3.7 Perpendicular Lines in the Coordinate Plane - PowerPoint PPT Presentation

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3.7 Perpendicular Lines in the Coordinate Plane

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3.7 Perpendicular Lines in the Coordinate Plane Standard/Objectives: Standard 3: Students will learn and apply geometric concepts. Objectives: Use slope to identify ... – PowerPoint PPT presentation

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Title: 3.7 Perpendicular Lines in the Coordinate Plane


1
3.7 Perpendicular Lines in the Coordinate Plane
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
Postulate 18 Slopes of Perpendicular Lines
  • 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

4
Ex. 1 Deciding whether lines are perpendicular
  • 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

5
Ex.2 Deciding whether lines are perpendicular
  • 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

The product of 2(-1/2) -1 so AC?DB
6
Ex.3 Deciding whether lines are perpendicular
  • 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.

7
Ex.4 Deciding whether lines are perpendicular
  • Line r 4x5y2
  • 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

The product of the slopes is NOT -1 so r and s
are NOT perpendicular.
-4 -5 1 5 4
8
Ex. 5 Writing the equation of a perpendicular
line.
  • 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

9
Ex. 6 Writing the equation of a perpendicular
line
  • 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 mirrors 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
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