Which equation is the equation of a line that passes through (–

Question

10, 3) and is perpendicular to y = 5x – 7?
(1 point)
y = 5x + 53
y =-1/5 –x – 7
y = -1/5 x + 1
y = 1/5x + 5

Answer 1

To find the equation of a line that is perpendicular to y = 5x - 7, we need to find the negative reciprocal of the slope of the given line.

In the equation y = 5x - 7, the slope is 5. The negative reciprocal of 5 is -1/5.

Now, we have the slope (-1/5) and the point (-10, 3) that the line passes through.

Using the point-slope form of a line, the equation of the line is:
y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope.

Plugging in the values, we get:
y - 3 = (-1/5)(x - (-10))
y - 3 = (-1/5)(x + 10)

Simplifying, we have:
y - 3 = (-1/5)x - 2

Adding 3 to both sides:
y = (-1/5)x + 1

Therefore, the equation of the line that passes through (-10, 3) and is perpendicular to y = 5x - 7 is y = (-1/5)x + 1.

Therefore, the correct equation from the given options is y = -1/5 x + 1.