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The lines \(y = \frac{5}{12} x\) and \(y = \frac{4}{3} x\) are drawn in the coordinate plane. Find the slope of the line that bisects the angle between these lines.

 

 Jun 22, 2019
 #1
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The lines y = (5/12)x and y = (4/3)x are drawn in the coordinate plane. Find the slope of the line that bisects the acute angle between these lines.

 

Here's one way, Logic.....but I don't know if it's the most elegant

 

Construct a circle centered at the origin with a radius of 1

 

The  equation  is

 

x^2 + y^2  = 1

 

We can find the x coordinate  of the intersection of the first line and this circle,thusly :

 

x^2  + [ (5/12)x ]^2  = 1

x^2 + (25/144)x^2   =1

[144 + 25] / 144  x^2  = 1

[169] /144 x^2  =1

x^2 = 144/169

x = 12/13

And y = (5/12)(12/13) = 5/13

So   ( 12/13, 5/13 )  is on the circle

 

Likewise...we can find the intersection of this circle with the  second line :

 

x^2  +[ (4/3)x\^2  =1

x^2  + (16/9)x^2  = 1

[9 + 16 ] / 9 * x^2  =1

25/9 x^2  =1

x^2  = 9/25

x = 3/5

And y = (4/3)(3/5)  = 4/5

So  (3/5, 4/5)  is on the circle

 

And a chord can be drawn on this circle with endpoints   ( 12/13, 5/13)  and (3/5, 4/5 )

And the midpoint of this chord is

  (  [12/13 + 3/5]  / 2,  [ 5/13 + 4/5 ] / 2 )  =

 

(99/130 , 77/130  )

 

And  the line drawn  from the origin to this point is the line we seek and it will have the slope

 

[77/130]  / [99/130]  =    77/99   =  7/9

 

 

  

 cool cool cool

 Jun 22, 2019
edited by CPhill  Jun 22, 2019

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