A line L passes through the point P(1, 2) and | vedclass

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(A) The line $L$ passes through $P(1, 2)$ with an angle of inclination $	heta = 60^{\circ}$.The coordinates of points $A$ and $B$ at a distance $r =

Vedclass · Accepted Answer

$4-2\sqrt{3}$

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(A) The line $L$ passes through $P(1, 2)$ with an angle of inclination $	heta = 60^{\circ}$.The coordinates of points $A$ and $B$ at a distance $r = 4$ from $P$ are given by $(x_1 + r \cos 	heta, y_1 + r \sin 	heta)$ and $(x_1 - r \cos 	heta, y_1 - r \sin 	heta)$.$A = (1 + 4 \cos 60^{\circ}, 2 + 4 \sin 60^{\circ}) = (1 + 4(1/2), 2 + 4(\sqrt{3}/2)) = (3, 2 + 2\sqrt{3})$.$B = (1 - 4 \cos 60^{\circ}, 2 - 4 \sin 60^{\circ}) = (1 - 4(1/2), 2 - 4(\sqrt{3}/2)) = (-1, 2 - 2\sqrt{3})$.The area of $	riangle OAB$ with vertices $O(0, 0)$,$A(x_A, y_A)$,and $B(x_B, y_B)$ is given by $\frac{1}{2} |x_A y_B - x_B y_A|$.Area $= \frac{1}{2} |(3)(2 - 2\sqrt{3}) - (-1)(2 + 2\sqrt{3})|$.Area $= \frac{1}{2} |6 - 6\sqrt{3} + 2 + 2\sqrt{3}| = \frac{1}{2} |8 - 4\sqrt{3}| = 4 - 2\sqrt{3}$.

$A$ line $L$ passes through the point $P(1, 2)$ and makes an angle of $60^{\circ}$ with the positive $X$-axis. $A$ and $B$ are two points lying on $L$ at a distance of $4$ units from $P$. If $O$ is the origin,then the area of $\triangle OAB$ is

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