Let a, b, and c be vectors each starting at the origin and terminating at the points (x, x + 1), (x + 2, x + 3), and (x + 3, 2x + 4), respectively.
Then the vectors a - b, a - c, and b - c are vectors that point in directions parallel to each of the legs formed by the triangle with these points as its vertices.
If this triangle is to contain a right angle, then exactly one of these pairs of vectors must be orthogonal. In other words, one of the following must be true:
(a - b) • (a - c) = 0
or
(a - b) • (b - c) = 0
or
(a - c) • (b - c) = 0
We have
a - b = (x, x + 1) - (x + 2, x + 3) = (-2, -2)
a - c = (x, x + 1) - (x + 3, 2x + 4) = (-3, -x - 3)
b - c = (x + 2, x + 3) - (x + 3, 2x + 4) = (-1, -x - 1)
Case 1: If (a - b) • (a - c) = 0, then
(-2, -2) • (-3, -x - 3) = (-2)×(-3) + (-2)×(-x - 3) = 2x + 12 = 0 ==> x = -6
which would make a - c = (-3, 3) and b - c = (-1, 5), and their dot product is not zero. Then the triangles vertices are at the points (-6, -5), (-4, -3), and (-3, -8).
Case 2: If (a - b) • (b - c) = 0, then
(-2, -2) • (-1, -x - 1) = (-2)×(-1) + (-2)×(-x - 1) = 2x + 4 = 0 ==> x = -2
which would make a - c = (-3, -1) and b = (-1, 1), and their dot product is also not zero. The vertices are the points (-2, -1), (0, 1), and (1, 0).
Case 3: If (a - c) • (b - c) = 0, then
(-3, -x - 3) • (-1, -x - 1) = (-3)×(-1) + (-x - 3)×(-x - 1) = x ² + 4x + 6 = 0
but the solutions to x here are non-real, so we throw out this case.
So there are two possible values of x that make a right triangle, x = -6 and x = -2.