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! :)