For the conditions stated, if one line segment (PX) is twice the length of the other (XQ), then point X must be 2/3 of the way towards Q because let s = the length of XQ and 2s = length of PX
2s / (2s +s) = 2s / 3s = 2/3
Therefore, if we go 2/3 the distance in the x direction from point P and 2/3 the distance in the y direction from point P, we will arrive at the location for point X.
So this turns out to be
(2/3) × (18-6) = 8 in the X direction
(2/3) × (8-2) = 4 in the Y direction
And our final answer is :
(X,Y) = (6+8, 2+4) = (14, 6)