SOLUTION: Find the equation of the line through point (−5,5) and perpendicular to y=59x−4. Use a forward slash (i.e. "/") for fractions (e.g. 1/2 for 12).and the next problem is Find the

Algebra ->  Linear-equations -> SOLUTION: Find the equation of the line through point (−5,5) and perpendicular to y=59x−4. Use a forward slash (i.e. "/") for fractions (e.g. 1/2 for 12).and the next problem is Find the      Log On


   



Question 1183225: Find the equation of the line through point (−5,5) and perpendicular to y=59x−4. Use a forward slash (i.e. "/") for fractions (e.g. 1/2 for 12).and the next problem is Find the equation of the line through point (−1,2) and perpendicular to x+3y=3. Use a forward slash (i.e. "/") for fractions (e.g. 1/2 for 12).
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
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equation of the line through point (−5,5) and perpendicular to y=59x−4. Use a forward slash (i.e. "/") for fractions (e.g. 1/2 for 12).and
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Negative reciprocal of coefficient on x term, because asked for "perpendicular" line.

y=-%281%2F59%29x%2Bb
y%2Bx%2F59=b
b=y%2Bx%2F59
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b=5%2B%28-5%29%2F59
b=4%2654%2F59
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highlight%28y=-x%2F59-4%2654%2F59%29

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
Find the equation of the line through point (−5,5) and perpendicular to y=59x−4. Use a forward slash (i.e. "/") for fractions (e.g. 1/2 for 12).and the next problem is Find the equation of the line through point (−1,2) and perpendicular to x+3y=3. Use a forward slash (i.e. "/") for fractions (e.g. 1/2 for 12).
highlight%28y=-x%2F59%2B54%2F59%29. He's WRONG, so: cross%28highlight%28y=-x%2F59%2B54%2F59%29%29
matrix%281%2C7%2C+y%2C+%22=%22%2C+59x+-+4%2C+%22====%3E%22%2C+59x+-+y%2C+%22=%22%2C+4%29
matrix%281%2C3%2C+x+%2B+59y%2C+%22=%22%2C+%28x%29+%2B+59%28y%29%29 ------ Switching coordinates, negating NEW y coordinate, and equating to a constant,
calculated using the coordinates of the requested equation

Same concept applies to the 2nd equation that's being sought.