
The slope of a line is a double of the slope of another line. If the tangent of the angle between them is $\dfrac{1}{3}$. Find the slopes of the line.
Answer
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Hint: In the above type of question, we will have to know about the formula of tangent of the angle between the two lines i.e, \[\tan \theta =\left| \dfrac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}{{m}_{2}}} \right|\],where \[{{m}_{1}},{{m}_{2}}\] are the slope of two lines and \[\theta \] is the angle between them.
Complete step-by-step answer:
It is said in the above question that the slope of a line is double of the slope of another line,so let’s assume that slope of one line is “m” then slope of another line will be “2m”. Also, it is given that the value of \[\tan \theta =\dfrac{1}{3}\].
Now, we will put these values in the above formula of tangent of the angle between the two lines and simplify it to obtain the value of “m”
So, we have \[{{m}_{1}}=m,{{m}_{2}}=2m\] and \[\tan \theta =\dfrac{1}{3}\].
\[\Rightarrow \dfrac{1}{3}=\left| \dfrac{2m-m}{1+2m\times m} \right|\]
On futher simplification, we will get;
\[\Rightarrow \dfrac{1}{3}=\left| \dfrac{m}{1+2{{m}^{2}}} \right|\]
Now, we have to simplify the modulus function in which we will solve the equation once with positive and another is with negative sign as shown below;
\[\Rightarrow \pm \dfrac{1}{3}=\dfrac{m}{1+2{{m}^{2}}}\]
Here, we will solve for the positive sign and we get;
\[\begin{align}
& \Rightarrow \dfrac{1}{3}=\dfrac{m}{1+2{{m}^{2}}} \\
& \Rightarrow 1+2{{m}^{2}}=3m \\
& \Rightarrow 2{{m}^{2}}-3m+1=0 \\
\end{align}\]
On, further solving the quadratic equation, we will get ;
m=1 and m= ${}^{1}/{}_{2}$
Now, we will solve for the negative sign and we get;
\[\begin{align}
& \Rightarrow -\dfrac{1}{3}=\dfrac{m}{1+2{{m}^{2}}} \\
& \Rightarrow 1+2{{m}^{2}}=-3m \\
& \Rightarrow 2{{m}^{2}}+3m+1=0 \\
\end{align}\]
On, further solving the quadratic equation,we will get;
m=-1 and m= \[-{}^{1}/{}_{2}\]
So, combining all the values of m from both cases will be the final answer.
Therefore, the slopes of the required line are ,\[m=1,{}^{1}/{}_{2},-1\] and \[-{}^{1}/{}_{2}\] .
NOTE: Be careful while solving the equation because it’s a great chance that you can make mistakes like most of us forget/skip the negative part of the solution which makes the solution incomplete.
Complete step-by-step answer:
It is said in the above question that the slope of a line is double of the slope of another line,so let’s assume that slope of one line is “m” then slope of another line will be “2m”. Also, it is given that the value of \[\tan \theta =\dfrac{1}{3}\].
Now, we will put these values in the above formula of tangent of the angle between the two lines and simplify it to obtain the value of “m”
So, we have \[{{m}_{1}}=m,{{m}_{2}}=2m\] and \[\tan \theta =\dfrac{1}{3}\].
\[\Rightarrow \dfrac{1}{3}=\left| \dfrac{2m-m}{1+2m\times m} \right|\]
On futher simplification, we will get;
\[\Rightarrow \dfrac{1}{3}=\left| \dfrac{m}{1+2{{m}^{2}}} \right|\]
Now, we have to simplify the modulus function in which we will solve the equation once with positive and another is with negative sign as shown below;
\[\Rightarrow \pm \dfrac{1}{3}=\dfrac{m}{1+2{{m}^{2}}}\]
Here, we will solve for the positive sign and we get;
\[\begin{align}
& \Rightarrow \dfrac{1}{3}=\dfrac{m}{1+2{{m}^{2}}} \\
& \Rightarrow 1+2{{m}^{2}}=3m \\
& \Rightarrow 2{{m}^{2}}-3m+1=0 \\
\end{align}\]
On, further solving the quadratic equation, we will get ;
m=1 and m= ${}^{1}/{}_{2}$
Now, we will solve for the negative sign and we get;
\[\begin{align}
& \Rightarrow -\dfrac{1}{3}=\dfrac{m}{1+2{{m}^{2}}} \\
& \Rightarrow 1+2{{m}^{2}}=-3m \\
& \Rightarrow 2{{m}^{2}}+3m+1=0 \\
\end{align}\]
On, further solving the quadratic equation,we will get;
m=-1 and m= \[-{}^{1}/{}_{2}\]
So, combining all the values of m from both cases will be the final answer.
Therefore, the slopes of the required line are ,\[m=1,{}^{1}/{}_{2},-1\] and \[-{}^{1}/{}_{2}\] .
NOTE: Be careful while solving the equation because it’s a great chance that you can make mistakes like most of us forget/skip the negative part of the solution which makes the solution incomplete.
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