# Equation solving

Perpendicular. To x-3y=2 and contains ( 2,4)

2). That is parallel to 2x+5y=4

I don't get how to solve and use y=Mx+b

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1. 1.
given: x - 3y = 2
-3y = -x + 2
y = (1/3)x - 2/3
so the slope of the given line is 1/3
then the slope of a perpendicular line is -3
new equation:
y = 3x + b
but (2,4) must lie on it, so
4 = 3(2) + b
4 = 6 + b
-2 = b

new equation: y = 3x - 2

2. I assume we are dealing with the same point (2,4)
Easiest way:
If the new line is parallel to 2x + 5y = 4
it must differ only in the constant.
New line: 2x + 5y = c
with (2,4) on it ...
2(2) + 5(4) = c = 24

new equation: 2x + 5y = 24
To change any equation to the form y = mx + b takes 2 steps:
1. take all terms except the y term to the right side
2x + 5y = 24 ------> 5y = -2x + 24
2. divide each term by the coefficient of the y term
5y = -2x + 24 ----> y = (-2/5)x - 24/5

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posted by Reiny

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