Question 283172: Find the equation of the line that passes through point (4,7) and that is perpendicular to the line whose equation is y = -2x + 1.
Found 2 solutions by Alan3354, nerdybill:
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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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 + 13
The slope, m = -3
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The slope of lines parallel have the same slope.
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.
Answer by nerdybill(7384)  (Show Source):
You can put this solution on YOUR website! Find the equation of the line that passes through point (4,7) and that is perpendicular to the line whose equation is y = -2x + 1.
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Slope of:
y = -2x + 1
is -2
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The new line has to be perpendicular so the two slopes are "negative reciprocals":
m(-2) = -1
m = 1/2
This slope along with the point (4,7) is plugged into the "point-slope" form:
y-y1 = m(x-x1)
y-7 = (1/2)(x-4)
y-7 = (1/2)x - 2
y = (1/2)x + 5 (this is what they're looking for)
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