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The table shows the proof of the relationship between the slopes of two parallel lines. What is the missing reason for step 2?

[tex]\[
\begin{tabular}{|l|l|}
\hline
\textbf{Statements} & \textbf{Reasons} \\
\hline
1. $r \parallel s$ & given \\
\hline
2. $m_{ r }=\frac{d-b}{c-0}=\frac{d-b}{c}$ & \textbf{?} \\
\hline
3. $m_{ s }=\frac{0-a}{c-0}=-\frac{a}{c}$ & \\
\hline
4. Distance from $(0, b)$ to $(0, a)$ equals the distance from $(c, d)$ to $(c, 0)$ & Definition of parallel lines \\
\hline
5. $d-0=b-a$ & Application of the distance formula \\
\hline
6. $m_{ r }=\frac{(b-a)-b}{c}$ & Substitution property of equality \\
\hline
7. $m_{ r }=\frac{a}{c}$ & Inverse property of addition \\
\hline
8. $m_{ r }=m_{ s }$ & Substitution property of equality \\
\hline
\end{tabular}
\][/tex]

What is the missing reason for step 2?

A. Pythagorean theorem
B. Application of the slope formula
C. Transitive property
D. Application of the distance formula



Answer :

GinnyAnswer
To solve this problem, let's work through the logic step-by-step and identify the missing reason for step 2:

1. Statement 1: [tex]\( r \| s \)[/tex] - Given.
2. Statement 2:
- For line [tex]\( r \)[/tex]: [tex]\( m_{r} = \frac{d - b}{c - 0} = \frac{d - b}{c} \)[/tex]
- For line [tex]\( s \)[/tex]: [tex]\( m_{s} = \frac{0 - a}{c - 0} = -\frac{a}{c} \)[/tex]

The reason for these formulas is the application of the slope formula. The slope of a line passing through two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by [tex]\( \frac{y_2 - y_1}{x_2 - x_1} \)[/tex].

Therefore, the missing reason for step 2 is:

B. application of the slope formula

The correct answer corresponds to the number 2.

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