# Math Help - Analytical Geometry

1. ## Analytical Geometry

Determine the equation of the straight bisector of the acute angle that the straight $y = 2x$ form with the axis $x$.

2. I have never done a problem like this but I will give it a try.

The angle between the x-axis and y=2x is
The bisector of an angle make half this angle with the x-axis, so .55355 radians
tan(.55355) = opp/adj = rise/run = m2 = .6180
So I think the formula for a bisector is y=.6180*x or y=(tan(atan(angle)/2))*x

3. Hello, Apprentice123!

We can solve this without a calculator
. . which is what they probably expected.

Determine the equation of the bisector of the acute angle
formed by the straight line $y \:=\: 2x$ and the $x$-axis.
Code:
        |      A
|     /          B
|    /         *
|   /       *
|  / ½θ  *
| /   *
|/ *  ½θ
O * - - - - - - - X
|

Line $OA$ has the equation: . $y = 2x$
. . Let $\theta \,=\,\angle AOX.$

Its slope is: . $m_1 \:=\:2\:=\:\tan \theta$

We want: . $m_2 \:=\:\tan\left(\tfrac{\theta}{2}\right)$

Since $\tan\theta \:=\:\frac{2}{1} \:=\:\frac{opp}{adj}$, $\theta$ is in a right triangle with: $opp = x,\;adj = 1$
And Pythagorus tell us that: . $hyp = \sqrt{1^2+2^2} \:=\:\sqrt{5}$
. . Hence: . $\cos\theta \:=\:\frac{adj}{hyp}\:=\:\frac{1}{\sqrt{5}}$

We have: . $\tan\tfrac{\theta}{2} \;=\;\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta} {2}} \;=\;\sqrt{\frac{1-\cos\theta}{1+\cos\theta}} \:=$ . $\sqrt{\frac{1-\frac{1}{\sqrt{5}}}{1 + \frac{1}{\sqrt{5}}}} \:=\:\sqrt{\frac{\sqrt{5}-1}{\sqrt{5}+1}}$

Rationalize: . $m_2 \;=\;\sqrt{\frac{\sqrt{5}-1}{\sqrt{5}+1}\cdot\frac{\sqrt{5}-1}{\sqrt{5}-1}}\:=\:\sqrt{\frac{(\sqrt{5}-1)^2}{4}} \:=\:\frac{\sqrt{5}-1}{2}$

Therefore, the equation of the bisector is: . $y \;=\;\left(\frac{\sqrt{5}-1}{2}\right)x$

4. Originally Posted by Soroban
Hello, Apprentice123!

We can solve this without a calculator
. . which is what they probably expected.

Code:
        |      A
|     /          B
|    /         *
|   /       *
|  / ½θ  *
| /   *
|/ *  ½θ
O * - - - - - - - X
|

Line $OA$ has the equation: . $y = 2x$
. . Let $\theta \,=\,\angle AOX.$

Its slope is: . $m_1 \:=\:2\:=\:\tan \theta$

We want: . $m_2 \:=\:\tan\left(\tfrac{\theta}{2}\right)$

Since $\tan\theta \:=\:\frac{2}{1} \:=\:\frac{opp}{adj}$, $\theta$ is in a right triangle with: $opp = x,\;adj = 1$
And Pythagorus tell us that: . $hyp = \sqrt{1^2+2^2} \:=\:\sqrt{5}$
. . Hence: . $\cos\theta \:=\:\frac{adj}{hyp}\:=\:\frac{1}{\sqrt{5}}$

We have: . $\tan\tfrac{\theta}{2} \;=\;\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta} {2}} \;=\;\sqrt{\frac{1-\cos\theta}{1+\cos\theta}} \:=$ . $\sqrt{\frac{1-\frac{1}{\sqrt{5}}}{1 + \frac{1}{\sqrt{5}}}} \:=\:\sqrt{\frac{\sqrt{5}-1}{\sqrt{5}+1}}$

Rationalize: . $m_2 \;=\;\sqrt{\frac{\sqrt{5}-1}{\sqrt{5}+1}\cdot\frac{\sqrt{5}-1}{\sqrt{5}-1}}\:=\:\sqrt{\frac{(\sqrt{5}-1)^2}{4}} \:=\:\frac{\sqrt{5}-1}{2}$

Therefore, the equation of the bisector is: . $y \;=\;\left(\frac{\sqrt{5}-1}{2}\right)x$

ok. Thank you