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# 3.7 Perpendicular Lines in the Coordinate Plane PowerPoint PPT Presentation

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.

3.7 Perpendicular Lines in the Coordinate Plane

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## 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.

### Assignment:

• Pp. 175-177 #7-45 and 47-50

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

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

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

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

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

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.

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

Chapter 3 Review – pp.180-182 #1-24.

Chapter 3 Exam

Chapter 4 Vocabulary

Chapter 4: Postulates/ Theorems