P(6,4) is a point on the line x-y-2=0. If A(a | vedclass

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(A) The line is $x-y-2=0$,which can be written as $x-y-2=0$. The slope $m$ of this line is $1$,so $	an 	heta = 1$,which implies $	heta = 45^{\circ}

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$136$

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(A) The line is $x-y-2=0$,which can be written as $x-y-2=0$. The slope $m$ of this line is $1$,so $	an 	heta = 1$,which implies $	heta = 45^{\circ}$.Thus,$\cos 	heta = \frac{1}{\sqrt{2}}$ and $\sin 	heta = \frac{1}{\sqrt{2}}$.The coordinates of points at a distance $r=4$ from $P(6,4)$ are given by $(x \pm r \cos 	heta, y \pm r \sin 	heta)$.For $A$,we have $(\alpha, \beta) = (6 + 4 \cdot \frac{1}{\sqrt{2}}, 4 + 4 \cdot \frac{1}{\sqrt{2}}) = (6 + 2\sqrt{2}, 4 + 2\sqrt{2})$.For $B$,we have $(\gamma, \delta) = (6 - 4 \cdot \frac{1}{\sqrt{2}}, 4 - 4 \cdot \frac{1}{\sqrt{2}}) = (6 - 2\sqrt{2}, 4 - 2\sqrt{2})$.Now,$\alpha^2 + \beta^2 = (6 + 2\sqrt{2})^2 + (4 + 2\sqrt{2})^2 = (36 + 8 + 24\sqrt{2}) + (16 + 8 + 16\sqrt{2}) = 68 + 40\sqrt{2}$.Similarly,$\gamma^2 + \delta^2 = (6 - 2\sqrt{2})^2 + (4 - 2\sqrt{2})^2 = (36 + 8 - 24\sqrt{2}) + (16 + 8 - 16\sqrt{2}) = 68 - 40\sqrt{2}$.Adding these,$\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = (68 + 40\sqrt{2}) + (68 - 40\sqrt{2}) = 136$.

$P(6,4)$ is a point on the line $x-y-2=0$. If $A(\alpha, \beta)$ and $B(\gamma, \delta)$ are two points on this line lying on either side of $P$ at a distance of $4$ units from $P$,then $\alpha^2+\beta^2+\gamma^2+\delta^2=$

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