Answer: \(D=(22,5)\) or \(D=(-18,13)\).

(a) At the points of intersection,

\(x^2+x-1=\frac{x^2+6x-2}{2}.\)

Multiplying by \(2\),

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

So

\(x^2-4x=0.\)

Therefore

\(x(x-4)=0,\)

giving \(x=0\) or \(x=4\).

Using \(y=x^2+x-1\), the corresponding points are

\((0,-1)\quad\text{and}\quad (4,19).\)

The mid-point of these points is

(b) The gradient of \(AB\) is

\(\frac{19-(-1)}{4-0}=5.\)

So the gradient of the perpendicular bisector is \(-\frac15\). Since it passes through \((2,9)\), its equation is

\(y-9=-\frac15(x-2).\)

Substituting \(C(12,7)\),

\(7-9=-\frac15(12-2).\)

This gives

\(-2=-2,\)

so \(C\) lies on \(l\).

(c) Since \(l\) is perpendicular to \(AB\), distance from \(AB\) along \(l\) is measured from the mid-point \((2,9)\). The vector from \((2,9)\) to \(C(12,7)\) is

The point \(D\) must be twice as far from \(AB\), so from \((2,9)\) the displacement is

in one direction, or the negative of this in the other direction.

Therefore

\(D=(2,9)+(20,-4)=(22,5),\)

or

\(D=(2,9)-(20,-4)=(-18,13).\)