Question

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 = –(x – 2)
b) y – 3 = – 3/8 ( x + 2)
c) y–2 =–(x + 3)
d) y+3=(x – 2)

Answer 1

To find the equation of a line that is perpendicular to the given line, we need to find its slope first. The given line can be rewritten in slope-intercept form (y = mx + b) by isolating y:

y - 3 = x + 2
y = x + 5

Comparing this equation to y = mx + b, we can see that the slope (m) of the given line is 1.

Since the line we are looking for is perpendicular to the given line, its slope will be the negative reciprocal of 1.

The negative reciprocal of 1 is -1, so the slope of the line we are looking for is -1.

Now, we have the slope (-1) and the point through which the line passes (-2, 3).

Using the point-slope form of the equation of a line (y - y1 = m(x - x1)), we can substitute the values and find the equation:

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

Therefore, the equation of the line that is perpendicular to the given line and passes through the point (-2, 3) is:
y = -x + 1

Therefore, the correct answer is option c) y - 2 = -(x + 3)