1. Region enclosed by the given curves/lines

The curve is $y=1+4x-x^2.$

The enclosed region is bounded by:

• $x=0$
• $y=0$
• $x=\frac{3}{2}$
• $y=1+4x-x^2$

Since on $0\le x\le \frac32$, the curve lies above the $x$-axis, the total area is $A=\int_0^{3/2}(1+4x-x^2)\,dx.$

2. Compute the total area

$A=\left[x+2x^2-\frac{x^3}{3}\right]_0^{3/2}$

At $x=\frac32$, $x=\frac32,\qquad 2x^2=2\cdot\frac94=\frac92,\qquad \frac{x^3}{3}=\frac{(3/2)^3}{3}=\frac{27/8}{3}=\frac98.$

So, $A=\frac32+\frac92-\frac98=6-\frac98=\frac{48-9}{8}=\frac{39}{8}.$

Hence half the area is $\frac{A}{2}=\frac{39}{16}.$

3. Equation of the bisecting line

The line is $y=mx.$ It passes through the origin. Since it bisects the enclosed area, the area of the part of the region below this line must be $\frac{39}{16}$.

For $0\le x\le \frac32$, the line lies below the parabola and above the $x$-axis. So the area below the line inside the region is simply the triangular area under $y=mx$ from $x=0$ to $x=\frac32$: $\int_0^{3/2} mx\,dx = m\left[\frac{x^2}{2}\right]_0^{3/2} = m\cdot \frac{1}{2}\cdot \frac94 = \frac{9m}{8}.$

Since this must equal half the total area, $\frac{9m}{8}=\frac{39}{16}.$

4. Solve for $m$

$9m=\frac{39}{2}$ $m=\frac{39}{18}=\frac{13}{6}.$

Therefore, $12m=12\cdot\frac{13}{6}=26.$

5. Comparison with stored answer

Derived answer is $26$, which matches the stored correct answer.