Question 280074
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<b><i>Read everything before you start doing anything</i></b>


Let *[tex \Large A\left(x_1,y_1\right)] be the right angle vertex of the given triangle.  Let *[tex \Large B\left(x_2,y_2\right)] and *[tex \Large C\left(x_3,y_3\right)]be the other two vertices of the given triangle.  Finally, let *[tex \Large D\left(x_4,y_4\right)] be the unknown point.


Use the slope formula to calculate the slope of the line containing the segment AB:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m_{AB}\ =\ \frac{y_1\ -\ y_2}{x_1\ -\ x_2}]


Opposite sides of a rectangle are parallel.  Parallel lines have equal slopes.  Using the point-slope form of the equation of a line write an equation of the line parallel to AB that passes through C.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ y_3\ =\ m_{AB}(x\ -\ x_3)]


Call that result Equation 1.


Now repeat the entire process, this time computing the slope of the line containing AC and deriving an equation for its parallel through B.



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m_{AC}\ =\ \frac{y_1\ -\ y_3}{x_1\ -\ x_3}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ y_2\ =\ m_{AC}(x\ -\ x_2)]


And call that Equation 2.


Finally, solve the system of equations consisting of Equation 1 and Equation 2.  The solution set will be *[tex \Large D\left(x_4,y_4\right)].


<b><i>The above is the General solution</i></b>


If you happen to have a situation where the two legs of your given triangle are parallel to the axes, then the solution is much simpler.  Take the *[tex \Large x]-coordinate of the given point that is on a vertical line above (or below) the unknown point as the *[tex \Large x]-coordinate of the unknown point.  Take the *[tex \Large y]-coordinate of the given point that is on a horizontal line to the right or left of the unknown point as the *[tex \Large y]-coordinate of the unknown point.  Done.


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