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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
 #2
avatar+111396 
+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   

 

 

cool cool cool

 May 13, 2019
 #3
avatar+25275 
+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.

 

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

 

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

 

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

 

laugh

 May 13, 2019

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