Question 289404
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In order to create an equation of a line, you need two pieces of information.  You either need the ordered pairs representing two points, or you need a point and the slope (from which you can calculate the coordinates of any other point).


You are given the equation of a line and you want a line parallel to that line.  The first thing to realize is that parallel lines have identical slopes.  That is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  L_1\ \parallel\ L_2 \ \ \Leftrightarrow\ \ m_1\ =\ m_2]


Two ways to find the slope given an equation.  You can either put the equation into slope-intercept form, which is to say solve it for *[tex \Large y] in terms of everything else.  Then the slope is the coefficient on *[tex \Large x]. Or, if the equation is in Standard form like the one in your problem, divide the negative of the coefficient on *[tex \Large x] by the coefficient on *[tex \Large y].


The point you need is the point where the other given equation intersects the vertical axis.  This is the point where *[tex \Large x] is zero.  So, substitute zero for *[tex \Large x] in *[tex \Large 5x\ -\ 5y\ =\ 6] and do the arithmetic to find the value of *[tex \Large y].  That will give you a point *[tex \Large (0, b)] where *[tex \Large b] is the value you just calculated for *[tex \Large y].


Now that you have the slope of the desired line and one point on the desired line, you can use the Point-Slope Form to create the desired equation.


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


Where *[tex \Large m] is the slope you calculated and *[tex \Large \left(x_1,y_1\right)] are the coordinates of the intercept point that you calculated.  In this case the coordinates will be *[tex \Large \left(0,y_1\right)] because it is the intercept.


You should be able to get to the end from here.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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