If the axes are rotated through an angle 45^{ | vedclass

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Question

(B) Let $(x, y)$ be the coordinates in the old system and $(X, Y)$ be the coordinates in the new system after rotation by an angle $	heta = 45^{\circ

Vedclass · Accepted Answer

$(-2, -10)$

Vedclass · Answer

(B) Let $(x, y)$ be the coordinates in the old system and $(X, Y)$ be the coordinates in the new system after rotation by an angle $	heta = 45^{\circ}$.The transformation equations are:$x = X \cos 	heta - Y \sin 	heta$$y = X \sin 	heta + Y \cos 	heta$Given $x = 4 \sqrt{2}$,$y = -6 \sqrt{2}$,and $	heta = 45^{\circ}$,we have $\cos 45^{\circ} = \sin 45^{\circ} = \frac{1}{\sqrt{2}}$.Substituting these values:$4 \sqrt{2} = X \left(\frac{1}{\sqrt{2}}ight) - Y \left(\frac{1}{\sqrt{2}}ight) \Rightarrow X - Y = 8$$-6 \sqrt{2} = X \left(\frac{1}{\sqrt{2}}ight) + Y \left(\frac{1}{\sqrt{2}}ight) \Rightarrow X + Y = -12$Adding the two equations: $2X = -4 \Rightarrow X = -2$.Subtracting the first from the second: $2Y = -20 \Rightarrow Y = -10$.Thus,the new coordinates are $(-2, -10)$.

If the axes are rotated through an angle $45^{\circ}$,then the coordinates of the point $(4 \sqrt{2}, -6 \sqrt{2})$ in the new system are . . . . . .

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