Question 182961: Write the equation for a line that is perpendicular to the line,
y = 5x + 10, and goes through the point (8, -1).
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! y = 5x + 10, and goes through the point (8, -1).
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The "slope-intercept" form of any line is:
y = mx + b
where
m is slope
b is the y-intercept
.
This is exactly the form you equation was given:
y = 5x + 10
(slope is 5)
Now, we know our NEW line has to have a slope that is the "negative reciprocal" for it to be perpendicular:
Let m = our new slope
then
5m = -1
m = -1/5 (slope of our NEW line)
.
The NEW slope along with the supplied point of (8, -1) can be stuffed into the "point-slope" form:
y - y1 = m(x-x1)
y - (-1) = (-1/5)(x-8)
y + 1 = (-1/5)x + 8/5
y = (-1/5)x + 8/5 - 1
y = (-1/5)x + 8/5 - 5/5
y = (-1/5)x + 3/5 (this is what they're looking for)
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