Write the equation of a line that is perpendicular to the given line and that passes through the given point. y – 3 = (x + 2); (–2, 7) (1 point) Responses y = 5x + 7 y = 5 x + 7 y = 5x + 17 y = 5 x + 17 y = x – 2 y = Image with alt text: one-fifth x – 2 y = –2x + 3

Answer 1

To find the equation of a line that is perpendicular to the given line, we need to determine the negative reciprocal of the slope of the given line.

The given line has a slope of 1, as the coefficient of x is 1. The negative reciprocal of 1 is -1.

Now, we can use the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the negative reciprocal of the slope.

Plugging in the values, we have:
y - 7 = -1(x - (-2))

Simplifying:
y - 7 = -x - 2
y = -x + 5

Therefore, the equation of the line that is perpendicular to y – 3 = (x + 2) and passes through the point (-2, 7) is: y = -x + 5.

Answer 2

what no?

Answer 3

Apologies for the confusion. Let's start over.

The given line, y - 3 = x + 2, can be rearranged to slope-intercept form, y = x + 5.

To find the equation of a line perpendicular to this, we need to determine the negative reciprocal of the slope, which is -1.

The negative reciprocal of 1 is -1.

Now, we can use the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the negative reciprocal of the slope.

Plugging in the values (-2, 7) and -1, we have:
y - 7 = -1(x - (-2))

Simplifying:
y - 7 = -x - 2
y = -x + 5

Therefore, the equation of the line that is perpendicular to y – 3 = (x + 2) and passes through the point (-2, 7) is: y = -x + 5.