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I had a few questions I needed help with but saw someone had already posted it, so I am re-posting it because I need help with the same questions. 

 

Here's the link: https://web2.0calc.com/questions/please-help-me-with-these-question-i-am-stuck-and

 

Thank you for your time.

 May 10, 2020
 #1
avatar+21000 
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Given two lines, one with a slope of 13/9 and the other with a slope of 1/3, find the slope of the bisector.

 

According to Wolfram (a very authoritative site), to find this slope, use this formula:

slope  =  [ m1·m2 - 1 + sqrt{ (m12 + 1)·(m22 + 1) } ]  /  ( m1 + m2 )

 

For this problem:  m1 = 13/9  and  m2 = 1/3.

 

Putting these numbers into the formula:  

  numerator  =  [ (13/9)·(1/3) - 1 + sqrt{ ( (13/9)2 + 1 ) · ( (1/3)2 + 1 ) }  =  [13/27 - 1 + 50/27 ]  =  36/27

  denominator  =  ( 13/9 + 1/3 )  =  16/9

slope  =  ( 36/27) / ( 16/9 )  =  3/4.

 

I checked these out using inverse tangent for 13/9, 1/3, and 3/4 and found that they work.

 May 10, 2020
 #2
avatar+21000 
+1

For the second question:  Find the largest x-value for the equation:  x2 + y2  =  4x + 4y.

 

This is the equation of a circle. 

I plan to find the center and radius of the circle by completing the square on both x and y.

 

x2 + y2  =  4x + 4y.      --->     (x2 - 4x       ) +  (y2 - 4y     )  =  0

complete the square   --->     (x2 - 4x + 4)  + (y2 - 4y + 4)  =  0 + 4 + 4

                                                  (x - 2)2      +     (y - 2)2      =  8

 

This is a circle with center  (2, 2)  and radius  sqrt(8).

 

So, the largest x-value is a distance of the sqrt(8) to the right of the x-value of the center:  2 + sqrt(8)

 May 10, 2020
 #3
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+1

Thank you geno3141! This problem seemed hard but your solution helped me figure it out. Thanks once again!

 May 10, 2020

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