SOLUTION: Which equation is the equation of a line that passes through (–10, 3) and is perpendicular to y = 5x – 7? (1 point) y = 5x + 53 y = –1/5x – 7 y = –1/5x + 1 y = 1/5x + 5

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Question 1002617: Which equation is the equation of a line that passes through (–10, 3) and is perpendicular to y = 5x – 7? (1 point)
y = 5x + 53
y = –1/5x – 7
y = –1/5x + 1
y = 1/5x + 5

Found 2 solutions by Cromlix, Edwin McCravy:
Answer by Cromlix(4381)   (Show Source): You can put this solution on YOUR website!
Hi there,
Lines that are perpendicular to
one another have slopes that multiply
together to give -1
m1 x m2 = -1
y = 5x – 7 has a slope = 5
5 x m2 = -1
m2 = -1/5
Using Line Equation:
y - b = m(x - a)
m = -1/5 and (a,b) = (-10,3)
y - 3 = -1/5(x - (-10))
y - 3 = -1/5(x + 10)
y - 3 = -1/5x - 2
y = -1/5x - 2 + 3
y = -1/5x + 1
Hope this helps :-)

Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
y = 5x – 7 has slope which is the coefficient of x or 5.

A line perpendicular to that line would have a slope which
is its negative reciprocal or -1/5, so that narrows the 
correct choice down to either y = –1/5x – 7 or y = –1/5x + 1.

So we need only to find out which one is satisfied by (-10,3).
Substituting x=-10 and y = 3 in

y = –1/5x – 7
3 = -1/5(-10) - 7
3 = 2-7
3 = -5
That's false, so that's not it.

So it can only be 

y = –1/5x + 1.   Check to make sure:
3 = -1/5(-10) + 1
3 = 2+1
3 = 3

That's true.  So the answer is y = –1/5x + 1

Edwin

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