Write the equation of a line that is perpendicular to the given line and passes through the given point: -x + 5y = 14; (-5, -2). A. y = -5x - 27 B. y = -(1/5)x - 27 C. y = -(1/5)x - 15 D. y = 5x - 27

To find the equation of a line that is perpendicular to a given line, you first need to find the slope of the given line. The slope-intercept form of a linear equation is y = mx + b, where m is the slope and b is the y-intercept.

The given line is -x + 5y = 14. Rearrange it into slope-intercept form:

5y = x + 14 y = (1/5)x + (14/5)

The slope of the given line is 1/5.

Perpendicular lines have slopes that are negative reciprocals of one another. Therefore, the slope of the line perpendicular to the given line will be the negative reciprocal of 1/5, which is -5.

Now that we know the slope of the desired line is -5, we can use the point-slope form to find the equation of the line passing through the given point (-5, -2).

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

We insert our point and slope into this equation:

y - (-2) = -5(x - (-5)) y + 2 = -5(x + 5)

Now, we want to simplify and put it in slope-intercept form (y = mx + b):

y + 2 = -5x - 25 y = -5x - 25 - 2 y = -5x - 27

The equation of the line that is perpendicular to the given line and passes through the point (-5, -2) is y = -5x - 27.

Among the provided options, the correct answer is:

A. y = -5x - 27