Answer: \(4x+3y-29=0\); \(p=3\); \(D=(-1,11)\).

(a) The midpoint of \(AB\) is

\(\left(\frac{-2+6}{2},\frac{4+10}{2}\right)=(2,7).\)

The gradient of \(AB\) is

\(\frac{10-4}{6-(-2)}=\frac68=\frac34.\)

So the gradient of the perpendicular bisector is

\(-\frac43.\)

Using point \((2,7)\),

\(y-7=-\frac43(x-2).\)

Multiply by 3:

\(3y-21=-4x+8.\)

Therefore

\(4x+3y-29=0.\)

(b) Since \(C=(5,p)\) lies on the perpendicular bisector, substitute \(x=5\), \(y=p\):

\(4(5)+3p-29=0.\)

So

\(3p=9,\)

and hence

\(p=3.\)

(c) Since \(AB\) bisects \(CD\), the midpoint of \(CD\) is the midpoint of \(AB\), which is \((2,7)\).

Let \(D=(x,y)\). With \(C=(5,3)\),

\(\left(\frac{5+x}{2},\frac{3+y}{2}\right)=(2,7).\)

Therefore

\(5+x=4,\qquad 3+y=14.\)

So

\(x=-1,\qquad y=11.\)

Hence

\(D=(-1,11).\)