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2. Use rigid transformations to transform [tex]\triangle MNO[/tex] to the congruent [tex]\triangle M^{\prime}N^{\prime}O^{\prime}[/tex] so that [tex]M^{\prime}[/tex] is at the origin and [tex]M^{\prime}O^{\prime}[/tex] lies on the [tex]x[/tex]-axis in the positive direction.

3. Any property that is true for [tex]\triangle M^{\prime}N^{\prime}O^{\prime}[/tex] will also be true for [tex]\triangle MNO[/tex].
- Definition of congruence

4. Let [tex]r[/tex], [tex]s[/tex], and [tex]t[/tex] be real numbers such that the vertices of [tex]\Delta M^{\prime}N^{\prime}O^{\prime}[/tex] are [tex]M(0,0)[/tex], [tex]N(2r, 2s)[/tex], and [tex]O(2t, 0)[/tex].
- Defining constants

5. Let [tex]P^{\prime}[/tex], [tex]Q^{\prime}[/tex], and [tex]R^{\prime}[/tex] be the midpoints of [tex]\overline{M^{\prime}N^{\prime}}[/tex], [tex]\overline{N^{\prime}O^{\prime}}[/tex], and [tex]\overline{M^{\prime}O^{\prime}}[/tex], respectively.
- Defining points

6. [tex]P^{\prime}=(r, s)[/tex], [tex]Q^{\prime}=(r+t, s)[/tex], [tex]R^{\prime}=(t, 0)[/tex]

7. Slope of [tex]\overline{M^{\prime}Q^{\prime}}=\frac{s}{r+t}[/tex]; Slope of [tex]\overline{N^{\prime}R^{\prime}}=\frac{2s}{2r-t}[/tex]; Slope of [tex]\overline{O^{\prime}P^{\prime}}=\frac{-s}{2t-r}[/tex]
- Definition of slope

8. Applying point-slope formula

9. [tex]M^{\prime}Q^{\prime}[/tex] and [tex]N^{\prime}R^{\prime}[/tex] intersect at point [tex]S\left(-\frac{2}{3}(r+t),-\frac{2}{3}s\right)[/tex].
- Algebra

10. The coordinates of [tex]S[/tex] satisfy the equation of [tex]O^{\prime}P^{\prime}[/tex].

11. All three lines share point [tex]S[/tex].
- Follows from 9 and 10

12. The medians are concurrent.
- Definition of concurrent lines

Which reason completes the proof for step 6?
A. Definition of slope
B. Definition of midsegment
C. Definition of midpoint
D. Definition of median


Sagot :

Step 6 of the question requires us to find the coordinates of the midpoints [tex]\( P^{\prime}, Q^{\prime}, \)[/tex] and [tex]\( R^{\prime} \)[/tex] of the line segments [tex]\( \overline{M^{\prime}N^{\prime}}, \overline{N^{\prime}O^{\prime}}, \)[/tex] and [tex]\( \overline{M^{\prime}O^{\prime}} \)[/tex], respectively. To do this, we use the definition of a midpoint, which states that the coordinates of the midpoint of a line segment whose endpoints are [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] can be found using the formula:

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

Let's apply this formula to find [tex]\( P^{\prime}, Q^{\prime}, \)[/tex] and [tex]\( R^{\prime} \)[/tex]:

1. [tex]\( P^{\prime} \)[/tex] is the midpoint of [tex]\( \overline{M^{\prime}N^{\prime}} \)[/tex] with endpoints [tex]\( M^{\prime}(0, 0) \)[/tex] and [tex]\( N^{\prime}(2r, 2s) \)[/tex]:

[tex]\[ P^{\prime} = \left( \frac{0 + 2r}{2}, \frac{0 + 2s}{2} \right) = \left( r, s \right) \][/tex]

2. [tex]\( Q^{\prime} \)[/tex] is the midpoint of [tex]\( \overline{N^{\prime}O^{\prime}} \)[/tex] with endpoints [tex]\( N^{\prime}(2r, 2s) \)[/tex] and [tex]\( O^{\prime}(2t, 0) \)[/tex]:

[tex]\[ Q^{\prime} = \left( \frac{2r + 2t}{2}, \frac{2s + 0}{2} \right) = \left( r + t, s \right) \][/tex]

3. [tex]\( R^{\prime} \)[/tex] is the midpoint of [tex]\( \overline{M^{\prime}O^{\prime}} \)[/tex] with endpoints [tex]\( M^{\prime}(0, 0) \)[/tex] and [tex]\( O^{\prime}(2t, 0) \)[/tex]:

[tex]\[ R^{\prime} = \left( \frac{0 + 2t}{2}, \frac{0 + 0}{2} \right) = \left( t, 0 \right) \][/tex]

Therefore, the reason that completes the proof for step 6 is the definition of midpoint.

So, the correct answer is:

C. Definition of midpoint