Question 656643
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Perpendicular lines have slopes that are negative reciprocals.  Start with the slope formula for the first two points:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m_1\ =\ \frac{8\ +\ 1}{4\ -\ 2} ]


Then the slope for the 2nd set of points:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m_2\ =\ \frac{2\ -\ 5}{x\ +\ 4} ]


Take the negative reciprocal of *[tex \LARGE m_2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -\frac{1}{m_2}\ =\ -\frac{x\ +\ 4}{2\ -\ 5}] 


Then set *[tex \LARGE m_1\ =\ -\frac{1}{m_2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{8\ +\ 1}{4\ -\ 2} =\ -\frac{x\ +\ 4}{2\ -\ 5}]


Solve for *[tex \LARGE x].  Be very careful with your signs.  Particularly the one in front of the RHS fraction.  Apply it to either the numerator or the denominator as suits your fancy, but distribute it across BOTH terms of whichever you choose.


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