+420 The line $y = mx$ bisects the angle between the two lines shown below. Find $m$.
The two lines are y = x and y = 2x.
+129630 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
