We first need to find the angle bounded by the two lines, and then apply some trig to find the slope of the red line that bisects the acute angle.
To calculate the slope between the lines: \(y=2x\) and \(y=0\), we can draw a triangle, with co-ordinates \((0,0)\), \((2,0)\) and \((2,4)\).
Let \(\angle {X}\) be bounded by the 2 lines.
We know that \(\tan = {\text{opposite} \over \text{adjacent}}\). This means that \(\tan(x) = 2\). Using the inverse function, we find that \(x \approx 63.435\)
Doing the same thing to the triangle bounded by the lines: \(y=4x\) and \(x=0\), we find that \(x \approx 14.036\)
This means the angle of the region bounded by the lines: \(y=2x\) and \(y=4x\) is \(\approx 12.529\)
This means that the angle from the red line to \(y=0\) is \(\approx 69.67\)
Now, form a right triangle with an angle of \(\approx 69.67\) and a base of 2.
Using the trig function mentioned above, we find that the height of the triangle is \(\approx 5.407\)
This means that the slope of the line is \({\text{rise} \over \text{run} } = {5.407 \over 2} = \color{brown}\boxed{\approx 2.703}\)
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