A question on co-ordinate geometry.

Question

4 sided of a quadrilateral is given by (xy)(x-2)(y-3)=0. Then equation of the line parallel to x-4y=0 that divides a quadrilateral into two equal halves is?

My thinking- the given line forms a rectangle of area 6 units. The slope of the required line is 1/4 so tan A=1/4 which is implies that the sides of the triangle containing A are 1,4,√17(useful for finding cos A and sin A in parametric form.) I thought of finding the diagonal which is equal to √13 and then use the parametric form (X +- rcosA, Y +- rsinA) and then use the two point form formula of a straight line to get the equation.But then I don't know what to take for X and Y. So that's why I need help. Please let me know if I can solve it in some other way too.

Answer 1

The quadrilateral is a rectangle of vertices $A=(0,0)$, $B=(2,0)$, $C=(0,3)$ and $D=(2,3)$. Take the baricenter $F$ of this rectangle, the line parallel to the given line $x-4y=0$ passing through the baricenter divides the rectangle in two part of equal area.

Answer 2

The quadrilateral is a rectangle. The line cutting the rectangle into two equal halves must pass through the centre of the rectangle, i.e. $\displaystyle \left(\frac{2}{2},\frac{3}{2}\right)=\left(1,\frac{3}{2}\right)$.

So the equation is

$$x-4y=(1)-4\left(\frac{3}{2}\right)$$

$$x-4y=-5$$