Question 1113244
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Put the given equation into Slope-Intercept Form, namely *[tex \Large y\ =\ mx\ +\ b]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2y\ =\ 7\ -\ x]


Divide by 2 and rearrange the RHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -\frac{1}{2}x\ +\ \frac{7}{2}]


Now the coefficient on *[tex \Large x] represents the slope of the straight-line graph of the equation.


The slopes of perpendicular lines are negative reciprocals of each other, so, since the slope of the given line is *[tex \Large -\frac{1}{2}], the slope of the desired line must be 2.


To derive an equation of a line given a slope and a point, use the Point-Slope form:


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


where *[tex \Large \(x_1,\,y_1\)] is the given point and *[tex \Large m] is the  given slope.  For *[tex \Large \(2,\,1\)] and *[tex \Large m\ =\ 2]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ 1\ =\ 2(x\ -\ 2)]


This is one possible equation that represents the desired line, however, note that it is impossible to find <i><b>the</b></i> equation of a line because for any given line the set of equations representing that line is uncountably infinite.  Since you don't specify a form, I'll let you manipulate the answer to satisfy the requirements of your Algebra course.


 *[illustration PointSlopePerpendicularCrop.jpg].


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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