Let P and Q be constants. The graphs of the lines x+5y=7 and 15x+Py=Q are perpendicular and intersect at the point (-8,3). Enter the ordered pair (P,Q).
First, we know that the line \(15x+Py = Q\) passes through \((-8,3)\). Thus, \(-120 + 3P = Q\) so \(3P-Q = 120\).
Also, the slope of \(x+5y=7\) is \(-\frac{1}{5}\) so thus the slope of \(15x+Py=Q\) must be the negative reciprocal of that, since it is perpendicular. Thus, its slope is 5. Moving the \(15x\) to the other side, we can see that its slope is also \(-\frac{15}{P}\). Thus, \(-\frac{15}{P} = 5\) so \(P=-3\).
We had that \(3P-Q = 120\), so since \(P=-3\), we have \(Q = -129\). Thus, the answer is \(\boxed{(-3,-129)}\).
