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.

Guest Mar 21, 2020

#1**+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

CPhill Mar 21, 2020

#2**+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