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 = (x – 2)
To find the equation of a line that is perpendicular to the given line, we need to find the negative reciprocal of the slope of the given line.
Starting with the equation of the given line, y - 3 = (x + 2), we can rewrite it in slope-intercept form (y = mx + b) by moving the constant term to the other side:
y = x + 2 + 3
y = x + 5
Since the slope of the given line is 1, the negative reciprocal of this slope is -1/1 = -1.
Now, we will use the point-slope form of a line to find the equation of the line perpendicular to the given line and passing through the point (-2, 3).
y - y1 = m(x - x1)
where (x1, y1) = (-2, 3) and m = -1 (slope of the line perpendicular to the given line).
Substituting the values into the equation:
y - 3 = -1(x - (-2))
y - 3 = -(x + 2)
y - 3 = -x - 2
Simplifying the equation gives:
y = -x - 2 + 3
y = -x + 1
Therefore, the equation of the line that is perpendicular to y - 3 = x + 2 and passes through (-2, 3) is D. y + 3 = (x - 2).