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

The two lines are y = x and y = 2x.

RedDragonl May 28, 2024

#1**+1 **

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

CPhill May 28, 2024