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Question Number 33218 by mondodotto@gmail.com last updated on 13/Apr/18

$$\mathbf{find}\mathbf{the}\mathbf{angle}\mathbf{bisector}$$$$3{\mathit{x}}^2+4\mathit{x}\mathit{y}-5{\mathit{y}}^2=0$$

Answered by MJS last updated on 13/Apr/18

$$y_1=-\frac{2+\sqrt{19}}{5}x$$$$y_2=\frac{2+\sqrt{19}}{5}x$$$$\mathrm{bisectors}\mathrm{are}y=0(=x-\mathrm{axis})\mathrm{and}$$$$x=0(=y-\mathrm{axis})$$

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