1 topic 4.4.2 perpendicular lines. 2 topic 4.4.2 perpendicular lines california standard: 8.0...

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Topic 4.4.2Topic 4.4.2

Perpendicular LinesPerpendicular Lines

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Topic4.4.2


California Standard:8.0 Students understand the concepts of parallel lines and perpendicular lines and how their slopes are related. Students are able to find the equation of a line perpendicular to a given line that passes through a given point.

What it means for you:You’ll work out the slopes of perpendicular lines and you’ll test if two lines are perpendicular.

Key words:• perpendicular• reciprocal

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Topic4.4.2

Math problems about parallel lines often deal with perpendicular lines too.


“Perpendicular” might sound like a difficult term, but it’s actually a really simple idea.

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Topic4.4.2

Perpendicular Lines Meet at Right Angles

Two lines are perpendicular if they intersect at 90° angles, like in the graphs below.


y

x

y

x

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Topic4.4.2

Slopes of Perpendicular Lines are Negative Reciprocals

To get the reciprocal of a number you divide 1 by it.

Two lines are perpendicular if the slope of one is

the negative reciprocal of the slope of the other.


The reciprocal of is = and so the negative reciprocal is – .1

yx

x y

y x

y x

For example, the reciprocal of x is and so the negative reciprocal is – . 1 x

1 x

6

–6 –4 –2 0 2 4 60

2

4

6y

x

–2

–4

–6

Topic4.4.2

Prove that lines A and B, shown on the graph, are perpendicular to each other.

Solution follows…

Example 1

Solution

Using the rise over run

formula, slope = :y x


Slope of A = m1 = =2 4

1 2

Slope of B = m2 = = –2–4 2

is the negative reciprocal of –2, so A and B must be perpendicular. 1 2

2

4

2

4

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Topic4.4.2

Guided Practice

Solution follows…

1. Perpendicular lines meet at angles.……………………….


5. Use the graph to prove that A and B are perpendicular.

90° or right

2. Find the negative reciprocal of 3. 13

–

4. Find the negative reciprocal of – .45

54

3. Find the negative reciprocal of – .14

4

mA = and mB = –23

32

The gradients are the negative reciprocals of each other, so A and B are perpendicular.

–6 –4 –2 0 2 4 60

2

4

6y

x

–2

–4

–6

A

B

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Topic4.4.2

Perpendicular Lines: m1 × m2 = –1

When you multiply a number by its reciprocal, you always get 1.


So because two perpendicular slopes are negative reciprocals of each other, their product is always –1. Here’s the same thing written in math-speak:

If two lines l1 and l2 have slopes m1 and m2, l1 l2 if and only if m1 × m2 = –1.

means “is perpendicular to.”

For example,3

2×

2

3= 1 .and

1

5× 5 =

5

5= 1

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Topic4.4.2

P and Q are two straight lines and P Q. P has a slope of –4. What is the slope of Q?

Solution follows…

Example 2

Solution


mP × mQ = –1

–4 × mQ = –1

mQ = = 1

4

–1

–4

So the slope of Q is . 1

4

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Topic4.4.2

Guided Practice

Solution follows…

6. Lines l1 and l2 are perpendicular.

If the slope of l1 is , find the slope of l2.

7. Lines A and B are perpendicular.

If the slope of A is – , find the slope of B.

8. Lines R and T are perpendicular.

If R has slope – , what is the slope of T?

9. The slope of l1 is –0.8. The slope of l2 is 1.25. Determine whether l1 and l2 are perpendicular.

85


7

11

1

5

5

8

–5

117

m1 × m2 = –0.8 × 1.25 = –1, so the lines are perpendicular.

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Topic4.4.2

Determine the equation of the line passing through (3, 1) that is perpendicular to the straight line through (2, –1) and (4, 2).

Example 3

Solution follows…

Solution

Step 1: Find slope m1 of the line through (2, –1) and (4, 2):

m1 =y2 – y1

x2 – x1

=2 – (–1)

4 – 2=

3

2

Step 2: Find the slope m2 of a line perpendicular to that line:

m1 × m2 = –1

Solution continues…


× m2 = –132

m2 = – 23

You can show that lines are perpendicular by finding their slopes.

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Topic4.4.2

Determine the equation of the line passing through (3, 1) that is perpendicular to the straight line through (2, –1) and (4, 2).

Solution (continued)

Example 3

y – y1 = m(x – x1)

y – 1 = – (x – 3)2

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3y – 3 = –2(x – 3)

3y – 3 = –2x + 6

Equation: 3y + 2x = 9


Step 3: Now use the point-slope formula to find the equation of

the line through (3, 1) with slope – . 2

3

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Topic4.4.2

Guided Practice

Solution follows…

10. Show that the line through the points (5, –3) and (–8, 1) is perpendicular to the line through (4, 6) and (8, 19).


11. Show that the line through (0, 6) and (5, 1) is perpendicular to the line through (4, 8) and (–1, 3).

m1 = = –4

131 – (–3)–8 – 5

m2 = =134

19 – 68 – 4

Since m1 × m2 = –1, the lines are perpendicular.


m1 = = –1 – 65 – 0

55

= –1

m2 = =–5–5

3 – 8–1 – 4

= 1

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Topic4.4.2

Guided Practice

Solution follows…

12. Show that the line through (4, 3) and (2, 2) is perpendicular to the line through (1, 3) and (3, –1).


13. Determine the equation of the line through (3, –4) that is perpendicular to the line through the points (–7, –3) and (–3, 8).


m1 = =–1–2

2 – 32 – 4

=12

m2 = =–42

–1 – 33 – 1

= –2

m1 = =114

8 – (–3)–3 – (–7)

So, m2 = –4

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y – (–4) = – (x – 3) 11y + 44 = –4x + 12 11y + 4x = –324

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Topic4.4.2

Guided Practice

Solution follows…

14. Determine the equation of the line through (6, –7) that is perpendicular to the line through the points (8, 2) and (–1, 8).


15. Find the equation of the line through (4, 5) that is perpendicular to the line –3y + 4x = 6.

So, m2 =32

m1 = =6

–98 – 2

–1 – 8= –

23

y – (–7) = (x – 6) 2y + 14 = 3x – 18 2y – 3x = –3232

Find two points on the line –3y + 4x = 6. For example, (0, –2) and (3, 2).Use these points to find m1.

m1 = = 2 – –2

3 – 0

4

3y – 5 = – (x – 4) 4y – 20 = –3x + 12

4y + 3x = 32

3

4So, m2 = –

3

4

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Topic4.4.2

Independent Practice

Solution follows…

In Exercises 1–8, J and K are perpendicular lines. The slope of J is given. Find the slope of K.

1

3


1. mJ = –3 2. mJ = –14

7. mJ = –0.18 8. mJ = 0.45

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3. mJ = 4. mJ =67

5. mJ = –83

6. mJ = –1415

–2

5

3

8

5.5555…

1

14

–7

6

15

14

–2.2222…

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Topic4.4.2


Solution follows…

9. Show that the line through (2, 7) and (–2, 8) is perpendicular to the line through (–3, –3) and (–2, 1).


10. Show that the line through (–4, 3) and (3, –2) is perpendicular to the line through (–7, –1) and (–2, 6).

m1 × m2 = –1, so the lines are perpendicular.

m2 = =41

1 – (–3)–2 – (–3)

= 4

m1 =8 – 7

–2 – 2= –

14

m1 × m2 = –1, so the lines are perpendicular.

m1 =–2 – 3

3 – (– 4)= –

57

m2 = =6 – (–1)

–2 – (–7)75

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Topic4.4.2


Solution follows…


12. Determine the equation of the line through (3, –5) that is perpendicular to the line through the points (–3, 2) and (–6, –4).

11. Determine the equation of the line through (5, 9) that is

perpendicular to a line with slope .1

3y + 3x = 24

2y + x = –7

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Topic4.4.2

Round UpRound Up

“Perpendicular” is just a special math word to describe lines that are at right angles to each other.


Remember that the best way to show that two lines are at right angles is to calculate their slopes — if they multiply together to make –1, then the lines are perpendicular.

1 topic 4.4.2 perpendicular lines. 2 topic 4.4.2 perpendicular lines california standard: 8.0...

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