Question 975071
<font face="Times New Roman" size="+2">


First construct the given line, call that *[tex \Large l_1] and the given point, call that *[tex \Large O] (because it will be the center of the circle later in the problem).


Calculate the slope of *[tex \Large l_1] and then using the idea that a perpendicular has a negative reciprocal slope, determine the slope of the line that passes through *[tex \Large O] and is perpendicular to *[tex \Large l_1].  Use the Point-Slope form of a line to write the equation of the perpendicular through *[tex \Large O], which we will call *[tex \Large l_2].


Solve the 2X2 system formed by the equations of *[tex \Large l_1] and *[tex \Large l_2] to find their point of intersection which we will call *[tex \Large P_0].  Then use the distance formula to calculate the distance between *[tex \Large O] and *[tex \Large P_0], a number we will call *[tex \Large r_0].


Using *[tex \Large r_0] as a radius and *[tex \Large P_0] as the center, write an equation for a circle centered at *[tex \Large P_0] that passes through *[tex \Large 0].  This circle will intersect *[tex \Large l_1] in two places, which we will call *[tex \Large V_1] and *[tex \Large V_2], the order is arbitrary.  Since the distance from *[tex \Large O] to *[tex \Large P_0] is half of the measure of our square and one side of the square lies in *[tex \Large l_1], segment *[tex \Large V_1V_2] represents one side of the circle.


Solve the 2X2 non-linear system consisting of the equation for *[tex \Large l_1] and the equation for circle *[tex \Large P_0] to determine the coordinates for *[tex \Large V_1] and *[tex \Large V_2].


Using the distance formula, calculate the distance from *[tex \Large O] to either *[tex \Large V_1] or *[tex \Large V_2].  (It doesn't matter which.  Given correctly performed arithmetic, the answer will be the same either way and the arithmetic will be just as ugly).  Call this distance *[tex \Large r].


Write the equation of a circle centered at *[tex \Large O] with radius *[tex \Large r].  This will be the circle that passes through all four vertices of the square.


Using the slope of line *[tex \Large l_2] and point *[tex \Large V_1] and the Point-Slope form of the equation of a line, derive the equation for the line containing the second side of the square, call this line *[tex \Large l_3].  You already have one point of intersection between *[tex \Large L_3] and circle *[tex \Large O]; use the two equations and solve the 2X2 non-linear system for the other point of intersection which we will call *[tex \Large V_4].  Use a similar process to find the equation of the line parallel to *[tex \Large l_2] through *[tex \Large V_2], and then solve the 2X2 system for the coordinates of *[tex \Large V_3].


The only thing remaining is to derive an equation for the line containing the fouth side of the triangle.  Use the slope of *[tex \Large l_1] and point *[tex \Large V_3] or *[tex \Large V_4]


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \