Answer: (a) \(-4\); (b) \((1,6)\) and \((19,-24)\).

Answer: (a) \(-4\); (b) \((1,6)\) and \((19,-24)\).

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

\(\left(\frac{5+15}{2},\frac{-12+(-6)}{2}\right)=(10,-9).\)

The gradient of \(PQ\) is

So the gradient of the perpendicular bisector \(l\) is

\(-\frac53.\)

Using the point \((10,-9)\), the equation of \(l\) is

\(y+9=-\frac53(x-10).\)

The \(x\)-coordinate of \(R\) is \(7\), so

\(y+9=-\frac53(7-10)=5.\)

Thus

\(y=-4.\)

(b) Let \(M=(10,-9)\), the midpoint of \(PQ\). Since \(l\) is perpendicular to \(PQ\), the distance of a point on \(l\) from \(PQ\) is its distance from \(M\) along the line \(l\).

Now

\(\overrightarrow{MR}=(7,-4)-(10,-9)=(-3,5).\)

The point \(S\) must be three times as far from \(PQ\) as \(R\) is, so

\(\overrightarrow{MS}=\pm3\overrightarrow{MR}.\)

Using \(+3\overrightarrow{MR}\),

\(S=(10,-9)+3(-3,5)=(1,6).\)

Using \(-3\overrightarrow{MR}\),

\(S=(10,-9)-3(-3,5)=(19,-24).\)

Therefore the two possible positions of \(S\) are

\((1,6)\quad\text{and}\quad(19,-24).\)