SOLUTION: find an equation in standard form of the line through the point (3, -5) that is perpendicular to the line with equation x + 3y = 6

Question 379233: find an equation in standard form of the line through the point (3, -5) that is perpendicular to the line with equation x + 3y = 6
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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