Parallel & Perpendicular Lines Homework 5: Linear Equations Slope-Intercept & Standard Form Directions: Determine if the equations are parallel, perpendicular, or neither. 11. y = -3x + 6 and y = 1/3 x – 8 12. y = 5/4 x + 1 and y = 5/4 x – 7 13. 3x + 2y = 6 and y = -3/2 x + 5 14. 3y = 4x + 15 and 9x + 12y = 12 15. 10x – 2y = 16 and x + 5y = -20 16. 5x + 3y = 3 and 3x + 5y = -25 17. -9y = -3x + 18 and 2x – 6y = 30 18. x – y = 5 and x + y = -10 Given Point Slope: 2 : (-5,1); Slope = 3 Directions: Determine if the equations are parallel, perpendicular, or neither. 11. y = -3x + 6 and y = -8 Neither 12. y = 5/4 x + 1 and y = 5/4 x – 7 Parallel 13. 3x + 2y = 6 and y = -3/2 x + 5 Parallel 14. 3y = 4x + 15 and 9x + 12y = 12 Perpendicular 15. 10x – 2y = 16 and x + 5y = -20 Parallel 16. 5x + 3y = 3 and 3x + 5y = -25 Perpendicular 17. -9y = -3x + 18 and 2x – 6y = 30 Parallel 18. x – y = 5 and x + y = -10 Neither
The Correct Answer and Explanation is:
Key Concepts:
- Parallel lines have equal slopes.
- Perpendicular lines have slopes that are negative reciprocals of each other (e.g., mmm and −1/m-1/m−1/m).
- If the slopes are neither equal nor negative reciprocals, the lines are neither.
Problem Analysis:
11. y=−3x+6y = -3x + 6y=−3x+6 and y=13x−8y = \frac{1}{3}x – 8y=31x−8
→ Slopes: -3 and 1/3 → Negative reciprocals → Perpendicular
12. y=54x+1y = \frac{5}{4}x + 1y=45x+1 and y=54x−7y = \frac{5}{4}x – 7y=45x−7
→ Slopes: same (5/4) → Parallel
13. 3x+2y=63x + 2y = 63x+2y=6 → Solve: y=−32x+3y = -\frac{3}{2}x + 3y=−23x+3
Compare to: y=−32x+5y = -\frac{3}{2}x + 5y=−23x+5
→ Same slope: -3/2 → Parallel
14. 3y=4x+153y = 4x + 153y=4x+15 → y=43x+5y = \frac{4}{3}x + 5y=34x+5
9x+12y=129x + 12y = 129x+12y=12 → Solve: y=−34x+1y = -\frac{3}{4}x + 1y=−43x+1
→ Slopes: 4/3 and -3/4 → Negative reciprocals → Perpendicular
15. 10x−2y=1610x – 2y = 1610x−2y=16 → y=5x−8y = 5x – 8y=5x−8
x+5y=−20x + 5y = -20x+5y=−20 → y=−15x−4y = -\frac{1}{5}x – 4y=−51x−4
→ Slopes: 5 and -1/5 → Negative reciprocals → Perpendicular
16. 5x+3y=35x + 3y = 35x+3y=3 → y=−53x+1y = -\frac{5}{3}x + 1y=−35x+1
3x+5y=−253x + 5y = -253x+5y=−25 → y=−35x−5y = -\frac{3}{5}x – 5y=−53x−5
→ Slopes: -5/3 and -3/5 → Not equal or reciprocals → Neither
17. −9y=−3x+18-9y = -3x + 18−9y=−3x+18 → y=13x−2y = \frac{1}{3}x – 2y=31x−2
2x−6y=302x – 6y = 302x−6y=30 → y=13x−5y = \frac{1}{3}x – 5y=31x−5
→ Slopes: both 1/3 → Parallel
18. x−y=5x – y = 5x−y=5 → y=x−5y = x – 5y=x−5
x+y=−10x + y = -10x+y=−10 → y=−x−10y = -x – 10y=−x−10
→ Slopes: 1 and -1 → Negative reciprocals → Perpendicular
✅ Corrected Answers:
| # | Equations | Answer |
|---|---|---|
| 11 | y=−3x+6y = -3x + 6y=−3x+6 and y=13x−8y = \frac{1}{3}x – 8y=31x−8 | Perpendicular |
| 12 | y=54x+1y = \frac{5}{4}x + 1y=45x+1 and y=54x−7y = \frac{5}{4}x – 7y=45x−7 | Parallel |
| 13 | 3x+2y=63x + 2y = 63x+2y=6 and y=−32x+5y = -\frac{3}{2}x + 5y=−23x+5 | Parallel |
| 14 | 3y=4x+153y = 4x + 153y=4x+15 and 9x+12y=129x + 12y = 129x+12y=12 | Perpendicular |
| 15 | 10x−2y=1610x – 2y = 1610x−2y=16 and x+5y=−20x + 5y = -20x+5y=−20 | Perpendicular |
| 16 | 5x+3y=35x + 3y = 35x+3y=3 and 3x+5y=−253x + 5y = -253x+5y=−25 | Neither |
| 17 | −9y=−3x+18-9y = -3x + 18−9y=−3x+18 and 2x−6y=302x – 6y = 302x−6y=30 | Parallel |
| 18 | x−y=5x – y = 5x−y=5 and x+y=−10x + y = -10x+y=−10 | Perpendicular |
🔍 Explanation
To determine if two lines are parallel, perpendicular, or neither, we focus on their slopes. In slope-intercept form y=mx+by = mx + by=mx+b, the coefficient mmm is the slope. Two lines are parallel if they have the same slope, and perpendicular if their slopes are negative reciprocals (e.g., 2 and -1/2). If the slopes are neither equal nor negative reciprocals, the lines are neither.
For example, in #11, the slopes -3 and 1/3 are negative reciprocals, making the lines perpendicular. In contrast, in #12, both lines have slope 5/4, so they are parallel. In #13, even though one is in standard form, rewriting it shows it has the same slope as the second equation, thus parallel.
In #14, we simplify both equations and find their slopes are 4/3 and -3/4 — negative reciprocals, hence perpendicular. In #15, the slopes 5 and -1/5 also show a perpendicular relationship. But in #16, after simplifying, we get slopes -5/3 and -3/5 — not reciprocals or equal, so the lines are neither.
In #17, both simplified equations have slope 1/3 → parallel. Finally, in #18, one slope is 1 and the other -1 — classic negative reciprocals — so perpendicular.
Understanding these relationships is crucial in geometry and algebra, especially when analyzing graphs, solving systems, or working in coordinate geometry.
