Answer:
Option B is perpendicular to y=-3x+4 and goes through point (-1,5)
Explanation:
We will look for an equation of the form y=mx+b, where m is the slope and b the y-intercept. The reference line, y=-3x+4, has a slope of -3. A line perpendicular to this line will have a slope that is the "negative inverse" of the refence lines slope. In this case, it's slope would be +(1/3).
We could graph each of the equation options and look for one that is perpendicular, but it might be faster to simply rearrnaged the options into standard format of y=mx+b. which we'll do here:
Option Rearranged Equation Slope
A) 3y=-x+16
y = -(1/3)x+(16/3) -(1/3)
B) 3y=x+16
y = (1/3)x + (16/3) (1/3)
C) y=-3x+8 -3
D) y=-3x+2 -3
There is only one option that has the required slope to be parallel to the reference equation (y=-3x+4), option B: y = (1/3)x + (16/3)
Does it go through point (-1,5)?
y = (1/3)x + (16/3)
Enter x = -1 and find 5:
y = (1/3)(-1) + (16/3)
y = (16/3)-(1/3)
y = (15/3) or 5
Yes, it goes through point (-1,5)
Option B is perpendicular to y=-3x+4 and goes through point (-1,5).