What is the equation of a line that passes through the point (10, 5) and is perpendicular to the line whose equation is y=5/4x−2
?
1 answer:
The slope of the given line is 5/4. The slope of a line perpendicular to the given line is the negative reciprocal of 5/4, or -4/5.
Using the point-slope form,
y - 5 = (-4/5)(x - 10).
This is the desired new line.
Note that mult. both sides by 5 removes the fractions:
5y - 25 = -4(x - 10), or 5y - 25 = -4x + 40.
Solving this for y: 5y = 25 - 4x + 40, or 5y = -4x + 65
Then y = (-4/5)x + 13 (answer)
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