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.
15
Reasoning:
4y = -3-14 : line1
y = -3/4x - 7/2
line2: y = 4/3x+b
line 2 passes thru (-5,7) plug in
7 = 4/3 (-5) + b
simplify
7 = -20/3 + b
b = 13 2/3
m + b = 4/3 + 13 2/3 = 15
A: 15
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But here is the answer that AoPS has:
Since lines l_1 and l_2 are perpendicular (and neither is vertical), the product of their slopes is -1. To find the slope of l_1, we write 3x + 4y = -14 in slope-intercept form by solving for y. Solving for y, we find
y = -3/4x - 7/2. Therefore, the slope of l_1 is -3/4
Since line l_2 is perpendicular to line l_1, the slope of l_2 is 4/3. The line l_2 passes through the point (-5,7), so its equation is given by
y - 7 = 4/3(x + 5). Isolating y, we find
y = 4/3x + 41/3, so m + b = 4/3 + 41/3 = 15
i used to know but now i forgot but i guess i may help you on that
if two lines are perpendicular their slope product = -1
line 1:y=-5/8 x-9
a1 x a2 =-1
-5/8 x a2 =-1
a2=8/5
line2: y= 8/5 x +b
Thanks a lot!
and then you substitute the point and u'll find b
then m+b=*
hope im correct
@Reiny I got that too but it said it was incorrect.
Your new line is perpendicular to 5x + 8y = -9
So it must have the form 8x - 5y = c
sub in the point (10,10)
8(10) - 5(10) = c = 30
new equation is 8x - 5y = 30
or y = (8/5)x - 15/4
then m+b = 8/5 - 15/4 = -43/20