SOLUTION: help me with this: write an equation in standard form of the line that contains the point (-1, 2) and is perpendicular to the line y=3x-1

Question 376082: help me with this:
write an equation in standard form of the line that contains the point (-1, 2) and is perpendicular to the line y=3x-1
Found 2 solutions by ankor@dixie-net.com, Alan3354:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
write an equation in standard form of the line that contains the
point (-1, 2) and is perpendicular to the line y=3x-1
:
slope of the equation, m1 = 3
:
The relationship of slopes of perpendicular lines: m1*m2 = -1
3*m2 = -1
m2 = is the slope of the perpendicular line
:
Use point slope form to find the equation of the line, y - y1 = m(x - x1)
where m2=, x1=-1, y1=2
:
y - 2 = (x - (-1))
y - 2 = (x + 1)
y - 2 = x -
y = x - + 2
y = x - +
y = x + ; is the equation of the perpendicular line
But they want it in the standard form
Multiply equation by 3, to get rid of the denominators
3y = -x + 5
x + 3y = 5; is the standard form
:
looks like this

Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
A line and a point example.
Write in standard form the eqation of a line that satisfies the given conditions. Perpendicular to 9x+3y=36, through (1,2)
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Find the slope of the line. Do that by putting the equation in slope-intercept form, y = mx + b. That means solve for y.
9x+3y = 36
3y= - 9x + 36
y = -3x + 12
The slope, m = -3
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The slope of lines parallel is the same.
The slope of lines perpendicular is the negative inverse, m = +1/3
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Use y = mx + b and the point (1,2) to find b.
2 = (1/3)*1 + b
b = 5/3
The equation is y = (1/3)x + 5/3 (slope-intercept form)
x - 3y = -5 (standard form)
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