For all values of theta,the line (2 cos theta | vedclass

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(B) The given equation is $(2 \cos 	heta + 3 \sin 	heta) x + (3 \cos 	heta - 5 \sin 	heta) y - (5 \cos 	heta - 2 \sin 	heta) = 0$. Rearranging t

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$(1, 1)$

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(B) The given equation is $(2 \cos 	heta + 3 \sin 	heta) x + (3 \cos 	heta - 5 \sin 	heta) y - (5 \cos 	heta - 2 \sin 	heta) = 0$. Rearranging the terms by grouping $\cos 	heta$ and $\sin 	heta$: $\cos 	heta (2x + 3y - 5) + \sin 	heta (3x - 5y + 2) = 0$. For this equation to hold for all values of $	heta$,the coefficients of $\cos 	heta$ and $\sin 	heta$ must be zero independently: $2x + 3y - 5 = 0$ (Equation $1$) $3x - 5y + 2 = 0$ (Equation $2$) Multiplying Equation $1$ by $5$ and Equation $2$ by $3$: $10x + 15y - 25 = 0$ $9x - 15y + 6 = 0$ Adding these two equations: $19x - 19 = 0 \implies x = 1$. Substituting $x = 1$ into Equation $1$: $2(1) + 3y - 5 = 0 \implies 3y = 3 \implies y = 1$. Thus,the fixed point is $(1, 1)$.

For all values of $\theta$,the line $(2 \cos \theta + 3 \sin \theta) x + (3 \cos \theta - 5 \sin \theta) y - (5 \cos \theta - 2 \sin \theta) = 0$ passes through which fixed point?

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