SOLUTION: Find the equation of the line containing the point (2,-5) and perpendicular to the line y=5/2x-4.

Question 267302: Find the equation of the line containing the point (2,-5) and perpendicular to the line y=5/2x-4.
Found 2 solutions by Alan3354, checkley77:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Do it like this:
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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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Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
(2,-5) and perpendicular to the line
y=5x/2-4 (red line)
Slope=5/2.
The perpendicular line will have a slope=-2/5.
-5=2*-2/5+b
-5=-4/5+b
b=-5+4/5
b=-25/5-4/5
b=-29/5 ans. for the Y intercept.
y=-2x/5-29/5 (green line)
(graph 300x200 pixels, x from -6 to 5, y from -10 to 10, of TWO functions 5x/2 -4 and -2x/5 -29/5).
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