Perpendicular Lines Topic 4.4.2

Perpendicular Lines 4.4.2 Topic 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 work out the slopes of perpendicular lines and you’ll test if two lines are perpendicular. Key words: perpendicular reciprocal

Perpendicular Lines 4.4.2 Topic Math problems about parallel lines often deal with perpendicular lines too. “Perpendicular” might sound like a difficult term, but it’s actually a really simple idea.

Perpendicular Lines 4.4.2 Topic Perpendicular Lines Meet at Right Angles Two lines are perpendicular if they intersect at 90° angles, like in the graphs below. y x y x

Perpendicular Lines 4.4.2 Topic Slopes of Perpendicular Lines are Negative Reciprocals Two lines are perpendicular if the slope of one is the negative reciprocal of the slope of the other. To get the reciprocal of a number you divide 1 by it. For example, the reciprocal of x is and so the negative reciprocal is – . 1 x The reciprocal of is = and so the negative reciprocal is – . 1 y x

Perpendicular Lines 4.4.2 Topic Solution Example 1 Prove that lines A and B, shown on the graph, are perpendicular to each other. –6 –4 –2 2 4 6 y x Solution 2 4 Using the rise over run formula, slope = : Dy Dx Slope of A = m1 = = 2 4 1 2 4 Slope of B = m2 = = –2 –4 2 is the negative reciprocal of –2, so A and B must be perpendicular. 1 2 Solution follows…

Perpendicular Lines 4.4.2 Topic Guided Practice 1. Perpendicular lines meet at angles. ………………………. 90° or right 1 3 – 2. Find the negative reciprocal of 3. 3. Find the negative reciprocal of – . 1 4 4 –6 –4 –2 2 4 6 y x 4. Find the negative reciprocal of – . 4 5 5 4 A 5. Use the graph to prove that A and B are perpendicular. B mA = and mB = – 2 3 The gradients are the negative reciprocals of each other, so A and B are perpendicular. Solution follows…

Perpendicular Lines 4.4.2 Topic Perpendicular Lines: m1 × m2 = –1 When you multiply a number by its reciprocal, you always get 1. For example, 3 2 × = 1 . and 1 5 × 5 = = 1 So because two perpendicular slopes are negative reciprocals of each other, their product is always –1. Here’s the same thing written in math-speak: If two lines l1 and l2 have slopes m1 and m2, l1 ^ l2 if and only if m1 × m2 = –1. ^ means “is perpendicular to.”

Perpendicular Lines 4.4.2 Topic Solution Example 2 P and Q are two straight lines and P ^ Q. P has a slope of –4. What is the slope of Q? Solution mP × mQ = –1 Þ –4 × mQ = –1 Þ mQ = = 1 4 –1 –4 So the slope of Q is . 1 4 Solution follows…

Perpendicular Lines 4.4.2 Topic Guided Practice 6. Lines l1 and l2 are perpendicular. If the slope of l1 is , find the slope of l2. 7. Lines A and B are perpendicular. If the slope of A is – , find the slope of B. 8. Lines R and T are perpendicular. If R has slope – , what is the slope of T? 9. The slope of l1 is –0.8. The slope of l2 is 1.25. Determine whether l1 and l2 are perpendicular. 1 5 –5 5 8 8 5 7 11 11 7 m1 × m2 = –0.8 × 1.25 = –1, so the lines are perpendicular. Solution follows…

Perpendicular Lines 4.4.2 Topic Solution Example 3 Determine the equation of the line passing through (3, 1) that is perpendicular to the straight line through (2, –1) and (4, 2). Solution You can show that lines are perpendicular by finding their slopes. Step 1: Find slope m1 of the line through (2, –1) and (4, 2): m1 = y2 – y1 x2 – x1 = 2 – (–1) 4 – 2 = 3 2 Step 2: Find the slope m2 of a line perpendicular to that line: m1 × m2 = –1 × m2 = –1 3 2 Þ m2 = – 2 3 Solution continues… Solution follows…

Perpendicular Lines 4.4.2 Topic Þ y – 1 = – (x – 3) Example 3 Determine the equation of the line passing through (3, 1) that is perpendicular to the straight line through (2, –1) and (4, 2). Solution (continued) Step 3: Now use the point-slope formula to find the equation of the line through (3, 1) with slope – . 2 3 y – y1 = m(x – x1) Þ y – 1 = – (x – 3) 2 3 Þ 3y – 3 = –2(x – 3) Þ 3y – 3 = –2x + 6 Equation: 3y + 2x = 9

Perpendicular Lines 4.4.2 Topic Guided Practice 10. Show that the line through the points (5, –3) and (–8, 1) is perpendicular to the line through (4, 6) and (8, 19). m1 = = – 4 13 1 – (–3) –8 – 5 m2 = = 19 – 6 8 – 4 Since m1 × m2 = –1, the lines are perpendicular. 11. Show that the line through (0, 6) and (5, 1) is perpendicular to the line through (4, 8) and (–1, 3). Since m1 × m2 = –1, the lines are perpendicular. m1 = = – 1 – 6 5 – 0 5 = –1 m2 = = –5 3 – 8 –1 – 4 = 1 Solution follows…

Perpendicular Lines 4.4.2 Topic Guided Practice 12. Show that the line through (4, 3) and (2, 2) is perpendicular to the line through (1, 3) and (3, –1). Since m1 × m2 = –1, the lines are perpendicular. m1 = = –1 –2 2 – 3 2 – 4 1 2 m2 = –4 –1 – 3 3 – 1 = –2 13. Determine the equation of the line through (3, –4) that is perpendicular to the line through the points (–7, –3) and (–3, 8). m1 = = 11 4 8 – (–3) –3 – (–7) So, m2 = – y – (–4) = – (x – 3) Þ 11y + 44 = –4x + 12 Þ 11y + 4x = –32 4 11 Solution follows…

Perpendicular Lines 4.4.2 Topic Guided Practice 14. Determine the equation of the line through (6, –7) that is perpendicular to the line through the points (8, 2) and (–1, 8). So, m2 = 3 2 m1 = = 6 –9 8 – 2 –1 – 8 = – y – (–7) = (x – 6) Þ 2y + 14 = 3x – 18 Þ 2y – 3x = –32 15. Find the equation of the line through (4, 5) that is perpendicular to the line –3y + 4x = 6. Find two points on the line –3y + 4x = 6. For example, (0, –2) and (3, 2). Use these points to find m1. m1 = = 2 – –2 3 – 0 4 3 y – 5 = – (x – 4) Þ 4y – 20 = –3x + 12 Þ 4y + 3x = 32 So, m2 = – Solution follows…

Perpendicular Lines 4.4.2 Topic Independent Practice In Exercises 1–8, J and K are perpendicular lines. The slope of J is given. Find the slope of K. 1 3 1 14 1. mJ = –3 2. mJ = –14 5 2 3. mJ = – 2 5 4. mJ = 6 7 – 7 6 5. mJ = – 8 3 3 8 6. mJ = – 14 15 15 14 7. mJ = –0.18 5.5555… 8. mJ = 0.45 –2.2222… Solution follows…

Perpendicular Lines 4.4.2 Topic Independent Practice 9. Show that the line through (2, 7) and (–2, 8) is perpendicular to the line through (–3, –3) and (–2, 1). m1 × m2 = –1, so the lines are perpendicular. m2 = = 4 1 1 – (–3) –2 – (–3) = 4 m1 = 8 – 7 –2 – 2 = – 10. Show that the line through (–4, 3) and (3, –2) is perpendicular to the line through (–7, –1) and (–2, 6). m1 × m2 = –1, so the lines are perpendicular. m1 = –2 – 3 3 – (– 4) = – 5 7 m2 = = 6 – (–1) –2 – (–7) Solution follows…

Perpendicular Lines 4.4.2 Topic Independent Practice 11. Determine the equation of the line through (5, 9) that is perpendicular to a line with slope . 1 3 y + 3x = 24 12. Determine the equation of the line through (3, –5) that is perpendicular to the line through the points (–3, 2) and (–6, –4). 2y + x = –7 Solution follows…

Perpendicular Lines 4.4.2 Topic Round Up “Perpendicular” is just a special math word to describe lines that are at right angles to each other. Remember that the best way to show that two lines are at right angles is to calculate their slopes — if they multiply together to make –1, then the lines are perpendicular.