Answer: \(E=(7,2)\), area \(=40\).

The gradient of \(AB\) is

\(\frac{-9-3}{6-(-4)}=-\frac65.\)

So the gradient of the perpendicular bisector is

\(\frac56.\)

The midpoint of \(AB\) is

\(\left(\frac{-4+6}{2},\frac{3+(-9)}{2}\right)=(1,-3).\)

Therefore the perpendicular bisector of \(AB\) has equation

\(y+3=\frac56(x-1).\)

This simplifies to

\(y=\frac56x-\frac{23}{6}.\)

At \(E\), this line meets \(L:y=11x-75\). Hence

\(11x-75=\frac56x-\frac{23}{6}.\)

Multiplying by \(6\),

\(66x-450=5x-23,\)

so \(61x=427\), giving \(x=7\). Then \(y=11(7)-75=2\), so

\(E=(7,2).\)

Now use the determinant formula for the area of triangle \(CDE\). With \(C=(15,10)\), \(D=(14,-1)\) and \(E=(7,2)\),

\(\text{area}=\frac12\left|(-1)(-8)-(-11)(-8)\right|=40.\)

Answer: \(\dfrac{121}{4}\).

We start with the main method. Use gradients: perpendicular lines have gradients whose product is \(-1\).

The midpoint of \(AB\) is

\(\left(\frac{1+5}{2},\frac{4+6}{2}\right)=(3,5).\)

The gradient of \(AB\) is

\(\frac{6-4}{5-1}=\frac12,\)

so the perpendicular bisector has gradient \(-2\).

Its equation through \((3,5)\) is

\(y-5=-2(x-3).\)

Hence \(y=-2x+11\).

The intercepts are \(D=(0,11)\) and \(C=\left(\frac{11}{2},0\right)\).

Therefore

\(\text{area of } \triangle OCD=\frac12\cdot\frac{11}{2}\cdot11=\frac{121}{4}.\)

This completes the solution and gives the required result.