Answer: (a) \(C=(-2,2)\); (b) \(y=\frac34x+\frac72\); (c) \((2,5)\) and \((-6,-1)\).

Answer: (a) \(C=(-2,2)\); (b) \(y=\frac34x+\frac72\); (c) \((2,5)\) and \((-6,-1)\).

(a) Since

\(\frac{AC}{CB}=\frac12,\)

we have \(AC:CB=1:2\). So \(C\) is one third of the way from \(A\) to \(B\).

The vector from \(A\) to \(B\) is

\((4-(-5),-6-6)=(9,-12).\)

One third of this vector is

\((3,-4).\)

Therefore

\(C=(-5,6)+(3,-4)=(-2,2).\)

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

Since \(CD\) is perpendicular to \(AB\), its gradient is

\(\frac34.\)

The line \(CD\) passes through \((-2,2)\), so

\(y-2=\frac34(x+2).\)

Hence

\(y=\frac34x+\frac72.\)

(c) Let \(D=(x,y)\). Since \(BD=\sqrt{125}\),

\((x-4)^2+(y+6)^2=125.\)

Also \(D\) lies on \(CD\), so

\(y=\frac34x+\frac72.\)

Substitute this into the distance equation:

\((x-4)^2+\left(\frac34x+\frac72+6\right)^2=125.\)

This simplifies to

\((x-4)^2+\left(\frac34x+\frac{19}{2}\right)^2=125.\)

Multiplying out and simplifying gives

\(25x^2+100x-300=0.\)

Divide by \(25\):

\(x^2+4x-12=0.\)

Factorising,

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

Thus

\(x=2\quad\text{or}\quad x=-6.\)

If \(x=2\), then

\(y=\frac34(2)+\frac72=5.\)

If \(x=-6\), then

\(y=\frac34(-6)+\frac72=-1.\)

So the two possible positions of \(D\) are

\((2,5)\quad\text{and}\quad(-6,-1).\)