The line perpendicular to 4x+3y=26 through (2,-3) is
y = 3/4 (x-2) - 3 = 3/4 x - 9/2 or 3x-4y=18
The lines intersect at (128,25, 6/25)
the distance from (2,-3) to the line 4x+3y-26=0 is
|4*2 - 3*3 - 26|/√(3^2+4^2) = 27/5
That is 1/2 the diagonal, so each side has length 54/(5√2)
Now you can find the other end of the diagonal, and the vertices. But it'll be messy
Find the coordinate vertexes of a square whose one side lies on the li e 4x+3y=26 with the intersection of the diagonals of the square ( 2,-3)
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