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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
avatar+111321 
+1

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

 

 

cool cool cool

 Mar 21, 2020
 #2
avatar+1956 
+2

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

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