Which equation is the equation of a line that passes through (–10, 3) and is perpendicular to y = 5x – 7?%0D%0A(1 point)%0D%0AResponses%0D%0A%0D%0Ay = 5x + 53%0D%0Ay = 5 x + 53%0D%0A%0D%0Ay = –x – 7%0D%0Ay = – Image with alt text: one-fifth x – 7%0D%0A%0D%0Ay = –x + 1%0D%0Ay = – Image with alt text: one-fifth x + 1%0D%0A%0D%0Ay = x + 5

The equation of a line that is perpendicular to y = 5x - 7 will have a slope that is the negative reciprocal of 5, which is -1/5.

We can use the point-slope form of a linear equation to find the equation of the line: y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope.

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

Simplifying the equation, we get:
y - 3 = (-1/5)x - 2
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.