The line $y=mx$ bisects the area enclosed by the lines $x=0,y=0,x=\frac{3}{2}$ and the curve $y=1+4x-x^2,$ then $m$ is equal to 

Here, the curve is $x^2-4x=1-y,ie,{(x-2)}^2=-(y-5)$

It is a parabola whose vertex is $(2,5),$ axis is $x-2=0$ and $y\le 5$ at every point of the curve. 

Also, when $x=0,y=1+4x-x^2$ gives $y=1$

So, the parabola cuts the $y-$ axis at $(0,1)$ when $y=0,y=1+4x-x^2$ gives; $1+4x-x^2=0$ ie, $x^2-4x-1=0$

$\therefore x=\frac{4\pm \sqrt{20}}{2}=2\pm \sqrt{5}$

The parabola cut’s $x-$ axes $A(2-\sqrt{5},0)\&B(2+\sqrt{5},0)$

$\therefore$The parabola cuts the $x-$ axis at $A(2-\sqrt{5},0)$ and $B(2+\sqrt{5},0).$

The area enclosed by the lines $x=0,y=0,x=3/2$ and the curve 

$={\int }_0^{3/2}ydx={\int }_0^{3/2}(1+4x-x^2)dx={[x+2x^2-\frac{x^3}{3}]}_0^{3/2}=(\frac{3}{2}+2.\frac{9}{4}-\frac{1}{3}.\frac{27}{8})=\frac{39}{8}$

The area bounded by the lines $(y=mx,y=0andx=3/2)={\int }_0^{3/2}ydx={\int }_0^{3/2}mxdx=\frac{9m}{8}$

According to questions, $\frac{9}{8}m=\frac{1}{2}.\frac{39}{8}\Rightarrow m=13/6$