3.7 Perpendicular Lines in the Coordinate Plane

GeometryMs. Reser

6

4

2

-2

-4

-5 5

g x = -1

2 x+2

f x = 2x-1

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.

Assignment:

• Pp. 175-177 #7-45 and 47-50

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

6

4

2

-2

-4

-5 5

g x = -1

2 x+2

f x = 2x-1

Ex. 1: Deciding whether lines are perpendicular• Find each slope.• Slope of j1

3-1 = - 21-3 3

• Slope of j23-(-3) = 6 = 3

0-(-4) 4 2Multiply the two slopes.

The product of -2 ∙ 3 = -1, so j1 j2

3 2

4

2

-2

-4

-5 5

h x = 3

2 x+3

g x = -2

3 x+3

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

4

2

-2

-4

-5 5

The product of 2(-1/2) = -1; so ACDB

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.

4

2

-2

-4

-5 5

g x = -4

3 x-3

f x = 3

4 x+2

Ex.4: Deciding whether lines are perpendicular

• 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

-4 ∙ -5 = 1

5 4

The product of the slopes is NOT -1; so r and s are NOT perpendicular.

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 = -1m2 = 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. 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 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) + b0 = 4/3 + b-4/3 = b

So, an equation for p is y = -2/3 x – 4/3

4

2

-2

-4

-5 5

g x = -2

3 x-

4

3f x =

3

2 x+3

Coming Attractions:

• Chapter 3 Review – pp.180-182 #1-24.

• Chapter 3 Exam• Chapter 4

Vocabulary• Chapter 4:

Postulates/ Theorems