Question 391317
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Determine the slope of the line that passes through *[tex \Large (t,t\,+\,1)] and *[tex \Large (3t,t\,+\,3)]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m\ =\ \frac{t\,+\,1\ -\ (t\,+\,3)}{t\ -\ 3t}\ =  \frac{-2}{-2t}\ =\ \frac{1}{t}]


The negative reciprocal of the slope is the slope of the perpendicular, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m_\perp\ =\ -t]


Determine the midpoint of the given segment:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_m\ = \frac{t + 3t}{2}\ =\ 2t] and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y_m\ = \frac{t\,+\,1\ +\ t\,+\,3}{2}\ =\ t\ +\ 2]


Write the desired equation using the point-slope form:


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


Simplify to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ +\ tx\ =\ 2t^2\ +\ t\ +\ 2]


Substitute (5,2) for x and y:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\ +\ 5t\ =\ 2t^2\ +\ t\ +\ 2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2t^2\ -\ 4t\ =\ 0]


Solve the quadratic for the two possible values for *[tex \Large t]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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