Question 264700
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You are very close.

You started correctly by determining that the slope of the given line is *[tex \Large \frac{2}{9}], therefore the slope of the perpendicular must be *[tex \Large -\frac{9}{2}].  But where you went astray is when you assumed that the point of intersection of the two lines is (18,0) just because it was given that the two lines intersect at *[tex \Large x\ =\ 18].

If the two lines intersect at some point where *[tex \Large x\ =\ 18], then that point must be *[tex \Large \left(18,\,y\\right)] such that the values 18 and *[tex \Large y] make your original given equation (and the equation we are about to derive for that matter) a true statement.  Hence:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\left(18\right)\ -\ 9y\ =\ -45]

And solve for *[tex \Large y]

I get *[tex \Large y\ =\ 9].  (Verification left as an exercise for the student)

And that tells us that the actual point of intersection is (18,9).  Now that we know that and the fact that the line you want to derive has a slope of  *[tex \Large -\frac{9}{2}], we can write:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -\frac{9}{2}\left(x\right)\ +\ b]

And then substitute the coordinate values from the point of intersection:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9\ =\ -\frac{9}{2}\left(18\right)\ +\ b]

Finally, solve for *[tex \Large b] which can then be substituted to create your final derived equation.  I get *[tex \Large b\ =\ 90], but you should check it for yourself.

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