Answer: \(A=(-6,8)\), \(B=(-14,12)\); possible \(D\) coordinates are \((10,50)\) and \((-20,-10)\).

The equation \(|x+y|=2\) represents the two lines

\(x+y=2 \quad\text{and}\quad x+y=-2.\)

Intersect \(x+2y=10\) with \(x+y=2\):

\((x+2y)-(x+y)=10-2,\)

so

\(y=8.\)

Then \(x+8=2\), so \(x=-6\). One point is

\((-6,8).\)

Intersect \(x+2y=10\) with \(x+y=-2\):

\(y=12.\)

Then \(x+12=-2\), so \(x=-14\). The other point is

\((-14,12).\)

The mid-point of \(AB\) is

The gradient of \(AB\) is

\(\frac{12-8}{-14-(-6)} =\frac4{-8} =-\frac12.\)

Therefore the perpendicular bisector has gradient \(2\).

The line through \(M(-10,10)\) with gradient \(2\) is

\(y-10=2(x+10).\)

For \(C(-5,20)\),

\(20-10=10\)

and

\(2(-5+10)=10.\)

So \(C\) lies on the perpendicular bisector of \(AB\).

(b) The vector from \(C\) to \(M\) is

Since \(D\) lies on the same line and \(CD=3CM\), there are two possible directions from \(C\).

One possibility is

\(D=C+3\overrightarrow{CM}.\)

So

\(D=(-5,20)+3(-5,-10)=(-20,-10).\)

The other possibility is

\(D=C-3\overrightarrow{CM}.\)

So

\(D=(-5,20)-3(-5,-10)=(10,50).\)

Therefore the possible coordinates of \(D\) are

\((-20,-10)\quad\text{and}\quad(10,50).\)