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If the lines $y=3 x+1$ and $2 y=x+3$ are equally inclined to the line $y=m x+4$, find the value of $m$.
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Verified Answer
Slope of the line $P Q: y=3 x+1$ is 3 .
The angle $\theta$ between $P Q$ and $Q R$ is given by
$\tan \theta=\left|\frac{\mathrm{m}-3}{1+3 \mathrm{~m}}\right|$
Slope of the line $P R: 2 y=x+3$ is $\frac{1}{2}$ and slope of $Q R$ is $m$ The angle between $P R$ and $Q R$ is given by
$\begin{aligned}
&\tan \theta=\left|\frac{m-\frac{1}{2}}{1+\frac{1}{2} \cdot m}\right|=\pm \frac{2 m-1}{m+2} \\
&\Rightarrow \quad \frac{m-3}{1+3 m}=\pm \frac{2 m-1}{m+2}
\end{aligned}$
For +ve sign, $\frac{m-3}{1+3 m}=\frac{2 m-1}{m+2} \Rightarrow m^2=-1$
For -ve sign, $\frac{m-3}{1+3 m}=-\frac{2 m-1}{m+2}$
$\Rightarrow m=\frac{1 \pm 5 \sqrt{2}}{7}$
The angle $\theta$ between $P Q$ and $Q R$ is given by
$\tan \theta=\left|\frac{\mathrm{m}-3}{1+3 \mathrm{~m}}\right|$
Slope of the line $P R: 2 y=x+3$ is $\frac{1}{2}$ and slope of $Q R$ is $m$ The angle between $P R$ and $Q R$ is given by
$\begin{aligned}
&\tan \theta=\left|\frac{m-\frac{1}{2}}{1+\frac{1}{2} \cdot m}\right|=\pm \frac{2 m-1}{m+2} \\
&\Rightarrow \quad \frac{m-3}{1+3 m}=\pm \frac{2 m-1}{m+2}
\end{aligned}$
For +ve sign, $\frac{m-3}{1+3 m}=\frac{2 m-1}{m+2} \Rightarrow m^2=-1$
For -ve sign, $\frac{m-3}{1+3 m}=-\frac{2 m-1}{m+2}$
$\Rightarrow m=\frac{1 \pm 5 \sqrt{2}}{7}$
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