Question 824660: find the equation of the line perpendicular to 5x+2y-1=0, and passing through the point (-2,7)
Answer by TimothyLamb(4379)   (Show Source):
You can put this solution on YOUR website! ---
given:
5x + 2y - 1 = 0
(-2,7)
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put in slope-intercept form:
2y= -5x + 1
y= (-5/2)x + 1/2
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perpendicular lines have negative reciprocal slopes,
so the perpendicular slope = 2/5 = 0.4
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to solve the linear equation, copy and paste this (given point and perpendicular slope):
-2,7,0.4
into the "Point-Slope form: x1 y1 m" input box here: https://sooeet.com/math/linear-equation-solver.php
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answer:
x-intercept = (-19.5, 0)
y-intercept = (0, 7.8)
slope = 0.4
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the perpendicular line's slope-intercept form:
y = (0.4)x + 7.8
y = (2/5)x + 7+4/5
y = (2/5)x + 39/5
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NOTE TO STUDENT:
The problem statement asks for an equation of the line, it doesn't ask for the equation of the line in standard form. So this is one answer to the problem statement:
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the perpendicular line's slope-intercept form:
y = (0.4)x + 7.8
y = (2/5)x + 7+4/5
y = (2/5)x + 39/5
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If you want the standard form of the equation, it's simple algebra to convert the slope-intercept form to standard form:
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y = (2/5)x + 39/5
(2/5)x - y + 39/5 = 0
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