The point P(alpha, beta) with alpha 0, beta | vedclass

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Question

(A) Let the initial point be $P_0 = (\alpha, \beta)$.Step $1$: Translation by $3$ units in the positive $x$-direction gives $P_1 = (\alpha + 3, \beta)

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) Let the initial point be $P_0 = (\alpha, \beta)$.Step $1$: Translation by $3$ units in the positive $x$-direction gives $P_1 = (\alpha + 3, \beta)$.Step $2$: Reflection about $y = -x$ maps $(x, y)$ to $(-y, -x)$. Thus,$P_2 = (-\beta, -(\alpha + 3)) = (-\beta, -\alpha - 3)$.Step $3$: Rotation of axes by $	heta = \frac{\pi}{4}$ in the positive direction. The new coordinates $(x', y')$ are related to old coordinates $(x, y)$ by $x = x' \cos 	heta - y' \sin 	heta$ and $y = x' \sin 	heta + y' \cos 	heta$. Given $(x', y') = (-4\sqrt{2}, -2\sqrt{2})$,we have:$x = (-4\sqrt{2}) \cos \frac{\pi}{4} - (-2\sqrt{2}) \sin \frac{\pi}{4} = (-4\sqrt{2}) \cdot \frac{1}{\sqrt{2}} + (2\sqrt{2}) \cdot \frac{1}{\sqrt{2}} = -4 + 2 = -2$.$y = (-4\sqrt{2}) \sin \frac{\pi}{4} + (-2\sqrt{2}) \cos \frac{\pi}{4} = (-4\sqrt{2}) \cdot \frac{1}{\sqrt{2}} - (2\sqrt{2}) \cdot \frac{1}{\sqrt{2}} = -4 - 2 = -6$.Equating $P_2 = (x, y)$,we get $-\beta = -2 \implies \beta = 2$ and $-\alpha - 3 = -6 \implies \alpha = 3$.Thus,$\alpha + \beta = 3 + 2 = 5$.

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