We have the lines y=3x and y=2x. Let y=mx be the angle bisector of these two lines. How can I find "m"?

dreadnought Jun 21, 2024

#1**0 **

To find the slope \( m \) of the angle bisector of the lines \( y = 3x \) and \( y = 2x \), we can use the formula for the slope of the angle bisector between two lines given their slopes \( m_1 \) and \( m_2 \):

Given the lines \( y = m_1 x \) and \( y = m_2 x \), the slope \( m \) of the angle bisector is given by:

\[

m = \frac{m_1 + m_2}{1 + m_1 m_2}

\]

Here, \( m_1 = 3 \) and \( m_2 = 2 \). Plugging these values into the formula, we get:

\[

m = \frac{3 + 2}{1 + 3 \cdot 2} = \frac{5}{1 + 6} = \frac{5}{7}

\]

Thus, the slope \( m \) of the angle bisector is \( \frac{5}{7} \).

gnistory Jun 21, 2024

#2**0 **

Correct me if I'm wrong but looking around, I saw the formula

(A1x + B1y + C1)/√(A1^2 + B12) = + (A2 x+ B2y + C2)/√(A2^2 + B2^2)

with

L1 : A1x + B1y + C1 = 0

L2 : A2x + B2y + C2 = 0 .

Does this do the same thing? or is it a different formula

dreadnought
Jun 21, 2024

#3**+1 **

There is a formula similar to the one you showed.

The equation states that we have \(\frac{Ax+By+C}{\sqrt{A^2+B^2}} = \pm \frac{ax+by+c}{\sqrt{a^2+b^2}}\) for two lines in the form of \(Ax + By + C =0, ax + by + c =0\)

It's similar, but the two lines given are in standard form rather than slope-intercept form.

I'm not too familiar on this equation, but I hope this clarified this a bit.

Feel free to ask if you're still confused.

Thanks! :)

NotThatSmart Jun 21, 2024

#5**+1 **

Unfortunately, I don't think it's used to calculate the angle bisector of a line.

It is mainly used to calculate if two lines are parrallel to eachother.

Since \(\frac{Ax+By+C}{\sqrt{A^2+B^2}} \) calculates the distance from the origin for the line \(Ax + By + C =0\) and the same applies for the left hand side, if the two are equal, they are parrallel lines.

However, I do believe I know what formula was used.

If we set the slopes of the two lines given in the question to \(m_1\) and \(m_2\), the slope of the angle bisector is

\(m=\sqrt{\frac{m_1m_2+1}{m_1+m_2}}\)

I hope this helps! If you have anymore questions, feel free to ask!

Thanks! :)

NotThatSmart
Jun 21, 2024

#7**0 **

Finally got it. The formula you entered was actually correct and the answer was y=(sqrt2 + 1)x. I watched a video and it explained it quite well.

dreadnought
Jun 21, 2024

#9**+1 **

I'm not sure if \(y=(\sqrt2 + 1)x\) is correct though.

Through my understanding, the answer must be \(y=\sqrt{\frac{7}{5}}x\) is the full equation of the line.

Am I mistaken with this judgement?

I entered the problem with an AI math bot...it confirmed my answer...

Thanks! :)

NotThatSmart
Jun 21, 2024

#6**+1 **

I'll tackle this problem.

First, we have two lines given, \(y=3x\) and \(y=2x\)

Let's note the slopes of the two line. The slope for the first line is 3, and the slope for the second line is 2.

Let's set \(m_1 = 3\) and \(m_2 = 2\)

Using the angle bisector formula, which states that we have

\(m=\sqrt{\frac{m_1m_2+1}{m_1+m_2}}\) where m1 and m2 are the slopes of the two lines, we have

\(m = \pm \sqrt{\frac{3*2+1}{3+2}}\\ m = \pm \sqrt{\frac{7}{5}} \)

So our final answer is \(m = \pm \sqrt{\frac{7}{5}} \)

Thanks! :)

NotThatSmart Jun 21, 2024