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Question 370666: Here is my second question.
Find the slope of a line perpendicular to the line that passes through the following points.
(3,9) and (7,15)
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Given 2 points example.
Find the equation of the line thru the points (2,1) and (3,5)
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This is a 2 step process. First find the slope of the line thru the points.
slope, m = diffy/diffx
m = (5-1)/(3-2)
m = 4
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Now use y = mx + b with either point to find b, the y-intercept.
y = mx + b
5 = 4*3 + b
b = -7
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y = mx + b
y = 4x - 7 is the answer.
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Another approach using determinants:
|x y 1|
|2 1 1| = 0
|3 5 1|
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x(1-5) - y(2-3) + 1(10-3) = 0
-4x + y + 7 = 0
y = 4x - 7
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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
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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
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