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# PLS HELP CPhill Or ANYONE

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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...... Jun 11, 2021