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Question 413564: Hi, can anyone help figure out these two coordinate geometry problems?:
1) How do I find the equation of the line passing through (12,4) and perpendicular to y=3x+6? Leave answer in general form.
2) How do I find the equation of the line passing through (6,10) and perpendicular to 3x-4y=12? Leave answer in general form.
Thank you so much for your time! I have test on these tomorrow.
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
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