There is a line through the origin that divides the region bounded by the parabola
y=4x−3x^2 and the x-axis into two regions with equal area. what is the slope of that line?

First, solve f(x)=4x-3x^2=0,
or
x(4-3x)=0
=>
x=0, x=4/3
The area enclosed by the parabola over the x-axis is therefore
A=integral f(x)dx from 0 to 4/3=[2x^2-x^3] from 0 to 4/3 = 32/27
Let the line intersect the parabola at a point (a,f(a)) such that the area bounded by the line, the parabola and the x-axis is half of A, or A/2, then the area consists of a triangle and a section below the parabola, the area is therefore
a*f(a)/2 + integral f(x)dx  from a to 4/3  =  A/2 = 16/27
=>
2a^2-3a^3/2+a^3-2a^2+32/27=16/27
=>
(1/2)a^3=16/27
a=(32/27)^(1/3)
=(2/3)(4^(1/3))
=1.058267368...

Slope of line is therefore
m=y/x=f(a)/a=4-2(4^(1/3))
=0.825197896... (approx.)