Question 438937
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Parallel lines have identical slopes.  When a two-variable linear equation is in the form *[tex \Large y\ =\ mx\ +\ b], then the slope of the graph is the coefficient on *[tex \Large x].  Use the point-slope form of an equation of a line to derive the desired equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ y_1\ =\ m(x\ -\ x_1) ]


where *[tex \Large \left(x_1,y_1\right)] are the coordinates of the given point and *[tex \Large m] is the value of the slope you determined by inspection of the given equation.


Given that you have not provided any particular form for your desired equaation, simple substitution of the values in the point-slope form is sufficient.  However, if there is a desired form, such as standard form (*[tex \Large Ax\ +\ By\ =\ C]) or slope-intercept form (*[tex \Large y\ =\ mx\ +\ b]) then you will have to do the appropriate algebraic manipulations to put it into the proper/desired form.


Once you have your equation in any form:


1.  Pick a value for *[tex \Large x].  It can be anything you like, but I like to use small integers, evenly divisible by any denominator that you might have in your slope number.


2.  Replace *[tex \Large x] with the value selected in step 1, and then do the indicated arithmetic to determine the value of *[tex \Large y]


3.  Form an ordered pair *[tex \Large (x, y)] using the value selected in step 1 and the value calculated in step 2.


4.  Plot this ordered pair on a set of coordinate axes.


5.  Repeat steps 1 through 4 using a different value for *[tex \Large x]


6.  Draw a line across your coordinate axes through the two points.


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