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# help

-1
215
2
+144

Let line $$l_1$$ be the graph of $$5x+8y=-9$$. Line $$l_2$$ is perpendicular to line $$l_1$$ and passes through the point $$(10,10)$$. If line $$l_2$$ is the graph of the equation $$y=mx+b$$, then find $$m+b$$

Aug 11, 2019

#1
+22062
0

l1  is re-written as   y = -5/8 x -9/8     slope m is  -5/8

l2 perpindicular slope is     -1/(-5/8) = 8/5

y = 8/5 x + b     sub in point   10,10

10 = 8/5 (10) + b

b = 10 - 80/5 = -30/5

-30/5 +8/5=  -22/5

Aug 11, 2019
#2
+1

The answer is $$-\frac{22}{5}$$ but if you want the full explantion you get get that here

https://www.algebra.com/algebra/homework/Graphs/Graphs.faq.question.986526.html

I don't know if this is the correct way to post a answer because I have not done it.

Hope it Helps!!

Aug 11, 2019