SOLUTION: Find an equation for the line that passes through (0,-5) and is perpendicular to the graph of x+3y=4. Write in Slope-Intercept form:
So far I have
x+3y-3y-4=4-3y-4
x-4=-3y
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Question 730616: Find an equation for the line that passes through (0,-5) and is perpendicular to the graph of x+3y=4. Write in Slope-Intercept form:
So far I have
x+3y-3y-4=4-3y-4
x-4=-3y
then divide each side by -3
-1x/3-(-4/3)=y or y=-1x/3+3/4
-5=-1(0)/3+3/4 it gets mixed up from here, am I even doing this correctly? Thanks
Found 2 solutions by mananth, Alan3354:
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
good job
slope = -1/3
the line perpendicular will have a slope of 3 ( negative reciprocal)
point (0,-5)
equation will be
(y-(-5))=3(x-0)
y+5=3x
y=3x-5 is the equation
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Find an equation for the line that passes through (0,-5) and is perpendicular to the graph of x+3y=4. Write in Slope-Intercept form:
----------------
Find the slope of the line.
x+3y=4
Solve for y
y = (-1/3)x + 4/3
The slope = -1/3
The slope of lines perpendicular is the negative inverse = 3
-----------
(0,-5) is b, the y-intercept
--> y = 3x - 5
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