If a line L passing through a point A(2, 3) i | vedclass

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(D) Let the angle made by line $L$ with the positive $X$-axis be $	heta$. The coordinates of point $B$ can be written as $(2 + 4 \cos 	heta, 3 + 4 \

Vedclass · Accepted Answer

$\pi - \operatorname{Tan}^{-1}\left(\frac{3}{4}\right)$

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(D) Let the angle made by line $L$ with the positive $X$-axis be $	heta$. The coordinates of point $B$ can be written as $(2 + 4 \cos 	heta, 3 + 4 \sin 	heta)$.Since $B$ lies on the line $4x - 3y - 19 = 0$,we substitute these coordinates into the equation:$4(2 + 4 \cos 	heta) - 3(3 + 4 \sin 	heta) - 19 = 0$$8 + 16 \cos 	heta - 9 - 12 \sin 	heta - 19 = 0$$16 \cos 	heta - 12 \sin 	heta = 20$Dividing by $4$,we get $4 \cos 	heta - 3 \sin 	heta = 5$.This can be written as $4 \cos 	heta - 3 \sin 	heta = 5(\cos^2 	heta + \sin^2 	heta)$.Alternatively,using $R \cos(	heta + \alpha) = 5$ where $R = \sqrt{4^2 + (-3)^2} = 5$,we have $5 \cos(	heta + \alpha) = 5$,where $\cos \alpha = 4/5$ and $\sin \alpha = 3/5$.Thus,$\cos(	heta + \alpha) = 1$,which implies $	heta + \alpha = 0$ or $2\pi$.Since $	an \alpha = 3/4$,$	heta = -\alpha = -\operatorname{Tan}^{-1}(3/4)$.In the anti-clockwise direction,the angle is $2\pi - \operatorname{Tan}^{-1}(3/4)$ or simply the angle corresponding to the slope $m = 	an 	heta$. Solving $4 \cos 	heta - 3 \sin 	heta = 5$ gives $	an 	heta = -3/4$. The angle is $\pi - \operatorname{Tan}^{-1}(3/4)$.

If a line $L$ passing through a point $A(2, 3)$ intersects another line $4x - 3y - 19 = 0$ at the point $B$ such that $AB = 4$,then the angle made by the line $L$ with the positive $X$-axis in the anti-clockwise direction is

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