+335 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 $(P,Q).$
+781 Write x+5y=7 in slope intercept form.
x+5y=7
5y=-x+7
y=-x/5+7/5
The slope of the perpendicular line, 15x+Py=Q must be 5.
15x+Py=Q
Py=-15x+Q
y=-15/Px+Q/P
-15/P must equal to 5 so P=-3. We substitute the given point into 15x-3y=Q.
15x-3y=Q
15(-8)-3(3)=Q
-120-9=Q
Q=-129
(-3, -129)
