If a straight line is at a distance of 10 uni | vedclass

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(D) The normal form of a line is given by $x \cos \alpha + y \sin \alpha = p$,where $p$ is the perpendicular distance from the origin and $\alpha$ is

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$x-y+10 \sqrt{2}=0$

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(D) The normal form of a line is given by $x \cos \alpha + y \sin \alpha = p$,where $p$ is the perpendicular distance from the origin and $\alpha$ is the angle the normal makes with the positive $X$-axis.Given $p = 10$.The normal makes an angle of $\frac{\pi}{4}$ with the negative $X$-axis in the negative direction (clockwise).The negative $X$-axis corresponds to an angle of $\pi$.Moving $\frac{\pi}{4}$ in the negative direction (clockwise) from $\pi$ gives $\alpha = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.Substituting these values into the normal form equation:$x \cos(\frac{3\pi}{4}) + y \sin(\frac{3\pi}{4}) = 10$$x(-\frac{1}{\sqrt{2}}) + y(\frac{1}{\sqrt{2}}) = 10$$-x + y = 10\sqrt{2}$$x - y + 10\sqrt{2} = 0$.

If a straight line is at a distance of $10$ units from the origin and the perpendicular drawn from the origin to it makes an angle of $\frac{\pi}{4}$ with the negative $X$-axis in the negative direction,then the equation of that line is:

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