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