To write an equation of a line that is perpendicular to line CD and passes through the point [tex]\((-1, 6)\)[/tex], we will follow these steps:
1. Determine the slope of line CD:
The equation of line CD is given as [tex]\( y = -0.5x - 5.5 \)[/tex].
The standard form of a linear equation is [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] is the slope and [tex]\( b \)[/tex] is the y-intercept. From the equation [tex]\( y = -0.5x - 5.5 \)[/tex], we see the slope [tex]\( m \)[/tex] of line CD is [tex]\( -0.5 \)[/tex].
2. Find the slope of the perpendicular line:
Lines that are perpendicular to each other have slopes that are negative reciprocals. The negative reciprocal of [tex]\( -0.5 \)[/tex] is [tex]\( 2 \)[/tex].
3. Use the point-slope form to find the y-intercept (b) of the new line:
The point-slope form of a line is [tex]\( y - y_1 = m(x - x_1) \)[/tex], where [tex]\((x_1, y_1)\)[/tex] is a point on the line, and [tex]\( m \)[/tex] is the slope.
We are given the point [tex]\((-1, 6)\)[/tex], and the slope [tex]\( m \)[/tex] of the perpendicular line is [tex]\( 2 \)[/tex]. Plugging in these values, we get:
[tex]\[
y - 6 = 2(x + 1)
\][/tex]
4. Solve for [tex]\( y \)[/tex] to write the equation in slope-intercept form:
Expand and rearrange the equation to solve for [tex]\( y \)[/tex]:
[tex]\[
y - 6 = 2x + 2
\][/tex]
Add 6 to both sides to isolate [tex]\( y \)[/tex]:
[tex]\[
y = 2x + 2 + 6
\][/tex]
Simplify:
[tex]\[
y = 2x + 8
\][/tex]
Thus, the equation of the line perpendicular to line CD that passes through the point [tex]\((-1,6)\)[/tex] is:
[tex]\[
\boxed{y = 2x + 8}
\][/tex]
Among the given options, none directly match the correct equation [tex]\( y = 2x + 8 \)[/tex]. Therefore, there may have been an error in the provided options. However, the equation we found based on the detailed steps and conditions given is indeed [tex]\( y = 2x + 8 \)[/tex].
