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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

 Oct 5, 2020
 #1
avatar+2 
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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  ?

 Oct 5, 2020
 #2
avatar+1328 
+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

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