Algebra and Trigonometry: Structure and Method, Book 2
Algebra and Trigonometry: Structure and Method, Book 2
2000th Edition
ISBN: 9780395977255
Author: MCDOUGAL LITTEL
Publisher: McDougal Littell
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Question
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Chapter 9.1, Problem 14P

a)

To determine

To prove: T1=(r+1,s+m1)&T2=(r+1,s+m2)

a)

Expert Solution
Check Mark

Explanation of Solution

Given information: P(r,s) is the point, where L1:y=m1x+b1  &  L2:y=m2x+b2 intersects each other.

  T1&T2 be the points where the line x=r+1 intersects L1&L2 , respectively.

Proof: Substitute P(r,s) for x and y

  L1:y=m1x+b1  ∴s=m1(r)+b1 L1:rm1=s−b1

Substitute x=r+1 ,

  L1:y=m1(r+1)+b1  L1:y=rm1+m1+b1

Now, solve further by taking both above equations,

  ⇒y=s−b1+m1+b1⇒y=s+m1∴T1=(r+1,s+m1)

Similarly, substitute P(r,s) for x and y

  L2:s=m2(r)+b2  L2:rm2=s−b2

Substitute x=r+1 ,

  L2:y=m2(r+1)+b2  L2:y=rm2+m2+b2

Now, solve further by taking both above equations,

  ⇒y=s−b2+m2+b2⇒y=s+m2∴T2=(r+1,s+m21)

Therefore, T1=(r+1,s+m1)&T2=(r+1,s+m2) .

Hence, proved.

b)

To determine

To find: The distance T1T2,PT1&PT2

b)

Expert Solution
Check Mark

Answer to Problem 14P

The required distances of T1T2,PT1&PT2 are (m1−m2),1+(m1)2&1+(m2)2 , respectively.

Explanation of Solution

Given information: P(r,s) is the point, where L1:y=m1x+b1  &  L2:y=m2x+b2 intersects each other.

  T1&T2 be the points where the line x=r+1 intersects L1&L2 , respectively.

Calculation: Get the distance T1T2 ,

  T1T2=(r+1−r−1)2+(s+m1−s−m2)2=(m1−m2)2T1T2=(m1−m2)

Get the distance PT1

  PT1=(r−r−1)2+(s−s−m1)2PT1=1+(m1)2

Now, get the distance PT2

  PT2=(r−r−1)2+(s−s−m2)2PT2=1+(m2)2

Therefore, the required distances are of T1T2,PT1&PT2 are (m1−m2),1+(m1)2&1+(m2)2 , respectively.

c)

To determine

To prove: (T1T2)2=(PT1)2+(PT2)2

c)

Expert Solution
Check Mark

Explanation of Solution

Given information: m1m2=−1 and triangle T1PT2 has a right angle at P. And L1&L2 are perpendicular.

Proof: Find the (T1T2)2

  (T1T2)2=(m1−m2)2          {∵T1T2=m1−m2}=(m1−m2)2=(m1)2+(m2)2−2m1m2=(m1)2+(m2)2+2

Now, get the (PT1)2

  (PT1)2=1+((m1)2)2     {∵PT1=1+(m1)2}=1+(m1)2

Now, get the (PT2)2

  (PT2)2=1+((m2)2)2     {∵PT2=1+(m2)2}=1+(m2)2

Hence, (PT1)2+(PT2)2

  (PT1)2+(PT2)2=1+(m1)2+1+(m2)2=(m1)2+(m2)2+2=(T1T2)2

Therefore, (T1T2)2=(PT1)2+(PT2)2 .

Hence, proved.

d)

To determine

To prove: m1m2=−1

d)

Expert Solution
Check Mark

Explanation of Solution

Given information: Triangle T1PT2 has a right angle at P. And L1&L2 are perpendicular.

Proof: Find the (T1T2)2

  (T1T2)2=(m1−m2)2          {∵T1T2=m1−m2}=(m1−m2)2=(m1)2+(m2)2−2m1m2

Now, get the (PT1)2

  (PT1)2=1+((m1)2)2     {∵PT1=1+(m1)2}=1+(m1)2

Now, get the (PT2)2

  (PT2)2=1+((m2)2)2     {∵PT2=1+(m2)2}=1+(m2)2

Hence, (T1T2)2=(PT1)2+(PT2)2

  (m1)2+(m2)2−2m1m2=1+(m1)2+1+(m2)2−2m1m2=2∴m1m2=−1

Therefore, m1m2=−1 .

Hence, proved.

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Chapter 9.1, Problem 13P
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Chapter 9.1, Problem 1MRE

Chapter 9 Solutions

Algebra and Trigonometry: Structure and Method, Book 2

Ch. 9.1 - Prob. 1OECh. 9.1 - Prob. 2OECh. 9.1 - Prob. 3OECh. 9.1 - Prob. 4OECh. 9.1 - Prob. 5OECh. 9.1 - Prob. 6OECh. 9.1 - Prob. 7OECh. 9.1 - Prob. 8OECh. 9.1 - Prob. 9OECh. 9.1 - Prob. 10OE
Ch. 9.1 - Prob. 11OECh. 9.1 - Prob. 12OECh. 9.1 - Prob. 13OECh. 9.1 - Prob. 14OECh. 9.1 - Prob. 15OECh. 9.1 - Prob. 1WECh. 9.1 - Prob. 2WECh. 9.1 - Prob. 3WECh. 9.1 - Prob. 4WECh. 9.1 - Prob. 5WECh. 9.1 - Prob. 6WECh. 9.1 - Prob. 7WECh. 9.1 - Prob. 8WECh. 9.1 - Prob. 9WECh. 9.1 - Prob. 10WECh. 9.1 - Prob. 11WECh. 9.1 - Prob. 12WECh. 9.1 - Prob. 13WECh. 9.1 - Prob. 14WECh. 9.1 - Prob. 15WECh. 9.1 - Prob. 16WECh. 9.1 - Prob. 17WECh. 9.1 - Prob. 18WECh. 9.1 - Prob. 19WECh. 9.1 - Prob. 20WECh. 9.1 - Prob. 21WECh. 9.1 - Prob. 22WECh. 9.1 - Prob. 23WECh. 9.1 - Prob. 24WECh. 9.1 - Prob. 25WECh. 9.1 - Prob. 26WECh. 9.1 - Prob. 27WECh. 9.1 - Prob. 28WECh. 9.1 - Prob. 29WECh. 9.1 - Prob. 30WECh. 9.1 - Prob. 31WECh. 9.1 - Prob. 32WECh. 9.1 - Prob. 1PCh. 9.1 - Prob. 2PCh. 9.1 - Prob. 3PCh. 9.1 - Prob. 4PCh. 9.1 - Prob. 5PCh. 9.1 - Prob. 6PCh. 9.1 - Prob. 7PCh. 9.1 - Prob. 8PCh. 9.1 - Prob. 9PCh. 9.1 - Prob. 10PCh. 9.1 - Prob. 11PCh. 9.1 - Prob. 12PCh. 9.1 - Prob. 13PCh. 9.1 - Prob. 14PCh. 9.1 - Prob. 1MRECh. 9.1 - 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