Slope-y

The lines of division must pass through the line y=4 (the horizontal midline) at values 10/3 and 20/3 for the areas to be equal.

Since the lines intersect above the rectangle the left line must have a positive slope. Therefore it is in the form y = 3x + b and pass through the point
(10/3,4). Solving for b after inserting the point value gives a y-intercept of -6. This yields the equation y=3x-6

Because the slopes have reflectional symmetry, we can expect that the intersection of the points will be along the vertical midline or x=5.

Putting 5 into our equation for x yields the point (5,9)

line 1: $f(x) = 3(x-a)$ , line 2: $g(x) = -3(x-b)$

if $a = 0$ , then left part of rectangle is a triangle, and the area is: $\\ S = \dfrac 12 \times 8 \times \dfrac 83 = \dfrac {32}{3} < \dfrac {S_{rectangle}}{3}$

So line 1 must move some distance to right, i.e. $a > 0$ . Then the area of left part is: $\\ S = \dfrac {32}{3} + 8a = \dfrac {8 \times 10}{3} \\ \implies a = 2 \\ \implies f(x) = 3(x-2)$

similarly $b = 8$ (10-2=8). So the intersection point is $P(5,9)$