1. 1 Topic 4.4.4 More About Slopes

2. 2 California Standard: 8.0 Students understand the concepts of parallel lines and perpendicular lines and how their slopes are related. Students are able to find the equation of a line perpendicular to a given line that passes through a given point. What it means for you: You’ll learn how to tell whether lines are parallel or perpendicular by looking at the slope-intercept form. Key Words: parallel perpendicular reciprocal Topic 4.4.4

3. 3 More About Slopes This Topic carries on from the material on parallel and perpendicular lines that you learned earlier in this Section. Topic 4.4.4 parallel perpendicular

4. 4 More About Slopes Values of m Tell You if Lines are Parallel Parallel lines all have the same slope, so the slope-intercept forms of their equations all have the same value of m. Topic 4.4.4 For example, the lines y = 3 x + 2, y = 3 x – 1 and y = 3 x – 6 are all parallel.

5. 5 More About Slopes Topic 4.4.4 If two lines have the same slope and pass through the same point, they’re said to be collinear. This is just the math way of saying that they’re on the same straight line. For example, the lines y = 4 x + 2, y – 4 x = 2 and 4 x – y = –2 are all collinear — if you rearrange them, you’ll see that the equations are equivalent.

6. 6 More About Slopes Example 1 Find the equation of the line through (4, –4) that is parallel to the line 2 x – 3 y = 6. Solution follows… Solution Topic 4.4.4 Step 1: Write 2 x – 3 y = 6 in the slope-intercept form — that is, solve the equation for y. 2 x – 3 y = 6  –3 y = –2 x + 6  y = x – 2 2 3 Step 2: Get the slope from the equation. The slope of the line y = x – 2 is. Since the required line through (4, –4) is parallel to the line y = x – 2, its slope is also. 2 3 2 3 2 3 2 3 Solution continues…

7. 7 More About Slopes Example 1 Find the equation of the line through (4, –4) that is parallel to the line 2 x – 3 y = 6. Solution (continued) Topic 4.4.4 y – y 1 = m ( x – x 1 )  y – (–4) = ( x – 4) 2 3 Step 2: Slope is. 2 3  3( y + 4) = 2( x – 4)  3 y + 12 = 2 x – 8  3 y – 2 x = –20 Step 3: Now write the equation of the line through (4, –4) with a slope of. 2 3

8. 8 More About Slopes Guided Practice 1. Give an example of a line that is parallel to y = x + 1. 2. Is the line y = x – 2 parallel to the line y = x + 6? Explain. Topic 4.4.4 Solution follows… 1 2 4 5 4 5 Any line with m = 1 2 Yes — they have the same slope

9. 9 More About Slopes Guided Practice 3. Find the equation of the line through (–4, 3) that is parallel to the line y = 3 x + 9. 4. Find the equation of the line through (3, 8) that is parallel to the line 3 x + y = 1. Topic 4.4.4 Solution follows… m 1 = 3, so m 2 = 3 Substitute (–4, 3): y – 3 = 3( x – (–4)) y – 3 = 3 x + 12  y – 3 x = 15 Rearrange the equation given: y = –3 x + 1 m 1 = –3, so m 2 = –3 Substitute (3, 8): y – 8 = –3( x – 3) y – 8 = –3 x + 9  y + 3 x = 17

10. 10 More About Slopes The slope-intercept forms of equations of perpendicular lines have values of m that are negative reciprocals of each other. Topic 4.4.4 For example, the lines y = 3 x + 2 and y = – x – 1 must be perpendicular,because the negative reciprocal of 3 is –. 3 1 3 1 Values of m Also Tell You if Lines are Perpendicular

11. 11 –3 y – x = 5  –3 y = x + 5  y = – x – 5 3 1 3 More About Slopes Example 2 Find the equation of the line through (2, –4) that is perpendicular to the line –3 y – x = 5. Solution follows… Solution Topic 4.4.4 Step 1: Write –3 y – x = 5 in the slope-intercept form (that is, solve the equation for y ). Step 2: Get the slope ( m 1 ) of the line y = – x – and determine the slope ( m 2 ) of the required line through (2, –4). 5 3 1 3 Solution continues… 1 3 1 3 Since m 1 = –, and m 2 is the negative reciprocal of –, m 2 must be 3.

12. 12 More About Slopes Example 2 Find the equation of the line through (2, –4) that is perpendicular to the line –3 y – x = 5. Solution (continued) Topic 4.4.4 Step 2: m 2 = 3 y – y 1 = m ( x – x 1 )  y – (–4) = 3( x – 2)  y + 4 = 3 x – 6  y – 3 x = –10 Step 3: Write the equation of the line through (2, –4) with a slope of 3. Use the point-slope formula here:

13. 13 More About Slopes Guided Practice 5. Give an example of a line that’s perpendicular to the line y = 6 x. 6. Is the line y = 4 x + 2 perpendicular to the line y = – x + 4? Explain your answer. Topic 4.4.4 Solution follows… 1 4 Any line with m = – 1 6 Yes — the gradients are negative reciprocals of each other

14. 14 More About Slopes Guided Practice 7. Find the equation of the line through (–2, 0) that is perpendicular to the line y = –2 x – 4. 8. Find the equation of the line through (–4, 6) that is perpendicular to the line 3 x – 4 y = 24. Topic 4.4.4 Solution follows… m 1 = –2, so m 2 = 0.5 Substitute (–2, 0): y – 0 = 0.5( x – (–2)) y = 0.5 x + 1  2 y – x = 2 Rearrange the equation given: y = 0.75 x – 6 m 1 = 0.75, so m 2 = – Substitute (–4, 6): y – 6 = ( x – (–4))  3 y – 18 = 4 x + 16  3 y + 4 x = 2 4 3 4 3

15. 15 Independent Practice Solution follows… perpendicular More About Slopes Topic 4.4.4 In Exercises 1–8, determine whether the pairs of lines are parallel, perpendicular, or collinear. 1. y = 2 x + 1 and y = – x – 6 2. y = x + 5 and y = –3 x – 4 3. y = 4 x – 8 and y = – x + 2 perpendicular parallel 1 4 1 2 1 3 4. y = 6 and y = 3 5. x = 2 and y = –4 6. 5 x – 2 y = –10 and 10 x – 4 y = –20 7. 3 x + y = 6 and 6 x + 2 y = –4 8. 2 x – y = –4 and 6 x – 3 y = –12 perpendicular collinear parallel collinear

16. 16 Independent Practice Solution follows… More About Slopes Topic 4.4.4 In Exercises 9–12, find the equations of the lines. 9. The line through (5, 2) that’s parallel to a line with slope. 10. The line through (3, –3) that’s parallel to a line with slope. 1 2 2 y – x = –1 5 y – 2 x = –21 3 y – x = –29 2 5 5 8 5 y + 8 x = –19 12. The line through (–3, 1) that’s perpendicular to a line with slope. 11. The line through (2, –9) that’s perpendicular to a line with slope –3.

17. 17 13. The line through (0, 0) that’s parallel to 3 x + y = 18. 14. The line through (3, 5) that’s parallel to 3 x – 7 y = –21. 15. The line through (4, –3) that’s parallel to 3 x – 4 y = 16. 16. The line through (–2, 6) that’s parallel to 6 x – 10 y = –20. Independent Practice Solution follows… More About Slopes Topic 4.4.4 In Exercises 13–16, find the equations of the lines. y + 3 x = 0 7 y – 3 x = 26 4 y – 3 x = –24 5 y – 3 x = 36

18. 18 17. The line through (0, 6) that’s perpendicular to 2 x + y = 18. 18. The line through (–3, –5) that’s perpendicular to 3 x – 6 y = –24. 19. The line through (6, –2) that’s perpendicular to 3 x – 5 y = –10. 20. The line through (8, 2) that’s perpendicular to the line joining the points (–6, 3) and (–2, 6). Independent Practice Solution follows… More About Slopes Topic 4.4.4 In Exercises 17–20, find the equations of the lines. y + 2 x = –11 3 y + 5 x = 24 3 y + 4 x = 38 y = x + 6 2 1

19. 19 21. Use slopes to decide whether the points (–3, –8), (3, –2), and (8, 3) are collinear (on the same line) or noncollinear. 22. Use slopes to decide whether the points (4, 5), (3, –2), and (8, 3) are collinear or noncollinear. Independent Practice Solution follows… More About Slopes Topic 4.4.4 noncollinear collinear

20. 20 23. Use slopes to decide whether the points B (2, 10), K (–1, 3), and J (5, –3) are vertices of a right triangle. 24. Show that the points M (3, 11), A (–4, 4), and T (3, –3) are vertices of a right triangle. Independent Practice Solution follows… More About Slopes Topic 4.4.4 Line MA has slope 1 and Line AT has slope –1, so they are perpendicular. This means the points do make a right triangle. Line BK has slope 3 7 Line BJ has slope – 3 13 Line KJ has slope –1 None of the lines are perpendicular, so the points do not make a right triangle.

21. 21 Round Up Hopefully you’ll see now why the slope-intercept form of a line is so useful — you can just glance at the equations to see whether lines are parallel or perpendicular, without having to plot the graphs. More About Slopes Topic 4.4.4