The coordinates of the endpoints of the hypotenuse of a right triangle are (7, 5) and (3, 1). Find the other vertex. There are two possible solutions.

Answer 1

actually, there is an infinite number of solutions

one of the properties of a circle is that a triangle is formed with the diameter as a base and the other point on the circle, you will always have a right angle.

so the centre would be (5,3) and the radius would be 2√2
equation:
(x-5)^2 + (y-3)^2 = 8

so now pick any x
e.g.
x = 4
1 + (y-3)^2 = 8
(y-3)^2 = 7
y = 3 ± √7 ----> 2 points , (4, 3+√7) and (4, 3-√7)

x = 5
0 + (y-3)^2 = 8
y-3 ± 2√2
y = 3 ± 2√2 ---> 2 more points (5,3+2√2) and (5, 3-2√2)
....
x = 6.5
y -3 = ±√5.75
y = 3 ± √5.75)
---- > 2 more points (6.5 , 3 + √5.75) and (6.5, 3-√5.75)

So far i have 6 such points,
btw, use your calculator to check that the slope of the 2 sides are negative recriprocals of each other, (or their product is -1)
It works for all of them, I did it.