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1. The lines y=x/3 and y=13x/9 are drawn in the coordinate plane. Find the slope of the line that bisects the angle between these lines.

 

2. The line y=(3x+15)/4 intersects circle x^2+y^2=36 at points Q and R. Find the length of QR.

 

For question number one, right triangles seem useful because slope is "rise over run", which, if you graph, creates a right triangle. But, otherwise, I'm very confused.

 Apr 28, 2020
 #1
avatar+23550 
+1

Y= mx + b   is the form of a line where m = slope

y = 1/3 x   +0     is of this form    slope = 1/3

y=  13/9  x  + 0                            slope = 13/9

 

 

 

we need the angle between these two       [arctan (1/3) + arctan (13/9)]/2 = 36.8699o   

tan ( 36.8688) = slope of line bisecting other two = .75   = 3/4        

so y = 3/4 x   will bisect the angle of the other two ~  cheeky

 

Here is a verification graph:

https://www.desmos.com/calculator/7gf4hgooic

 Apr 28, 2020
edited by ElectricPavlov  Apr 28, 2020
 #2
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Thanks, EP!

Guest Apr 29, 2020
 #3
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Does anyone have anything on number 2?

Guest Apr 29, 2020
 #4
avatar+20786 
+1

2)    x2 + y2  =  36         and        y  =  (3x + 15)/4

       x2 + [ (3x + 15) / 4 ]2  =  36

       x2 + (9x2 + 90x + 225) / 16  =  36

   16x2 + (9x2 + 90x + 225)  =  576

   25x2 + 90x - 351  =  0

 

Quadratic formula:  x1  =  [ -90 + sqrt[( 43 200 ) ] / 50              or         x2  =   [ -90 - sqrt[( 43 200 ) ] / 50   

                                y1  =  [ 480 + 3·sqrt( 43 200 ) ] / 200                     y2  =   [ 480 - 3·sqrt( 43 200 ) ] / 200 

 

Distance formula:  sqrt[ ( x2 - x1 )2 + ( y2 - y1 )2 ]

                             =  10.392                      (calculator time)

 Apr 29, 2020
 #5
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is this an approximation?
it says simplify as an improper fraction.

Guest Apr 30, 2020
 #6
avatar+109484 
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You can adjust that yourself.

Melody  Apr 30, 2020
 #7
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alright thanks

Guest Apr 30, 2020

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