Angle bisector from two lines of graph

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 \( \frac{5}{7} \).

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! :)

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! :)