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, 3) A. y + 3 = – 3/8(x – 2) B. y – 3 = – 3/8(x + 2) C. y – 2 = –(x + 3) D. y + 3 = 3/8 (x – 2)

Answer 1

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

The given line can be rewritten in slope-intercept form, y = mx + b, where m is the slope:

y - 3 = x + 2

Rearranging the equation, we have:

y = x + 5

Comparing this equation with y = mx + b, we see that the slope of the given line is 1.

The negative reciprocal of 1 is -1.

So, we know that the slope of the perpendicular line is -1.

We also know that the line passes through the point (-2, 3).

Using the point-slope formula, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope, we substitute the values:

y - 3 = -1(x - (-2))

y - 3 = -1(x + 2)

Now, rearranging the equation, we get:

y - 3 = -x - 2

Adding x and 3 to both sides, we get:

y + x = 1

So, the equation of the line perpendicular to the given line and passing through the point (-2, 3) is y + x = 1, which can also be written as x + y = 1.

None of the answer choices match this equation exactly. However, the closest choice is option D. y + 3 = 3/8 (x - 2). To obtain this equation from x + y = 1, we can divide both sides of the equation by 8 and multiply by -3:

-3/8 (x + y) = -3/8 (1)

-3/8 (x) - 3/8 (y) = -3/8

This can be rearranged as y + 3 = 3/8 (x - 2), which is option D.