geometry section 3-6 1112

Section 3-6Perpendiculars and Distance

Thursday, January 5, 2012

Essential Questions

n How do you find the distance between a point and a line?

n How do you find the distance between parallel lines?


Vocabulary1. Equidistant:

2. Distance Between a Point and a Line:

3. Distance Between Parallel Lines:


Vocabulary1. Equidistant: The distance between any two lines as

measured along a perpendicular is the same; this occurs with parallel lines

2. Distance Between a Point and a Line:





2. Distance Between a Point and a Line: The length of the segment perpendicular to the line with the point one endpoint on the segment





2. Distance Between a Point and a Line: The length of the segment perpendicular to the line with the point one endpoint on the segment

3. Distance Between Parallel Lines: The length of the segment perpendicular to the two parallel lines with the endpoints on either of the parallel lines


Postulates & Theorems1. Perpendicular Postulate:

2. Two Lines Equidistant from a Third:


Postulates & Theorems1. Perpendicular Postulate: If given a line and a point

not on the line, then there exists exactly one line through the point that is perpendicular to the given line

2. Two Lines Equidistant from a Third:


Postulates & Theorems1. Perpendicular Postulate: If given a line and a point

not on the line, then there exists exactly one line through the point that is perpendicular to the given line

2. Two Lines Equidistant from a Third: In a plane, if two lines are each equidistant from a third line, then the two lines are parallel to each other


Example 1The line a contains the points T(0, 0) and U(−5, 5).

Find the distance between line a and the point V(1, 5).




1. Find the equation of the original line





m =

0 − 50 + 5





m =

0 − 50 + 5

=−55





m =

0 − 50 + 5

=−55

= −1





m =

0 − 50 + 5

=−55

= −1 T(0, 0)





m =

0 − 50 + 5

=−55

= −1 T(0, 0) y = mx + b





m =

0 − 50 + 5

=−55

= −1 T(0, 0) y = −x

y = mx + b





m =

0 − 50 + 5

=−55

= −1 T(0, 0) y = −x

2. Find the equation of the perpendicular line through the other point

y = mx + b





m =

0 − 50 + 5

=−55

= −1 T(0, 0) y = −x


m = 1

y = mx + b





m =

0 − 50 + 5

=−55

= −1 T(0, 0) y = −x


m = 1

V(1, 5)

y = mx + b





m =

0 − 50 + 5

=−55

= −1 T(0, 0) y = −x


m = 1

V(1, 5) y − y

1= m(x − x

1)

y = mx + b





m =

0 − 50 + 5

=−55

= −1 T(0, 0) y = −x


m = 1

V(1, 5) y − y

1= m(x − x

1)

y = mx + b

y − 5 = 1(x −1)Thursday, January 5, 2012




m =

0 − 50 + 5

=−55

= −1 T(0, 0) y = −x


m = 1

V(1, 5) y − y

1= m(x − x

1)

y = mx + b

y − 5 = 1(x −1) y − 5 = x −1





m =

0 − 50 + 5

=−55

= −1 T(0, 0) y = −x


m = 1

V(1, 5) y − y

1= m(x − x

1)

y = mx + b

y − 5 = 1(x −1) y − 5 = x −1

y = x + 4Thursday, January 5, 2012

Example 1

3. Solve the system of these two equations.


Example 1


y = −x

y = x + 4

⎧⎨⎩


Example 1


y = −x

y = x + 4

⎧⎨⎩

−x = x + 4


Example 1


y = −x

y = x + 4

⎧⎨⎩

−x = x + 4

−2x = 4


Example 1


y = −x

y = x + 4

⎧⎨⎩

−x = x + 4

−2x = 4

x = −2


Example 1


y = −x

y = x + 4

⎧⎨⎩

−x = x + 4

−2x = 4

x = −2

y = −(−2)


Example 1


y = −x

y = x + 4

⎧⎨⎩

−x = x + 4

−2x = 4

x = −2

y = −(−2) = 2


Example 1


y = −x

y = x + 4

⎧⎨⎩

−x = x + 4

−2x = 4

x = −2

y = −(−2) = 2

2 = −2 + 4


Example 1


y = −x

y = x + 4

⎧⎨⎩

−x = x + 4

−2x = 4

x = −2

y = −(−2) = 2

2 = −2 + 4

(−2,2)


Example 1

4. Use the distance formula utilizing this point on the line and the point not on the line.


Example 1


(1, 5), (−2, 2)


Example 1


(1, 5), (−2, 2)

d = (x

2− x

1)2 + (y

2− y

1)2


Example 1


(1, 5), (−2, 2)

d = (x

2− x

1)2 + (y

2− y

1)2

= (−2 −1)2 + (2 − 5)2


Example 1


(1, 5), (−2, 2)

d = (x

2− x

1)2 + (y

2− y

1)2

= (−2 −1)2 + (2 − 5)2

= (−3)2 + (−3)2


Example 1


(1, 5), (−2, 2)

d = (x

2− x

1)2 + (y

2− y

1)2

= (−2 −1)2 + (2 − 5)2

= (−3)2 + (−3)2 = 9 + 9


Example 1


(1, 5), (−2, 2)

d = (x

2− x

1)2 + (y

2− y

1)2

= (−2 −1)2 + (2 − 5)2

= (−3)2 + (−3)2 = 9 + 9 = 18


Example 1


(1, 5), (−2, 2)

d = (x

2− x

1)2 + (y

2− y

1)2

= (−2 −1)2 + (2 − 5)2

= (−3)2 + (−3)2 = 9 + 9 = 18 ≈ 4.24


Example 1


(1, 5), (−2, 2)

d = (x

2− x

1)2 + (y

2− y

1)2

= (−2 −1)2 + (2 − 5)2

= (−3)2 + (−3)2 = 9 + 9 = 18 ≈ 4.24 units


Example 2Find the distance between the parallel lines m and n with

the following equations.

y = 2x + 3 y = 2x −1




y = 2x + 3 y = 2x −11. Find the equation of the perpendicular line.





y = mx + b





y = mx + b

m = −

12

,(0,3)





y = mx + b

y = −

12

x + 3

m = −

12

,(0,3)


Example 22. Find the intersection of the perpendicular line and the

other parallel line using a system.




y = 2x −1

y = −12

x + 3

⎧⎨⎪

⎩⎪




y = 2x −1

y = −12

x + 3

⎧⎨⎪

⎩⎪

2x −1 = −

12

x + 3




y = 2x −1

y = −12

x + 3

⎧⎨⎪

⎩⎪

2x −1 = −

12

x + 3

52

x = 4




y = 2x −1

y = −12

x + 3

⎧⎨⎪

⎩⎪

2x −1 = −

12

x + 3

52

x = 4 x = 1.6




y = 2x −1

y = −12

x + 3

⎧⎨⎪

⎩⎪

2x −1 = −

12

x + 3

52

x = 4 x = 1.6

y = 2(1.6)−1




y = 2x −1

y = −12

x + 3

⎧⎨⎪

⎩⎪

2x −1 = −

12

x + 3

52

x = 4 x = 1.6

y = 2(1.6)−1

y = 2.2




y = 2x −1

y = −12

x + 3

⎧⎨⎪

⎩⎪

2x −1 = −

12

x + 3

52

x = 4 x = 1.6

y = 2(1.6)−1

y = 2.2 y = −

12

(1.6)−1




y = 2x −1

y = −12

x + 3

⎧⎨⎪

⎩⎪

2x −1 = −

12

x + 3

52

x = 4 x = 1.6

y = 2(1.6)−1

y = 2.2 y = −

12

(1.6)−1

y = 2.2




y = 2x −1

y = −12

x + 3

⎧⎨⎪

⎩⎪

2x −1 = −

12

x + 3

52

x = 4 x = 1.6

y = 2(1.6)−1

y = 2.2 y = −

12

(1.6)−1

y = 2.2

(1.6, 2.2)


Example 23. Use the new point and original y-intercept you chose in

step 2 in the distance formula.




(0, 3), (1.6, 2.2)




(0, 3), (1.6, 2.2)

d = (x

2− x

1)2 + (y

2− y

1)2




(0, 3), (1.6, 2.2)

d = (x

2− x

1)2 + (y

2− y

1)2

= (1.6 − 0)2 + (2.2 − 3)2




(0, 3), (1.6, 2.2)

d = (x

2− x

1)2 + (y

2− y

1)2

= (1.6 − 0)2 + (2.2 − 3)2

= (1.6)2 + (−0.8)2




(0, 3), (1.6, 2.2)

d = (x

2− x

1)2 + (y

2− y

1)2

= (1.6 − 0)2 + (2.2 − 3)2

= (1.6)2 + (−0.8)2 = 2.56 + .64




(0, 3), (1.6, 2.2)

d = (x

2− x

1)2 + (y

2− y

1)2

= (1.6 − 0)2 + (2.2 − 3)2

= (1.6)2 + (−0.8)2 = 2.56 + .64

= 3.2Thursday, January 5, 2012



(0, 3), (1.6, 2.2)

d = (x

2− x

1)2 + (y

2− y

1)2

= (1.6 − 0)2 + (2.2 − 3)2

= (1.6)2 + (−0.8)2 = 2.56 + .64

= 3.2 ≈ 1.79Thursday, January 5, 2012



(0, 3), (1.6, 2.2)

d = (x

2− x

1)2 + (y

2− y

1)2

= (1.6 − 0)2 + (2.2 − 3)2

= (1.6)2 + (−0.8)2 = 2.56 + .64

= 3.2 ≈ 1.79 units


Example 3You try it out! Refer to the process in example 1.

Line h contains the points E(2, 4) and F(5, 1). Find the distance between line h and the point G(1, 1).




Solution:




Solution:

d = 8




Solution:

d = 8 ≈ 2.83




Solution:

d = 8 ≈ 2.83 units


Check Your Understanding

Review problems #1-8 on p. 218


Problem Set


Problem Set

p. 218 #13-33 odd, 53, 59, 63

“I’m a great believer in luck, and I find the harder I work the more I have of it.” - Thomas Jefferson


geometry section 3-6 1112

Education

perpendicular line

point v1

points t0

lines equidistant

perpendicular postulate

iftwo lines

mx x1 v1

original linethursday