Sure, let's determine the equation of the line that is the perpendicular bisector of the segment connecting points [tex]\( C(6, -12) \)[/tex] and [tex]\( D(10, -8) \)[/tex].
1. Calculate the Midpoint of Segment CD:
The midpoint [tex]\( M \)[/tex] of a line segment with endpoints [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] can be found using the midpoint formula:
[tex]\[
M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
\][/tex]
For points [tex]\( C(6, -12) \)[/tex] and [tex]\( D(10, -8) \)[/tex]:
[tex]\[
x_{\text{mid}} = \frac{6 + 10}{2} = \frac{16}{2} = 8
\][/tex]
[tex]\[
y_{\text{mid}} = \frac{-12 + (-8)}{2} = \frac{-20}{2} = -10
\][/tex]
Thus, the midpoint is [tex]\( M(8, -10) \)[/tex].
2. Calculate the Slope of Segment CD:
The slope [tex]\( m \)[/tex] of a line segment connecting [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by:
[tex]\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\][/tex]
For points [tex]\( C(6, -12) \)[/tex] and [tex]\( D(10, -8) \)[/tex]:
[tex]\[
m_{CD} = \frac{-8 - (-12)}{10 - 6} = \frac{-8 + 12}{10 - 6} = \frac{4}{4} = 1
\][/tex]
3. Slope of the Perpendicular Bisector:
The slope of the perpendicular bisector is the negative reciprocal of the slope of [tex]\( CD \)[/tex]. If the slope of [tex]\( CD \)[/tex] is [tex]\( m \)[/tex], then the slope of the perpendicular bisector is:
[tex]\[
m_{\text{perpendicular}} = -\frac{1}{m} = -\frac{1}{1} = -1
\][/tex]
4. Equation of the Perpendicular Bisector:
We use the point-slope form of the equation of a line: [tex]\( y - y_1 = m(x - x_1) \)[/tex].
Here, the slope [tex]\( m = -1 \)[/tex], and the bisector passes through the midpoint [tex]\( M(8, -10) \)[/tex].
Therefore, substituting [tex]\( x_1 = 8 \)[/tex], [tex]\( y_1 = -10 \)[/tex], and [tex]\( m = -1 \)[/tex]:
[tex]\[
y - (-10) = -1(x - 8)
\][/tex]
[tex]\[
y + 10 = -1(x - 8)
\][/tex]
[tex]\[
y + 10 = -x + 8
\][/tex]
[tex]\[
y = -x + 8 - 10
\][/tex]
[tex]\[
y = -x - 2
\][/tex]
Thus, the equation of the line that is the perpendicular bisector of the segment connecting [tex]\( C \)[/tex] and [tex]\( D \)[/tex] is:
[tex]\[ \boxed{y = -x - 2} \][/tex]
So, the correct answer is:
B) [tex]\( y = -x - 2 \)[/tex]
