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# Help

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Let line l1 be the graph of 3x+4y=-14. Line l2 is perpendicular to line l1 and passes through the point (-5,7). If line l2 is the graph of the equation y=mx+b, then find m+b.

Mar 21, 2020

#1
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The slope of the first line  can be found as

3x + 4y  = -14

4y   =  -3x  - 14      divide  both sides by  4

y = (-3/4)x - 14/4

y = (-3/4)x - 7/2

Since line 2  is perpendicular to this one then its slope =  4/3

And   since line 2 passes through  (-5, 7)   we have that

7 = (4/3)(-5) + b

7 = -20/3 + b

21/3  = -20/3  + b     add 20/3 to both sides

41/3  = b

So.... line 2  has the equation

y = (4/3)x + 41/3

And  m + b  =   (4/3) + (41/3)  =  45/3 =   15   Mar 21, 2020
#2
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Nice, Chris! We can also look at it like this--

First equation:   3x + 4y  = -14

y = (-3/4)x - 7/2

We know that Line 2  is perpendicular to Line 1.

Therefore, its slope is also -4/3

The question also says that Line 2 passes through (-5, 7).

Second equation: 7 = (4/3)(-5) + b

b= 41/3

Line 2 will have the equation

y = (4/3)x + 41/3

Where 4/3 is the slope and 41/3 is the y-intercept.

Therefore, Slope+y-intercept will give us 4/3) + (41/3)  =  45/3 = 15

CalTheGreat  Mar 21, 2020