Let's solve the problem step-by-step:
### Step 1: Calculate the Slope of Line AB
The coordinates of points [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are [tex]\( A(-3, 8) \)[/tex] and [tex]\( B(1, 15) \)[/tex].
The formula for the slope of a line passing through two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\][/tex]
For line AB:
[tex]\[
m(\overrightarrow{AB}) = \frac{15 - 8}{1 - (-3)} = \frac{15 - 8}{1 + 3} = \frac{7}{4} = 1.75
\][/tex]
### Step 2: Calculate the Slope of Line CD
The coordinates of points [tex]\( C \)[/tex] and [tex]\( D \)[/tex] are [tex]\( C(10, -9) \)[/tex] and [tex]\( D(3, -5) \)[/tex].
For line CD:
[tex]\[
m(\overrightarrow{CD}) = \frac{-5 - (-9)}{3 - 10} = \frac{-5 + 9}{3 - 10} = \frac{4}{-7} = -\frac{4}{7} \approx -0.5714285714285714
\][/tex]
### Step 3: Determine if the Lines are Perpendicular
Two lines are perpendicular if and only if the product of their slopes is [tex]\(-1\)[/tex].
Let's calculate the product of the slopes:
[tex]\[
m(\overrightarrow{AB}) \times m(\overrightarrow{CD}) = 1.75 \times -0.5714285714285714 = -1
\][/tex]
### Conclusion
Since the product of the slopes [tex]\(1.75\)[/tex] and [tex]\(-0.5714285714285714\)[/tex] is [tex]\(-1\)[/tex], the lines [tex]\( \overrightarrow{AB} \)[/tex] and [tex]\( \overrightarrow{CD} \)[/tex] are indeed perpendicular.
Hence, the completed table is:
[tex]\[
\begin{tabular}{|l|l|l|}
\hline
$m(\overrightarrow{A B})$ & $m(\overrightarrow{C D})$ & \text{Types of Lines} \\
\hline
1.75 & -0.5714285714285714 & \text{Perpendicular} \\
\hline
\end{tabular}
\][/tex]
Thus, lines AB and CD are perpendicular.
