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3.7 Perpendicular Lines in the Coordinate Plane

3.7 Perpendicular Lines in the Coordinate Plane. Geometry Mrs. Spitz Fall 2005. Standard/Objectives:. Standard 3: Students will learn and apply geometric concepts. Objectives: Use slope to identify perpendicular lines in a coordinate plane Write equations of perpendicular lines.

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3.7 Perpendicular Lines in the Coordinate Plane

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  1. 3.7 Perpendicular Lines in the Coordinate Plane Geometry Mrs. Spitz Fall 2005

  2. Standard/Objectives: Standard 3: Students will learn and apply geometric concepts. Objectives: • Use slope to identify perpendicular lines in a coordinate plane • Write equations of perpendicular lines.

  3. Assignment: • Pp. 175-177 #7-45 and 47-50

  4. In a coordinate plane, two non-vertical lines are perpendicular if and only if the product of their slopes is -1. Vertical and horizontal lines are perpendicular Postulate 18: Slopes of Perpendicular Lines

  5. Find each slope. Slope of j1 3-1 = - 2 1-3 3 Slope of j2 3-(-3) = 6 = 3 0-(-4) 4 2 Multiply the two slopes. The product of -2∙ 3 = -1, so j1 j2 3 2 Ex. 1: Deciding whether lines are perpendicular

  6. Decide whether AC and DB are perpendicular. Solution: Slope of AC= 2-(-4) = 6 = 2 4 – 1 3 Slope of DB= 2-(-1) = 3 = 1 -1 – 5 -6 -2 Ex.2 Deciding whether lines are perpendicular The product of 2(-1/2) = -1; so ACDB

  7. Line h: y = ¾ x +2 The slope of line h is ¾. Line j: y=-4/3 x – 3 The slope of line j is -4/3. The product of ¾ and -4/3 is -1, so the lines are perpendicular. Ex.3: Deciding whether lines are perpendicular

  8. Line r: 4x+5y=2 4x + 5y = 2 5y = -4x + 2 y = -4/5 x + 2/5 Slope of line r is -4/5 Line s: 5x + 4y = 3 5x + 4y = 3 4y = -5x + 3 y = -5/4 x + 3/5 Slope of line s is -4/5 Ex.4: Deciding whether lines are perpendicular The product of the slopes isNOT-1; so r and s areNOTperpendicular. -4∙ -5 = 1 5 4

  9. Line l1 has an equation of y = -2x + 1. Find the equation of a line l2 that passes through P(4, 0) and is perpendicular to l1. First you must find the slope, m2. m1∙ m2 = -1 -2∙ m2 = -1 m2 = ½ Then use m = ½ and (x, y) = (4, 0) to find b. y = mx + b 0 = ½(4) + b 0 = 2 + b -2 = b So, an equation of l2 is y = ½ x - 2 Ex. 5: Writing the equation of a perpendicular line.

  10. The equation y = 3/2 x + 3 represents a mirror. A ray of light hits the mirror at (-2, 0). What is the equation of the line p that is perpendicular to the mirror at this point? The mirror’s slope is 3/2, so the slope of p is -2/3. Use m = -2/3 and (x, y) = (-2, 0) to find b. 0 = -2/3(-2) + b 0 = 4/3 + b -4/3 = b So, an equation for p is y = -2/3 x – 4/3 Ex. 6: Writing the equation of a perpendicular line

  11. Chapter 3 Review – pp.180-182 #1-24. Chapter 3 Exam Chapter 4 Vocabulary Chapter 4: Postulates/ Theorems Coming Attractions:

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