Question 301408
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<b>Step 1:</b> Find the slope of the line represented by the given equation.  Since the given equation is in standard form, use the negative of the coefficient on *[tex \Large x] divided by the coefficient on *[tex \Large y].  Alternatively, you can solve the equation for *[tex \Large y] to put the given equation into slope intercept form.  Having done that, the slope of the line represented by the given equation is equal to the coefficient on *[tex \Large x].


<b>Step 2:</b> Determine the slope of the desired line.  Calculate the negative reciprocal of the slope of the given line because:

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


<b>Step 3:</b> Derive an equation of the desired line using 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)] is the given point and *[tex \Large m] is the slope calculated in Step 2.


<b>Step 4:</b> Solve the equation derived in Step 3 for *[tex \Large y].  That is to say rearrange the equation so that *[tex \Large y] is all by itself on the left side of the equals sign and everything else is on the right side.  The equation will then be in the *[tex \Large y\ =\ mx\ +\ b] form which is the slope intercept form.


I can't actually solve your problem for you because you didn't actually give us an equation of a line to work with.  You wrote *[tex \Large 7x\ -\ 2y\ +\ 3] which is not an equation.  I can't tell whether you really meant *[tex \Large 7x\ -\ 2y\ =\ 3] or *[tex \Large 7x\ -\ 2y\ +\ 3\ =\ 0].  Besides, I don't want to deprive you of the fun and excitement of getting the answer by yourself.


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