Then the original trapezoid becomes two quadrangles, with the graphic area S 1 on the left and the graphic area S2 on the right.
The whole trapezoidal area S=2*(2+4)/2=6.
S 1=(X+Y)*2/2=X+Y
S2=[(4-X)+(2-Y)]*2/2=6-(X+Y)
S 1=S2
X+Y=6-(X+Y)
X+Y =3
It can be seen that there is no need to pass through the fixed point. As long as the sum of upper x and lower y =3, the requirements can be met.