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\begin{tabular}{|c|c|}
\hline 1. Define the vertices & Statement & Reason \\
\hline \begin{tabular}{l}
1. Define the vertices of [tex]$\triangle ABC$[/tex] to have unique points [tex]$A \left( x_1, y_1 \right)$[/tex], [tex]$B \left( x_2, y_2 \right)$[/tex], and \\
[tex]$C \left( x_3, y_3 \right)$[/tex].
\end{tabular} & Given \\
\hline \begin{tabular}{l}
2. Use rigid transformations to transform [tex]$\triangle ABC$[/tex] to [tex]$\triangle A^{\prime}B^{\prime}C^{\prime}$[/tex] so that [tex]$A^{\prime}$[/tex] is at the \\
origin and [tex]$\overline{A^{\prime}C^{\prime}}$[/tex] lies on the [tex]$x$[/tex]-axis in the positive direction.
\end{tabular} & \begin{tabular}{l}
In the coordinate plane, any point can be moved to any other \\
point using rigid transformations, and any line can be moved \\
to any other line using rigid transformations.
\end{tabular} \\
\hline 3. Any property that is true for [tex]$\triangle A^{\prime}B^{\prime}C^{\prime}$[/tex] will also be true for [tex]$\triangle ABC$[/tex]. & Definition of congruence \\
\hline \begin{tabular}{l}
4. Let [tex]$r, s$[/tex], and [tex]$t$[/tex] be real numbers such that the vertices of [tex]$\triangle A^{\prime}B^{\prime}C^{\prime}$[/tex] are [tex]$A^{\prime}(0,0)$[/tex], \\
[tex]$B^{\prime}(2r, 2s)$[/tex], and [tex]$C^{\prime}(2t, 0)$[/tex].
\end{tabular} & Defining constants \\
\hline \begin{tabular}{l}
5. Let [tex]$D^{\prime}, E^{\prime}$[/tex], and [tex]$F^{\prime}$[/tex] be the midpoints of [tex]$\overline{A^{\prime}B^{\prime}}, \overline{B^{\prime}C^{\prime}}$[/tex], and [tex]$\overline{A^{\prime}C^{\prime}}$[/tex] respectively.
\end{tabular} & Defining points \\
\hline 6. [tex]$D^{\prime} = (r, s)$[/tex] & Definition of midpoint \\
\hline
\end{tabular}

What is the reason for statement 7 in the given proof?
A. Definition of midpoint
B. Definition of slope
C. Parallel lines have equal slopes.
D. Using point-slope formula


Sagot :

Certainly! Let's carefully examine the details and logic behind the given proof and reach the conclusion step-by-step to find the reason for statement 7.

Here's an overview of the proof provided:

1. Define the vertices of [tex]\(\triangle ABC\)[/tex] at specific points [tex]\(A (x_1, y_1)\)[/tex], [tex]\(B (x_2, y_2)\)[/tex], [tex]\(C(x_3, y_3)\)[/tex].
- Reason: Given.

2. Use rigid transformations to transform [tex]\(\triangle ABC\)[/tex] to [tex]\(\triangle A'B'C'\)[/tex] such that [tex]\(A'\)[/tex] is at the origin, and [tex]\(\overline{A'C'}\)[/tex] lies along the x-axis.
- Reason: Using rigid transformations, any point can move to any other point, and any line can move to any other line.

3. Any property true for [tex]\(\triangle A'B'C'\)[/tex] is also true for [tex]\(\triangle ABC\)[/tex].
- Reason: Definition of congruence.

4. Use constants [tex]\(r, s, t\)[/tex] to redefine the vertices [tex]\(A'(0,0)\)[/tex], [tex]\(B'(2r, 2s)\)[/tex], [tex]\(C'(2t, 0)\)[/tex].
- Reason: Defining constants.

5. Define midpoints of [tex]\(\overline{A'B'}\)[/tex], [tex]\(\overline{B'C'}\)[/tex], and [tex]\(\overline{A'C'}\)[/tex] as [tex]\(D', E', F'\)[/tex].
- Reason: Defining points.

6. Specify the coordinates for [tex]\(D'\)[/tex] and [tex]\(F'\)[/tex] as [tex]\(D'(r, s)\)[/tex] and [tex]\(F'(t, 0)\)[/tex].
- Reason: Definition of midpoints.

We need to determine the reasoning for statement 7, which is about [tex]\(\triangle A'B'C'\)[/tex] and its property related to slope and parallelism.

Given choices are:
A. Definition of midpoint
B. Definition of slope
C. Parallel lines have equal slopes.
D. Using point-slope formula

By examining the statements provided:

- Since we know the vertices positions and midpoints, the property we need to justify here is related to the slopes and properties of lines.
- Statement 7 is likely connecting two lines and referring to their slopes, asserting they are equal because they are parallel.

Thus, carefully analyzing the proof step-by-step and understanding what is needed to justify a statement about the slopes of lines being equal as a key property of parallel lines, we conclude:
The reason for statement 7 in the given proof is:

C. Parallel lines have equal slopes.