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.
+128474 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
+2094 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
