delta x\ =\ 2] and *[tex \Large \Delta y\ =\ -1]


Once you have the slope and the y-coordinate value of the y-intercept, writing an equation that represents the line is easy.  You have the information to write your equation in slope-intercept form, namely 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ mx\ +\ b]


where *[tex \Large m] is the slope number and *[tex \Large b] is the y-coordinate of the y-intercept.  For your problem, that is *[tex \Large m\ =\ -\frac{1}{2}] and *[tex \Large b\ =\ 1], hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -\frac{1}{2}x\ +\ 1]


From there you can use a little algebra to put the equation into Standard form, namely:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ Ax\ +\ By\ =\ C]


For you:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 2y\ =\ 2]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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