One side of a square is inclined at an acute | vedclass

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Question

$	herefore \,\,\,$ slope of line $AC =$ $\frac{{
ot 4\sin\, \alpha  - 
ot 4\cos\, \alpha }}{{
ot 4\cos\, \alpha  + 
ot 4\sin\, \al

Vedclass · Accepted Answer

$(cos\, \alpha - sin\, \alpha) x + (cos\, \alpha + sin\, \alpha) y = 4$

Vedclass · Answer

$	herefore \,\,\,$ slope of line $AC =$ $\frac{{
ot 4\sin\, \alpha  - 
ot 4\cos\, \alpha }}{{
ot 4\cos\, \alpha  + 
ot 4\sin\, \alpha }}$

$	herefore \,\,\,$ equation of line = $\frac{{\sin\, \alpha  - \sin\, \alpha }}{{\cos\, \alpha \sin\, \alpha }}$

$y - sin\, \alpha =$ $\frac{{\sin\, \alpha  - \sin\, \alpha }}{{\cos\, \alpha \sin\, \alpha }}$ $(x - 4cos\, \alpha)$

$y\, cos\, \alpha + y\, sin\, \alpha - 4 \, sin\, \alpha &middot; cos\, \alpha = x \,sin\, \alpha - x \,cos\, \alpha - 4 \,sin^2\, \alpha - sin\, \alpha cos\, \alpha + 4\, cos^2\, \alpha$

$x (cos\, \alpha - sin\, \alpha) + y (cos\, \alpha + sin\, \alpha) = 4\Rightarrow(D)$

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