Answer: \(k=-\frac45\). The possible coordinates of \(D\) are \(\left(\frac92,-2\right)\) and \(\left(-\frac{51}{10},\frac{14}{5}\right)\).
Work through the given conditions step by step, keeping exact values until the final answer is required.
Substitute \(y=2x+1\) into \(y+xy+3x^2=15\):
\(2x+1+x(2x+1)+3x^2=15.\)
This simplifies to
\(5x^2+3x-14=0.\)
Solving gives \(x=\frac75\) or \(x=-2\). The corresponding \(y\)-values are \(\frac{19}{5}\) and \(-3\).
The midpoint of \(AB\) is
\(\left(\frac{\frac75-2}{2},\frac{\frac{19}{5}-3}{2}\right)=\left(-\frac{3}{10},\frac25\right).\)
The line \(AB\) has gradient \(2\), so the perpendicular bisector has gradient \(-\frac12\). Its equation is
\(y-\frac25=-\frac12\left(x+\frac{3}{10}\right).\)
Since \(C=\left(\frac{21}{10},k\right)\) lies on this line,
\(k-\frac25=-\frac12\left(\frac{21}{10}+\frac{3}{10}\right)=-\frac65.\)
Thus \(k=-\frac45\).
The vector from the midpoint to \(C\) is
\(\left(\frac{21}{10},-\frac45\right)-\left(-\frac{3}{10},\frac25\right)=\left(\frac{12}{5},-\frac65\right).\)
A point twice as far from \(AB\) on the same perpendicular line can lie on either side of the midpoint. Hence
\(D=\left(-\frac{3}{10},\frac25\right)+2\left(\frac{12}{5},-\frac65\right)=\left(\frac92,-2\right)\),
or
\(D=\left(-\frac{3}{10},\frac25\right)-2\left(\frac{12}{5},-\frac65\right)=\left(-\frac{51}{10},\frac{14}{5}\right).\)
This gives the required answer: \(k=-\frac45\). The possible coordinates of \(D\) are \(\left(\frac92,-2\right)\) and \(\left(-\frac{51}{10},\frac{14}{5}\right)\).
