Two vertices of a triangle are (0,2) and (4,3). If its orthocentre is at the origin, then its third vertex lies in which quadrant?

Let A = (0, 2), B = (4, 3), and C = (a, b). The orthocenter is at the origin (0, 0).
The slope of AB is mAB = (3 - 2) / (4 - 0) = 1/4.
The altitude from C to AB has slope -4.
The equation of the altitude from C to AB is y - b = -4(x - a).
Since this altitude passes through (0, 0), we have -b = -4a, which simplifies to b = 4a. (i)
The slope of AC is mAC = (b - 2) / (a - 0) = (b - 2) / a.
The altitude from B to AC has slope a / (2 - b).
The equation of the altitude from B to AC is y - 3 = [a / (2 - b)](x - 4).
Since this altitude passes through (0, 0), we have -3 = a / (2 - b), which simplifies to 3(2 - b) = 4a.
Substituting b = 4a from (i), we get 3(2 - 4a) = 4a.
6 - 12a = 4a
6 = 16a
a = 6/16 = 3/8
b = 4a = 4(3/8) = 3/2
Thus, the coordinates of C are (3/8, 3/2). Since both coordinates are positive, the third vertex lies in the First quadrant.