If the origin is shifted to (1, -2) and the a | vedclass

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Question

(B) Let the new coordinates be $(x', y')$. The transformation equations are given by:$x = h + x' \cos \alpha - y' \sin \alpha$$y = k + x' \sin \alpha

Vedclass · Accepted Answer

$\left( \frac{3}{2}, \frac{3\sqrt{3}}{2} \right)$

Vedclass · Answer

(B) Let the new coordinates be $(x', y')$. The transformation equations are given by:$x = h + x' \cos \alpha - y' \sin \alpha$$y = k + x' \sin \alpha + y' \cos \alpha$Given $(x, y) = (1, 1)$,$(h, k) = (1, -2)$,and $\alpha = 30^{\circ}$.Substituting the values:$1 = 1 + x' \cos 30^{\circ} - y' \sin 30^{\circ} \implies x' \frac{\sqrt{3}}{2} - y' \frac{1}{2} = 0 \implies x' \sqrt{3} - y' = 0$ (Equation $1$)$1 = -2 + x' \sin 30^{\circ} + y' \cos 30^{\circ} \implies 3 = x' \frac{1}{2} + y' \frac{\sqrt{3}}{2} \implies x' + y' \sqrt{3} = 6$ (Equation $2$)From Equation $1$,$y' = x' \sqrt{3}$.Substituting into Equation $2$:$x' + (x' \sqrt{3}) \sqrt{3} = 6$$x' + 3x' = 6 \implies 4x' = 6 \implies x' = \frac{3}{2}$.Then $y' = \frac{3}{2} \sqrt{3} = \frac{3\sqrt{3}}{2}$.Thus,the new coordinates are $\left( \frac{3}{2}, \frac{3\sqrt{3}}{2} \right)$.

If the origin is shifted to $(1, -2)$ and the axes are rotated by an angle of $30^{\circ}$,what will be the new coordinates of $(1, 1)$?

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