Question 986526: Let line l1 be the graph of 5x + 8y = -9. Line l2 is perpendicular to line l1 and passes through the point (10,10). If line l2 is the graph of the equation y=mx +b, then find m+b.
This makes no sense to me. Help?
Found 2 solutions by Cromlix, Alan3354:
Answer by Cromlix(4381)   (Show Source):
You can put this solution on YOUR website! Hi there,
First of all rearrange 5x + 8y = -9
into the form:
y = mx + b
5x + 8y = -9
8y = -5x -9
y = -5/8x - 9/8
m is the gradient and equals
here -5/8
b is the y coordinate of the
line's intersection with the 'y' axis.
here -9/8
..................
Lines that are perpendicular to one another
have gradients that multiply together to give -1
m1 x m2 = -1
To find the gradient (m) of line 12
-5/8 x m2 = -1
m2 = 8/5
Now using the line equation:
y - b = m(x - a)
Where m = 8/5
(a,b) -> (10, 10)
y - 10 = 8/5(x - 10)
y - 10 = 8/5x - 80/5
y = 8/5x - 80/5 + 50/5 (10)
y = 8/5x - 30/5......this is line 12
Now you can report that m = 8/5
and b = -30/5
m + b
So 8/5 + (-30/5) = -22/5
Hope this helps :-)
Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! Let line l1 be the graph of 5x + 8y = -9. Line l2 is perpendicular to line l1 and passes through the point (10,10). If line l2 is the graph of the equation y=mx +b, then find m+b.
======================
Find the slope of L1 5x + 8y = -9
To do that, put it in slope-intercept form y = mx + b. That means solve for y.
5x + 8y = -9
8y = -5x - 9
y = (-5/8)x - 9/8
m is the slope = -5/8
b is the y-intercept = -9/8 but that's not relevant here.
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The slope of any line perpendicular is the negative inverse of -5/8 --> 8/5.
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Now find the equation of L2 thru (10,10) with a slope of 8/5
--
y-y1 = m*(x-x1) where (x1,y1) is the point (10,10)'
y-10 = (8/5)*(x-10) = (8/5)x - 80/5
y = (8/5)x - 6
--> for L2, m = 8/5 and b = -6
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