Question 279804
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JBarnum is right -- you have to tell us what you want.  Put in the entire question.  Having said that, I'm going to go out on a limb here and guess that you want the equation, in slope-intercept form, that is either parallel or perpendicular to the graph represented by *[tex \Large 2y\ =\ x\ -\ 1] and that passes through the point *[tex \Large \left(-3,\,0\right)].


First step is to find the slope of the given line.  The easiest way to do that is to solve the equation for *[tex \Large y] in terms of everything else and then examine the coefficient on *[tex \Large x]


Multiply both sides of the equation by *[tex \Large \frac{1}{2}]:


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


The coefficient on *[tex \Large x] is *[tex \Large \frac{1}{2}], so the slope of the given line is *[tex \Large \frac{1}{2}].


Now you need to determine the slope of the desired line.  If you are looking for a line parallel to the given line, then parallel lines have identical slopes.


In other words:


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


That means the slope of the line you are trying to find is *[tex \Large \frac{1}{2}].


On the other hand if you want a line that is perpendicular to the given line then the slope of the desired line is the negative reciprocal of the slope of the given line, thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  L_1\ \perp\ L_2 \ \ \Leftrightarrow\ \ m_1\ =\ -\frac{1}{m_2}\ \text{ and } m_1,\, m_2\, \neq\, 0]


That would mean that the slope of the desired line is *[tex \Large -\frac{\ 1\ }{\frac{1}{2}}\ =\ -2]


Next you need to use the point-slope form of the equation of a line:


*[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 slope of the desired line.  In this case, those values are: *[tex \Large \left(-3,\,0\right)] and either *[tex \Large \frac{1}{2}] or *[tex \Large -2]


Just substitute the values:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ 0\ =\ \frac{1}{2}(x\ -\ (-3))]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ 0\ =\ -2(x\ -\ (-3))]


Finally, do the algebra required to put your equation into slope-intercept form which is *[tex \Large y\ =\ mx\ +\ b]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \frac{1}{2}x\ +\ \frac{3}{2}]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -2x\ -\ 6]



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