Question
Question 5 (1 point) (04.02 MC) Quadrilateral WXYZ is on a coordinate plane. Segment \( X Y \) is on the line \( x-y=-3 \), and segment \( W Z \) is on the line \( x-y=1 \). Which statement proves how segments \( X Y \) and WZ are related? a They have slopes that are opposite reciprocals of 1 and -1 and are, therefore, perpendicular. d They have the same slope of 1 and are, therefore, parallel. They have the same slope of -1 and are, therefore, parallel. Thare opposite reciprocals of 0 and undefined and are, therefore, perpendicular.
Ask by Davies Page. in the United States
Jan 23,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
Segments \( XY \) and \( WZ \) are parallel because they have the same slope of 1.
Solution
To determine the relationship between segments \( XY \) and \( WZ \):
1. **Find the slopes of the lines:**
- For \( XY \) on the line \( x - y = -3 \):
\[
y = x + 3 \quad \Rightarrow \quad \text{Slope} = 1
\]
- For \( WZ \) on the line \( x - y = 1 \):
\[
y = x - 1 \quad \Rightarrow \quad \text{Slope} = 1
\]
2. **Analyze the slopes:**
- Both lines have the same slope of **1**.
3. **Conclusion:**
- Since \( XY \) and \( WZ \) have identical slopes, they are **parallel**.
**Correct Statement:**
**d. They have the same slope of 1 and are, therefore, parallel.**
Answer: d They have the same slope of 1 and are, therefore, parallel.
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Simplify this solution Extra Insights
When assessing the lines \( x - y = -3 \) and \( x - y = 1 \), you can rewrite them in slope-intercept form, \( y = mx + b \). For both lines, you’ll find that they have a slope of 1. Because they share the same slope, it proves that segments \( XY \) and \( WZ \) are parallel—never intersecting, just like two well-behaved train tracks! A common mistake is mistaking the slopes as being opposite reciprocals. When two lines have slopes that are equal, like in this case, they are parallel; however, lines with slopes that are true opposites, like 1 and -1, are perpendicular, meaning they intersect at a right angle. So, keep those definitions straight as you tackle more geometry problems!
