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Line $l_1$ represents the graph of $3x + 4y = -14$. Line $l_2$ passes through the point $(-5,7)$, and is perpendicular to line $l_1$. If line $l_2$ represents the graph of $y=mx +b$, then find $m+b$.

 Jun 10, 2021
edited by bossboy1335  Jun 10, 2021
 #1
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Isolate y in 3x + 4y = -14 ==> y = 3x/4 - 14/4

 

Slope of Perpendicular = -4/3, point-slope form of equation y - 7 = -4/3(x + 5)

==> 3y - 21 = -4(x + 5) = -4x + 20

==> 3y = -4x + 41

==> y = -4/3*x + 41/3

 

m + b = -4/3 + 41/3 = 37/3

 Jun 10, 2021
 #2
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incorrect

bossboy1335  Jun 10, 2021
 #3
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L1    3x+4y = - 14    is tha same as     y = - 3/4 x - 14/4       slope, m = - 3/4      perpindicular slope is  - 1/m = 4/3

 

Your new line L2  is thus   y = 4/3 x + b    sub in the point given to calculate 'b'

                                         7 = 4/3 (-5) + b

                                              so b = 13 2/3

 

 

y = 4/3 x + 13 2/3          Guest messed up with the slope ...it is not negative in the perpindicular line....

                   you can finish......    cheeky

 Jun 11, 2021

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