Answer
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Hint: We have to use the formula for the equation of the line in terms of the perpendicular distance from the origin and its slope. We need to first find the angle of the line joining the origin and the point on the line from which the equation of the line can be obtained.
Complete step-by-step answer:
We are given the perpendicular distance of the line from the origin and its slope. The equation of the line at a perpendicular distance p from the origin and where the perpendicular line from the origin to the line makes an angle $\theta $ with the x-axis is given by
$x\cos \left( \theta \right)+y\sin \left( \theta \right)=p\text{ }........\text{(1}\text{.1)}$
As shown in the figure below:
Now, as the line joining the origin and the line should be perpendicular to the original line, their slopes should satisfy
${{m}_{1}}\times {{m}_{2}}=-1\text{ }...........\text{(1}\text{.2)}$
where ${{m}_{1}}$ is the slope of the line joining the line from origin and ${{m}_{2}}$ is the slope of the original line (which is to be found out in the question).
We are given that slope of the original line = ${{m}_{2}}=-1$
Thus, by using this this value in equation (1.2), we obtain
${{m}_{1}}\times -1=-1\Rightarrow {{m}_{1}}\text{=1}...........\text{(1}\text{.3)}$
As ${{m}_{1}}$ is the slope of the line joining the origin and the original line, as as slope is the tangent of the angle with the x-axis, we should have
$\tan (\theta )={{m}_{1}}\Rightarrow \theta ={{\tan }^{-1}}({{m}_{1}})={{\tan }^{-1}}(1)={{45}^{\circ }}$
Using this value of $\theta $ in equation (1.1), we get
$\begin{align}
& x\cos \left( {{45}^{\circ }} \right)+y\sin \left( {{45}^{\circ }} \right)=5\text{ } \\
& \Rightarrow \dfrac{x}{\sqrt{2}}+\dfrac{y}{\sqrt{2}}=5\Rightarrow x+y=5\sqrt{2} \\
\end{align}$
Thus, the required equation of the line is $x+y=5\sqrt{2}$.
Note: We should be careful as $\theta $ represents the slope of the line joining the origin to the given line and not the slope of the given line itself. We should be careful to take the ${{\tan }^{-1}}$ of the slope to find the angle and then use it in the equation (1.1).
Complete step-by-step answer:
We are given the perpendicular distance of the line from the origin and its slope. The equation of the line at a perpendicular distance p from the origin and where the perpendicular line from the origin to the line makes an angle $\theta $ with the x-axis is given by
$x\cos \left( \theta \right)+y\sin \left( \theta \right)=p\text{ }........\text{(1}\text{.1)}$
As shown in the figure below:
Now, as the line joining the origin and the line should be perpendicular to the original line, their slopes should satisfy
${{m}_{1}}\times {{m}_{2}}=-1\text{ }...........\text{(1}\text{.2)}$
where ${{m}_{1}}$ is the slope of the line joining the line from origin and ${{m}_{2}}$ is the slope of the original line (which is to be found out in the question).
We are given that slope of the original line = ${{m}_{2}}=-1$
Thus, by using this this value in equation (1.2), we obtain
${{m}_{1}}\times -1=-1\Rightarrow {{m}_{1}}\text{=1}...........\text{(1}\text{.3)}$
As ${{m}_{1}}$ is the slope of the line joining the origin and the original line, as as slope is the tangent of the angle with the x-axis, we should have
$\tan (\theta )={{m}_{1}}\Rightarrow \theta ={{\tan }^{-1}}({{m}_{1}})={{\tan }^{-1}}(1)={{45}^{\circ }}$
Using this value of $\theta $ in equation (1.1), we get
$\begin{align}
& x\cos \left( {{45}^{\circ }} \right)+y\sin \left( {{45}^{\circ }} \right)=5\text{ } \\
& \Rightarrow \dfrac{x}{\sqrt{2}}+\dfrac{y}{\sqrt{2}}=5\Rightarrow x+y=5\sqrt{2} \\
\end{align}$
Thus, the required equation of the line is $x+y=5\sqrt{2}$.
Note: We should be careful as $\theta $ represents the slope of the line joining the origin to the given line and not the slope of the given line itself. We should be careful to take the ${{\tan }^{-1}}$ of the slope to find the angle and then use it in the equation (1.1).
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