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# HEEEELPPP!

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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).

Jun 5, 2020

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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 $$\boxed{(-3,-129)}$$.

Jun 5, 2020