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\begin{tabular}{|c|c|}
\hline 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 x-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]$D^{\prime}$[/tex], [tex]$E^{\prime}$[/tex], and [tex]$F^{\prime}$[/tex] be the midpoints of [tex]$\overline{A^{\prime}B^{\prime}}$[/tex], [tex]$\overline{B^{\prime}C^{\prime}}$[/tex], and [tex]$\overline{A^{\prime}C^{\prime}}$[/tex] \\
respectively.
\end{tabular} & Defining points \\
\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 :

GinnyAnswer
Let's walk through the question step-by-step to understand how we reach the reason for statement 7.

1. Define the vertices of [tex]\(\triangle ABC\)[/tex]:
- We have three unique points: [tex]\(A(x_1, y_1)\)[/tex], [tex]\(B(x_2, y_2)\)[/tex], and [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]:
- Move [tex]\(A'\)[/tex] to the origin [tex]\((0,0)\)[/tex] and place [tex]\(\overline{A'C'}\)[/tex] on the x-axis.
- Reason: In the coordinate plane, any point can be moved to any other point and any line can be moved to any other line using rigid transformations.

3. Property consistency between [tex]\(\triangle ABC\)[/tex] and [tex]\(\triangle A'B'C'\)[/tex]:
- If a property is true for [tex]\(\triangle A'B'C'\)[/tex], it is also true for [tex]\(\triangle ABC\)[/tex].
- Reason: Definition of congruence.

4. Vertices of [tex]\(\triangle A'B'C'\)[/tex]:
- Let [tex]\(r, s, t\)[/tex] be real numbers such that [tex]\(A'(0, 0)\)[/tex], [tex]\(B'(2r, 2s)\)[/tex], and [tex]\(C'(2t, 0)\)[/tex].
- Reason: Defining constants.

5. Midpoints of segments [tex]\(\overline{A'B'}\)[/tex], [tex]\(\overline{B'C'}\)[/tex], and [tex]\(\overline{A'C'}\)[/tex]:
- Let [tex]\(D'\)[/tex], [tex]\(E'\)[/tex], and [tex]\(F'\)[/tex] be the midpoints of [tex]\(\overline{A'B'}\)[/tex], [tex]\(\overline{B'C'}\)[/tex], and [tex]\(\overline{A'C'}\)[/tex], respectively.
- Reason: Defining points.

From these steps, we notice that statement 5 involves defining the midpoints, which directly involves the use of the midpoint formula. The midpoint formula states that the coordinates of the midpoint [tex]\(M\)[/tex] of a line segment connecting points [tex]\( (x_1, y_1) \)[/tex] and [tex]\( (x_2, y_2) \)[/tex] are:

[tex]\[M \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\][/tex]

Given that the setup and subsequent transformations place emphasis on the placement and properties of these midpoints in relation to their segments, the most relevant reason for statement 7 pertains to ensuring midpoint properties.

Thus, the reason for statement 7 in the given proof is indeed:
A. definition of midpoint
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