Find the equation of a line in slope-intercept form that passes through the point (-3, 1) and is vertical to the line 3x + 4y = 0.

a) y = -4/3x + 5

b) y = 3/4x + 5

c) y = -3/4x + 5

d) y = 4/3x + 5

SmartMathMan Oct 5, 2020

#1**+1 **

I don't like that they use the word "vertical" in the question. To me that would mean a line of the form x = -3.

I assume they mean "perpendicular" to.

Even " y = -4/3x + 5" is slightly ambiguous y = -4/(3x) + 5 or y = (-4/3)x + 5 I assume the latter.

Solutions

3x + 4y = 0 solve for y .

4y= -3x

y=-3/4*x gives k =-3/4 for the original line

k_{perpendicular} = -1/k = -1(1/(-3/4)) = -(-4/3) =** 4/3 the only answer that fits that is d**

Check does (-3, 1) exist on that line?

d) y = 4/3x + 5 = 4*(-3)/3 +5 = -4 +5 = 1 Yes

How did you get y = -4/3x -3 ?

jeffjeff Oct 5, 2020

#2**+1 **

THANK YOU!! I was trying to do what they taught me but it always came out weird. Maybe it was just a simple problem that I made too complicated.

SmartMathMan
Oct 6, 2020