+1438 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
+2 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
kperpendicular = -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 ?
+1438 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.
