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Help plz! Geometry!

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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 acute angle between these lines.

May 13, 2019

#1
+200
+1

If you call the angles of the lines to the x axis A and B

tan(A) = 4/3 and tan(B) = 5/12

The angle of the bisector will be (A+B)/2.

You can work this out using the following tan formulas

$$tan(A+B) = \frac{tan(A) + tan(B)}{1 - tan(A)tan(B)}$$

and

$$tan(\theta) = \frac{2tan(\frac{\theta}{2})}{1 - tan^2(\frac{\theta}{2})}$$

Also from $$\frac{9}{\sqrt{130}}y-\frac{7}{\sqrt{130}}x=0$$

from that you can find the slope

May 13, 2019
edited by Guest  May 13, 2019
edited by Guest  May 13, 2019
#2
+103982
+1

We can find the slope thusly :

tan ( [Arctan (4/3) + Arctan (5/12) ] / 2 )  =   7/9

See the graph here :  https://www.desmos.com/calculator/x0uxs0c1re

May 13, 2019
#3
+23170
+1

The lines  $$y = \dfrac{5}{12} x$$ and  $$y = \dfrac{4}{3} x$$ are drawn in the coordinate plane.

Find the slope of the line that bisects the acute angle between these lines.

$$\begin{array}{|rcll|} \hline y &=& \dfrac{5}{12} x \quad | \quad x=12 ~ \Rightarrow ~ y=5, \quad \vec{v}=\binom{12}{5}, \quad \vec{v_0}=\dfrac{1}{\sqrt{12^2+5^2} }\dbinom{12}{5}= \dbinom{\frac{12}{13}}{\frac{5}{13}}\\ y &=& \dfrac{4}{3} x \quad | \quad x=3 ~ \Rightarrow ~ y=4, \quad \vec{w}=\binom{3}{4}, \quad \vec{w_0}=\dfrac{1}{\sqrt{3^2+4^2} }\dbinom{3}{4}=\dbinom{\frac{3}{5}}{\frac{4}{5}}\\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \vec{v_0}+\vec{w_0} &=& \dbinom{\frac{12}{13}}{\frac{5}{13}} + \dbinom{\frac{3}{5}}{\frac{4}{5}} \\ \vec{v_0}+\vec{w_0} &=& \dbinom{\frac{12}{13}+\frac{3}{5}}{\frac{5}{13}+\frac{4}{5}} \\ \vec{v_0}+\vec{w_0} &=& \dbinom{\frac{99}{65}}{\frac{77}{65}} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \text{slope of the line that bisects} &=& \dfrac{y}{x} \\\\ &=& \dfrac{\frac{77}{65}}{\frac{99}{65}} \\\\ &=& \dfrac{77}{99} \\\\ &=& \dfrac{7}{9} \\ \hline \end{array}$$

May 13, 2019