A straight line L at a distance of 4 units fr | vedclass

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Question

(A) The normal form of a line is $x \cos \alpha + y \sin \alpha = p$,where $p = 4$ is the perpendicular distance from the origin and $\alpha$ is the a

Vedclass · Accepted Answer

$\left( {\sqrt 3 - 1} \right)x + \left( {\sqrt 3 + 1} \right)y = 8\sqrt 2 $

Vedclass · Answer

(A) The normal form of a line is $x \cos \alpha + y \sin \alpha = p$,where $p = 4$ is the perpendicular distance from the origin and $\alpha$ is the angle the normal makes with the positive $x$-axis.The line $x + y = 0$ has a slope of $-1$,which corresponds to an angle of $135^o$ with the positive $x$-axis.The perpendicular from the origin to line $L$ makes an angle of $60^o$ with $x + y = 0$. Thus,$\alpha = 135^o \pm 60^o$.Case $1$: $\alpha = 135^o - 60^o = 75^o$.$\cos 75^o = \cos(45^o + 30^o) = \frac{\sqrt 3 - 1}{2\sqrt 2}$ and $\sin 75^o = \sin(45^o + 30^o) = \frac{\sqrt 3 + 1}{2\sqrt 2}$.The equation is $x \left( \frac{\sqrt 3 - 1}{2\sqrt 2} \right) + y \left( \frac{\sqrt 3 + 1}{2\sqrt 2} \right) = 4$,which simplifies to $(\sqrt 3 - 1)x + (\sqrt 3 + 1)y = 8\sqrt 2$.Case $2$: $\alpha = 135^o + 60^o = 195^o$. Since the line makes positive intercepts,$\alpha$ must be in the first quadrant $(0^o Comparing with the options,the correct equation is $(\sqrt 3 - 1)x + (\sqrt 3 + 1)y = 8\sqrt 2$.

$A$ straight line $L$ at a distance of $4$ units from the origin makes positive intercepts on the coordinate axes and the perpendicular from the origin to this line makes an angle of $60^o$ with the line $x + y = 0$. Then an equation of the line $L$ is

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