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Question 341221: Write an equation that is perpendicular to 2x+3y=6 with point (3,5).
Found 2 solutions by Alan3354, stanbon:
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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For further assistance, or to check your work, email me via the thank you note, or at Moral Loophole@aol.com
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! Write an equation that is perpendicular to 2x+3y=6 with point (3,5).
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The given line: y = (-2/3)x+2
slope is -2/3
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A perpendicular line will have slope = 3/2 and pass thru (3,5)
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y = mx+b
5 = (3/2)3+b
5 = (9/2)+b
(10/2) = (9/2) + b
b = 1/2
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Equation:
y = (3/2)x + 1/2
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Graph:
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Cheers,
Stan H.
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