Write the equation of a line that is perpendicular to the given line and that passes through the given point.
A. y + 3 = –(x – 2)
B. y – 3 = –(x + 2)
C. y – 2 = –(x + 3)
D. y+3=3/8(x-2)
To find the equation of a line that is perpendicular to a given line, we must first determine the slope of the given line. The given line can be written in slope-intercept form (y = mx + b), where m represents the slope.
A. y + 3 = –(x – 2)
This equation can be rewritten as:
y = -x + 5
The slope of this line is -1.
Now, we must find a line that is perpendicular to this line and passes through a given point. Let's assume the given point is P(x1, y1). The equation of the perpendicular line is given by:
y - y1 = -1/m(x - x1)
Let's assume the given point is (4, 6). Plugging in the values, we get:
y - 6 = -1/(-1)(x - 4)
y - 6 = x - 4
y = x + 2
Therefore, the equation of the line that is perpendicular to the given line and passes through the point (4, 6) is y = x + 2.
B. y – 3 = –(x + 2)
This equation can be rewritten as:
y = -x - 1
The slope of this line is -1.
C. y – 2 = –(x + 3)
This equation can be rewritten as:
y = -x - 1
The slope of this line is -1.
D. y+3=3/8(x-2)
This equation is already in slope-intercept form (y = mx + b), where the slope is 3/8.
Therefore, the equation of the line that is perpendicular to the given line and passes through the point is not D (since the given line is already perpendicular to another line).
Therefore, the equation of the line that is perpendicular to the given line and passes through the point (4, 6) is represented by the equation y = x + 2, which can be found in option A.