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Question Number 67026 by TawaTawa last updated on 21/Aug/19

Commented by Tony Lin last updated on 22/Aug/19

$$\because slopeof{L}_{BC}=tan150°=-\frac{\sqrt{3}}{3}$$$$\therefore L_{BC}:y=-\frac{\sqrt{3}}{3}(x+4)$$$$\because L_{BC}\mathrm{\perp }{L}_{AB}$$$$\therefore L_{AB}:y=\sqrt{3}x$$$$L_{AC}:x=2$$$$letycoordinateofB-y$$$$\sqrt{3}y+\frac{y}{\sqrt{3}}=4$$$$y=-\sqrt{3}\Rightarrow x=-\frac{y}{\sqrt{3}}=-1$$$$\Rightarrow B(-1,-\sqrt{3})$$$$Aistheintersectionof{L}_{AB}\mathrm{\&}{L}_{AC}$$$$\Rightarrow A(2,2\sqrt{3})$$$$Cistheintersectionof{L}_{BC}\mathrm{\&}{L}_{AC}$$$$\Rightarrow C(2,-2\sqrt{3})$$$$theareaof△ABC$$$$=\frac{1}{2}\mid \left|\begin{array}{cccc}-1& 2& 2& -1\\ -\sqrt{3}& 2\sqrt{3}& -2\sqrt{3}& -\sqrt{3}\end{array}\right|\mid$$$$=6\sqrt{3}$$

Commented by TawaTawa last updated on 22/Aug/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}.$$

Commented by TawaTawa last updated on 22/Aug/19

$$\mathrm{please}\mathrm{sir}.\mathrm{how}\mathrm{is}{\mathrm{L}}_{\mathrm{AB}}:\mathrm{y}=\sqrt{3}\mathrm{x}$$$$\mathrm{And}\mathrm{from}\mathrm{let}\mathrm{y}\mathrm{coordinate}\mathrm{of}\mathrm{B}=-\mathrm{y}$$$$\mathrm{how}\mathrm{is},\sqrt{3}\mathrm{y}+\frac{\mathrm{y}}{\sqrt{3}}=4$$$$\mathrm{And}\mathrm{some}\mathrm{intersection}\mathrm{you}\mathrm{got}\mathrm{sir}.$$$$\mathrm{When}\mathrm{you}\mathrm{are}\mathrm{chanced}\mathrm{sir},\mathrm{please}\mathrm{help}\mathrm{me}\mathrm{explain}\mathrm{little}\mathrm{steps}$$

Commented by Tony Lin last updated on 22/Aug/19

Commented by TawaTawa last updated on 22/Aug/19

$$\mathrm{I}\mathrm{really}\mathrm{appreciate}\mathrm{sir}$$

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