Being (-2, 4) and B (3, -1) consecutive vertices of a square, determine the other two vertices How to solve this only algebraically?
Would point-to-point distance (relative position between points and circumference) help?
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$\begingroup$We find the coordianates of the other two vertices considering three facts.
1) The midpoints of the diagonals should match.
2) The adjecent sides should be perpendicular.
3) The adjacent sides should have the same length.
With these properties we have a system of four eqautions $$ a+3= c-2, d+4=b-1$$ and
$$ d+1=c-3, (c-3)^2+(d+1)^2 = 50$$ where the two missing vertices are $$ (a,b), (c,d)$$
After solving we get the other vertices to be either $$(8,4), (3,9)$$ or $$(-7,-1), (-2,-6)$$Thus we have two squares with the given data.
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