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The line $y = mx$ bisects the angle between the two lines shown below. Find $m$.
The two lines are y = x and y = 2x.

 May 28, 2024
 #1
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Let these two lines intersect a cirlce with a radius of 1

 

To find the point on y= x  that lies on the circle

 

x^2 + x^2  =1

2x^2 =1

x^2= 1/2

x =1/sqrt 2     and y= 1/sqrt 2

 

To find the point on y= 2x that lies on the circle

 

x^2 + (2x)^2 = 1

5x^2 =1

x^2 = 1/5

x = 1/sqrt5      y = 2/sqrt 5

 

The midpoint  between these two points is [  (1/sqrt 2 + 1/sqrt 5) / 2 , (1/sqrt 2 + 2/sqrt 5)/2 ] =

 

[ (sqrt 2 + sqrt 5)  / (2sqrt 10) ,  (2sqrt 2 + sqrt 5) / (2sqrt 10) ]

 

The slope of the line drawn from the origin through this point = m  =

 

[ 2sqrt 2 +sqrt 5 ] / [ sqrt 2 + sqrt 5] =  

 

[ sqrt 5 - sqrt 2] [ sqrt 5 + 2sqrt 2] / 3  =

 

[ 5 + sqrt 10 - 4 ] /  3  =

 

[ 1 + sqrt 10 ] /  3

 

cool cool cool

 May 28, 2024
edited by CPhill  May 28, 2024

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