Question 1122461
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Your answer in italics as far as you were correct:


<i>2x-3y=0 <=== I found a slope of 2/3

 
Then:

 
y-(-3)=2/3(x-2)


y+3=2/3(x-2) </i>


Then you started to go awry:


Your next step should have been to get rid of the fractional slope by multiplying by the denominator, namely 3:


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3y\ +\ 9\ =\ 2(x\ -\ 2)]


Distribute:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3y\ +\ 9\ =\ 2x\ -\ 4]


Put the variable terms in the LHS and the constants in the RHS


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -2x\ +\ 3y\ =\ -4\ -\ 9]


Collect


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -2x\ +\ 3y\ =\ -13]


I like my lead coefficient to be positive, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ -\ 3y\ =\ 13]
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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