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Jun 25, 2023 •352 likes •652 views

Section 3.1 d i E Lines a l l u d Dr. Abdulla Eid b A College of Science . r D MATHS 103: Mathematics for Business I Dr. Abdulla Eid (University of Bahrain) Lines 1 / 29 Introduction Recall: If we have a linear function y = ax +

  1. Section 3.1 d i E Lines a l l u d Dr. Abdulla Eid b A College of Science . r D MATHS 103: Mathematics for Business I Dr. Abdulla Eid (University of Bahrain) Lines 1 / 29

  2. Introduction Recall: If we have a linear function y = ax + b ,then by plotting two points on the graph of this function and connecting them, we get the graph of d i the function which is a line. (See Section 2.5) E a Goal: Given two points on the line (i.e., we are given ( x 1 , y 1 ) and l l ( x 2 , y 2 ) ). Find the equation of the line. u d Note: We will need only one point ( x 1 , y 1 ) and the slope m of the line. b A Topics: . 1 Slope. r D 2 Equation of a line (multiple forms). 3 Parallel and Perpendicular lines. Dr. Abdulla Eid (University of Bahrain) Lines 2 / 29

  3. 1 - The slope of a line d 1 The slope of a line is a number that measures how sloppy the line is i E (how hard to climb the stairs!). a l l u d b A . r D Dr. Abdulla Eid (University of Bahrain) Lines 3 / 29

  4. 1 Consider the two lines L 1 and L 2 (both of positive slope),but you can see that L 1 has slope greater than L 2 . d i E a l l u d 2 Slope has a clear relation with the angle between the line and the b A x –axis. if the slope rises, then θ rises too!. . r D Dr. Abdulla Eid (University of Bahrain) Lines 4 / 29

  5. Finding the slope of a line 1 From the equation of the line: Solve the equation for y , i.e., let y be alone. Then, you get d i y = mx + b E a and the slope is m . l l u 2 From the graph of the line: Choose any two points ( x 1 , y 1 ) and d b ( x 2 , y 2 ) on the line. Then, A m = y 2 − y 1 Vertical change . r = D x 2 − x 1 Horizontal change Special Case: The vertical line has no slope. Why? Dr. Abdulla Eid (University of Bahrain) Lines 5 / 29

  6. Example Find the slope of the line that passes through (1) ( 3, − 1 ) and ( 6, 9 ) . (2) ( − 6, 7 ) and ( 0, 1 ) . d Solution: i E (1) ( 3 , − 1 ) and ( 6 , 9 ) . ���� ���� ���� ���� a x 1 x 2 y 2 l y 1 l u d = 9 − ( − 1 ) m = y 2 − y 1 = 10 b 3 . x 2 − x 1 A 6 − 3 . Which means for every 3 steps to the right, we need to go 10 steps up. r D (2) ( − 6 , 7 ) and ( 0 , 1 ) . ���� ���� ���� ���� y 1 x 2 y 2 x 1 1 − 7 ) m = y 2 − y 1 0 − ( − 6 ) = − 6 6 = − 1 = 1 . x 2 − x 1 Which means for every one step to the right, we need to go one step down. Dr. Abdulla Eid (University of Bahrain) Lines 6 / 29

  7. Exercise Find the slope of the line that passes through (1) ( 5, 2 ) and ( 4, − 3 ) . (2) ( 1, 7 ) and ( − 9, 0 ) . (3) ( 5, 2 ) and ( 4, 2 ) . d (4) ( 3, 1 ) and ( 3, 3 ) . i E a l l u d b A . r D Dr. Abdulla Eid (University of Bahrain) Lines 7 / 29

  8. 2 - Equation of the line To get the equation of a line, you need to find One point on the line ( x 1 , y 1 ) and d The slope of the line m . i E Then, the equation of the line is a l l y − y 1 = m ( x − x 1 ) “point–slope form“ u − − − d b A Other forms: . r General Linear Form ax + by + c = 0, where a , b , and c have no common D factor. Slope–Intercept Form y = mx + b , where m is the slope of the line and ( 0, b ) is the y –intercept. Special Case: The equation of the vertical line is x = x 1 . Dr. Abdulla Eid (University of Bahrain) Lines 8 / 29

  9. Example Find a general linear equation ( ax + by + c = 0) of the line with the following properties: d (1) passes through ( 1, − 7 ) and has slope − 3. i E a Solution: l l u d y − y 1 = m ( x − x 1 ) b A y − ( − 7 ) = − 3 ( x − 1 ) . y + 7 = − 3 x + 3 r D y + 3 x + 7 − 3 = 0 y + 3 x + 4 = 0 Dr. Abdulla Eid (University of Bahrain) Lines 9 / 29

  10. Example Find a general linear equation ( ax + by + c = 0) of the line with the following properties: (2) passes through ( − 3, 4 ) and ( 6, − 4 ) . d Solution: First we find the slope m which is i E m = y 2 − y 1 6 − ( − 3 ) = − 8 − 4 − 4 = a x 2 − x 1 9 l l u d b y − y 1 = m ( x − x 1 ) A y − 4 = − 8 9 ( x − ( − 3 )) . r D y − 4 = − 8 9 ( x + 3 ) 9 ( y − 4 ) = − 8 ( x + 3 ) 9 y − 36 = − 8 x − 24 9 y + 8 x − 36 + 24 = 0 9 y + 8 x − 12 = 0 Dr. Abdulla Eid (University of Bahrain) Lines 10 / 29

  11. Example Find a general linear equation ( ax + by + c = 0) of the line with the following properties: d i (3) has slope 4 and y –intercept -4. E a Solution: l l u d b A y = mx + b . y = 4 x − 4 r D y − 4 x + 4 = 0 Dr. Abdulla Eid (University of Bahrain) Lines 11 / 29

  12. Example Find a general linear equation ( ax + by + c = 0) of the line with the following properties: d (4) is a vertical line passes through ( − 2, − 7 ) . i E a Solution: l l u d b A x = x 1 . x = − 2 r D x + 2 = 0 Dr. Abdulla Eid (University of Bahrain) Lines 12 / 29

  13. Exercise Find a general linear form equation of the line that (1) passes ( − 5, 5 ) and has slope − 1 3 . (2) passes ( 2, 4 ) and ( − 1, 3 ) . (3) has slope − 2 3 and y –intercept − 2. d (4) vertical line passes ( 8, 0 ) . i E (5) Horizontal line passes ( 2, 5 ) . a l l u d b A . r D Dr. Abdulla Eid (University of Bahrain) Lines 13 / 29

  14. Example Find the slope and the y –intercept of a line with equation y = 5 ( 2 − 3 x ) and sketch its graph. d Solution: We need to write the equation in the slope–intercept form (i.e., i E we isloate y ). a l y = 5 ( 2 − 3 x ) l u d y = 10 − 15 x b A y = − 15 x + 10 . r so we have D Slope = − 15 and y -intercept = ( 0, 10 ) 1 To sketch the graph of the line, we start at the point ( 0, 10 ) and for every one step to the right, we go 15 steps down. Dr. Abdulla Eid (University of Bahrain) Lines 14 / 29

  15. Exercise Find the slope and the y -intercept general linear form equation of the line that (1) 2 x = 5 − 3 y . (2) 3 ( x − 4 ) − 7 ( y + 1 ) = 2. d (3) y = 1 2 x + 8. i E 3 + 3 2 y = − 1 1 (4) − x 2 . a l l u d b A . r D Dr. Abdulla Eid (University of Bahrain) Lines 15 / 29

  16. Example Find a (a) general linear equation and (b) slope–intercept form of the line with an equation: (1) 2 x = 5 − 3 y . d Solution: (a) General Linear form: i E 2 x = 5 − 3 y a l 2 x + 3 y − 5 = 0 l u d b A (b) Slope–intercept form: . r 2 x = 5 − 3 y D 2 x − 5 = − 3 y − 3 x − 5 2 − 3 = y Dr. Abdulla Eid (University of Bahrain) Lines 16 / 29

  17. Example Find a (a) general linear equation and (b) slope–intercept form of the line with an equation: (2) 3 ( x − 4 ) − 7 ( y + 1 ) = 2. d Solution: (a) General Linear form: i E 3 ( x − 4 ) − 7 ( y + 1 ) = 2 a l 3 x − 12 − 7 y − 7 = 2 l u d 3 x − 7 y − 19 − 2 = 0 b A 3 x − 7 y − 21 = 0 . (b) Slope–intercept form: r D 3 x − 7 y − 21 = 0 3 x − 21 = 7 y 3 7 x − 21 7 = y 3 7 x − 3 = y Dr. Abdulla Eid (University of Bahrain) Lines 17 / 29

  18. Example Find a (a) general linear equation and (b) slope–intercept form of the line with an equation: (3) y = 1 2 x + 8. d Solution: (a) General Linear form: i E y = 1 a 2 x + 8 l l u d 2 y = x + 16 b 2 y − x − 16 = 0 A . r D (b) Slope–intercept form: y = 1 2 x + 8 Dr. Abdulla Eid (University of Bahrain) Lines 18 / 29

  19. Example Find a (a) general linear equation and (b) slope–intercept form of the line with an equation: 3 + 3 2 y = − 1 1 (4) − x 2 . d Solution: (a) General Linear form: i E 3 + 3 2 y = − 11 − x a 2 l l u − x + 9 2 y = − 31 d 2 b A − 2 x + 9 y = − 9 . − 2 x + 9 y + 9 = 0 r D (b) Slope–intercept form: − 2 x + 9 y + 9 = 0 9 y = 2 x − 9 y = 2 9 x − 9 9 y = 2 x − 1 Dr. Abdulla Eid (University of Bahrain) Lines 19 / 29

  20. 3 - Parallel and Perpendicular Lines d Definition i E Two lines are parallel if a m 1 = m 2 l l u d b A Two lines are perpendicular if . r m 1 m 2 = − 1 D Dr. Abdulla Eid (University of Bahrain) Lines 20 / 29

  21. Example d Determine whether the given lines are parallel, perpendicular, or neither? i E (1) y = − 5 x + 7 and y = − 5 x − 2. a l Solution: l u d b m 1 = − 5 and m 2 = − 5 A . r D So the two lines are parallel. Dr. Abdulla Eid (University of Bahrain) Lines 21 / 29

  22. Example Determine whether the given lines are parallel, perpendicular, or neither? (2) x + 3 y + 5 = 0 and y = 3 x . d i E Solution: a l l 3 y = − x + 5 and y = 3 x u d 3 y = − 1 3 x + 5 b 3 and y = 3 x A m 1 = − 1 . and m 2 = 3 r 3 D Now m 1 m 2 = ( − 1 3 ) 3 = − 1. Thus the two lines are perpendicular lines. Dr. Abdulla Eid (University of Bahrain) Lines 22 / 29

  23. Example Determine whether the given lines are parallel, perpendicular, or neither? (2) x − 2 = 3 and y = 2. d i E Solution: a l l x − 2 = 3 and y = 2 u d b x = 5 and y = 2 A m 1 = undefined and m 2 = 0 . r D Note that the line without slope is a vertical line. The line with a slope of zero is a horizontal line. Thus the two lines are perpendicular lines. Dr. Abdulla Eid (University of Bahrain) Lines 23 / 29

  24. Exercise Determine whether the given lines are parallel, perpendicular, or neither? (1) x + 2 y = 0 and x + 4 y − 7 = 0. (2) x = 3 and x = − 2. (3) y = 4 x + 7 and 4 x − y + 6 = 0. d i E a l l u d b A . r D Dr. Abdulla Eid (University of Bahrain) Lines 24 / 29

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