Three vertices of a square have coordinates A(2, 4), B(5, –1), and C(0, –4). Give the coordinates of the fourth vertex, D, in ordered pair (x, y) form.
AB = BC = CD by the definition of square ( I didn't forget DA ).
Thus we can use distance formula to calculate length AB, which has length sqrt(34) units, and slope -5/3. Thus CD should also have slope -5/3, and length sqrt(34) units.
Point D should be on line y = -5x/3 - 4. DA has equation y = 3x/5 + 2.8 (perpendicular).
Thus we can solve for the x coodinate of D, which is obtained by using transitive property:
Point D has coordinates (-3,1)
x = -3
y = 1
u could also use desmos lol