Their dot product should be zero since they are perpendicular.
(3 - a)(1 - a) + (4 - b)(-1 - b) = 0
Then use 4a + 10b = 23 to eliminate one of the variables.
This becomes an algebraic nightmare, if you do it by hand, Anybody got a better idea?
The other 2 vertices are (-1/2,5/2) and (9/2,1/2)
I used Wolfram Alpha
a = (23 - 10b)/4, (3 - a)(1 - a) - (4 -b)(1 + b) = 0 - Wolfram|Alpha
The plot is not to scale, should be a circle
Wolfram gave both vertices in 1 shot
You can confirm that these vertices are on a square by using distance formula to check length of sides AND check length of diagonals.
You can also construct vectors on the vertices and take dot products, also checking lenths of sides using distance formula
4a + 10b = 23
Is a line that runs along the diagonal of the missing vertices, I find that fascinating.
I have to admit, I like this problem very much.
You used the distance formula, set length = width, and got an equation for a line that is 45º to both.
A bit more "geometric": The diagonals of a square are congruent and cross at right angles at their midpoint.
The midpoint of the line between (3,4) and (1,-1) is ((3+1)/2,(4- 1)/2)= (2, 3/2). It has slope (4-(-1))/(3-1)= 5/2 and length .
The line through (2,3/2) perpendicular to that diagonal is [tex]y- 3/2= (-2/5)(x- 2)[/itex] or y= (-2/5)x+ 4/5+ 3/2= (-2/5)x+ 23/10. That can also be written as 10x- 4y= 23. Find the intersections of that line with the circle, centered on (2,3/2) of radius , .