Question 1194177: Write the equation of the line that is passing through (5,-2)
parallel to y=2/5x-2
perpendicular to y=2/5x-2
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
The slope of y = mx+b is m
For y = (2/5)x - 2, that slope is m = 2/5
Rule: Parallel lines have equal slopes, but different y intercepts.
Anything parallel to y = (2/5)x-2 must have a slope of m = 2/5, but a different y intercept.
Let's plug in the coordinates of (x,y) = (5,-2) and solve for b
y = mx + b
y = (2/5)x + b
-2 = (2/5)*5 + b
-2 = 2 + b
b = -2-2
b = -4
The parallel line passing through (5,-2) is y = (2/5)x - 4
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Given a slope of a/b, the perpendicular slope is -b/a
We do two things:
* flip the fraction
* flip the sign (from positive to negative or vice versa)
For 2/5, the perpendicular slope is -5/2
Note how the two slopes multiply to -1
Now let's find the y intercept
y = mx+b
y = (-5/2)x+b
-2 = (-5/2)*5+b
-2 = -12.5+b
b = -2+12.5
b = 10.5
b = 21/2
The perpendicular line passing through (5,-2) is y = (-5/2)x + 21/2
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