Answer: \(2x+6y+1=0\).

At the points where the line and curve meet, their \(y\)-values are equal:

\(3x+4=2x^2+8x+1\).

Rearrange to form a quadratic:

\(2x^2+5x-3=0\).

Factorising,

\((2x-1)(x+3)=0\).

So

\(x=\frac12\quad\text{or}\quad x=-3\).

Use the straight line \(y=3x+4\) to find the corresponding \(y\)-coordinates:

when \(x=\frac12\), \(y=3\left(\frac12\right)+4=\frac{11}{2}\);

when \(x=-3\), \(y=3(-3)+4=-5\).

Thus the two points are

\(A\left(\frac12,\frac{11}{2}\right)\quad\text{and}\quad B(-3,-5)\).

The midpoint of \(AB\) is

\(\left(\dfrac{\frac12+(-3)}{2},\dfrac{\frac{11}{2}+(-5)}{2}\right)=\left(-\frac54,\frac14\right)\).

The gradient of \(AB\) is the same as the gradient of the line \(y=3x+4\), so it is \(3\).

Therefore the perpendicular bisector has gradient

\(-\frac13\).

Using the midpoint \(\left(-\frac54,\frac14\right)\), the perpendicular bisector is

\(y-\frac14=-\frac13\left(x+\frac54\right)\).

Multiply by \(12\):

\(12y-3=-4x-5\).

Rearranging,

\(4x+12y+2=0\).

Dividing by \(2\),

\(2x+6y+1=0\).

Answer: \(A=\left(-\frac32,-9\right)\), \(B=(2,5)\), and the area of \(\triangle POQ\) is \(\dfrac{961}{128}\).

At the points of intersection, the two expressions for \(y\) are equal:

\(4x-3=3+5x-2x^2\).

Rearranging gives

\(2x^2-x-6=0\).

Factorising,

\((2x+3)(x-2)=0\).

Hence

\(x=-\frac32\quad\text{or}\quad x=2\).

Using the straight line \(y=4x-3\), the corresponding coordinates are

\(x=-\frac32:\quad y=4\left(-\frac32\right)-3=-9\),

and

\(x=2:\quad y=4(2)-3=5\).

So the two points are

\(A=\left(-\frac32,-9\right)\), \(B=(2,5)\).

The gradient of \(AB\) is

\(\dfrac{5-(-9)}{2-\left(-\frac32\right)}=\dfrac{14}{\frac72}=4\).

Therefore the gradient of the perpendicular bisector is \(-\dfrac14\).

The midpoint of \(AB\) is

\(\left(\dfrac{-\frac32+2}{2},\dfrac{-9+5}{2}\right)=\left(\frac14,-2\right)\).

So the perpendicular bisector is

\(y+2=-\dfrac14\left(x-\frac14\right)\).

Simplifying,

\(y=-\dfrac14x-\dfrac{31}{16}\).

When \(y=0\), \(x=-\dfrac{31}{4}\), so \(P=\left(-\dfrac{31}{4},0\right)\).

When \(x=0\), \(y=-\dfrac{31}{16}\), so \(Q=\left(0,-\dfrac{31}{16}\right)\).

The triangle \(POQ\) is right-angled at \(O\), hence

\(\text{area}=\dfrac12\left|OP\right|\left|OQ\right|=\dfrac12\cdot\dfrac{31}{4}\cdot\dfrac{31}{16}=\dfrac{961}{128}\).