Which equation is the equation of a line that passes through (-10,3) and is perpendicular to y=5x-7?
A. y=5x+53
B. y=-1/5x-7
C. y=-1/5x+1
D. y=1/5x+5
To find the equation of a line that is perpendicular to y=5x-7, we need to find the slope of the given line and then take the negative reciprocal of that slope.
The original equation, y=5x-7, is in the form y=mx+b, where m is the slope of the line. In this case, the slope is 5.
The negative reciprocal of 5 is -1/5.
Now we can use the point-slope form to find the equation of the perpendicular line.
Point-slope form: y - y1 = m(x - x1)
Using the point (-10,3) and the slope -1/5, we have:
y - 3 = -1/5(x - (-10))
y - 3 = -1/5(x + 10)
y - 3 = -1/5x - 2
y = -1/5x + 1
Therefore, the equation of the line that passes through (-10,3) and is perpendicular to y=5x-7 is y = -1/5x + 1.
The correct option is C.