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.
Please help! This is overdue and if I don't turn it in I fail!
in
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looks like I had a typo from
8x - 5y = 30 <---- ok here
so 5y = 8x - 30 , to
y = (8/5) x - 6
so m + b = 8/5 - 6
= -22/5
you should have caught my error instead of saying you got the same thing
L1 has slope -5/8
L2 must have slope 8/5
So, the point-slope form of L2 is
y-10 = 8/5 (x-10)
y = 8/5 x - 6
m+b = 8/5 - 6 = -22/5
To find the equation of the line perpendicular to line l_1, we first need to determine the slope of line l_1. The given equation of line l_1 is 5x + 8y = -9.
To get the slope of line l_1 in the form of y = mx + b, we can rearrange the equation by isolating y:
8y = -5x - 9
y = (-5/8)x - 9/8
The slope of line l_1, represented by m_1, is -5/8.
Perpendicular lines have slopes that are negative reciprocals of each other. So, the slope of line l_2, represented by m_2, will be the negative reciprocal of m_1:
m_2 = -1 / m_1
m_2 = -1 / (-5/8)
m_2 = 8/5
Now that we have the slope of line l_2, we can find its equation using the point (10, 10). Since line l_2 passes through this point, we can substitute x = 10 and y = 10 into the equation y = mx + b:
10 = (8/5)(10) + b
10 = 16 + b
b = 10 - 16
b = -6
The equation of line l_2 is y = (8/5)x - 6. Therefore, m + b equals (8/5) + (-6) = 8/5 - 30/5 = -22/5.