The perpendiculars are drawn to lines L_1 and | vedclass

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

Vedclass · Accepted Answer

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

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(C) The normal form of the equation of a line is $x \cos \alpha + y \sin \alpha = p$,where $p$ is the perpendicular distance from the origin and $\alpha$ is the angle the perpendicular makes with the positive $X$-axis.For line $L_1$,$\alpha = \frac{\pi}{4}$ and $p = 1$:$x \cos \frac{\pi}{4} + y \sin \frac{\pi}{4} = 1$$\Rightarrow \frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = 1$$\Rightarrow x + y - \sqrt{2} = 0$For line $L_2$,$\alpha = \frac{3 \pi}{4}$ and $p = 1$:$x \cos \frac{3 \pi}{4} + y \sin \frac{3 \pi}{4} = 1$$\Rightarrow -\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = 1$$\Rightarrow -x + y - \sqrt{2} = 0$$\Rightarrow x - y + \sqrt{2} = 0$The joint equation is the product of the two individual equations:$(x + y - \sqrt{2})(x - y + \sqrt{2}) = 0$$x(x - y + \sqrt{2}) + y(x - y + \sqrt{2}) - \sqrt{2}(x - y + \sqrt{2}) = 0$$x^2 - xy + \sqrt{2}x + xy - y^2 + \sqrt{2}y - \sqrt{2}x + \sqrt{2}y - 2 = 0$$x^2 - y^2 + 2\sqrt{2}y - 2 = 0$

The perpendiculars are drawn to lines $L_1$ and $L_2$ from the origin making an angle $\frac{\pi}{4}$ and $\frac{3 \pi}{4}$ respectively with the positive direction of the $X$-axis. If both the lines are at a unit distance from the origin,then their joint equation is

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