Question Number 33218 by mondodotto@gmail.com last updated on 13/Apr/18
$$\:\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{angle}}\:\boldsymbol{\mathrm{bisector}} \\ $$$$\:\mathrm{3}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{{xy}}−\mathrm{5}\boldsymbol{{y}}^{\mathrm{2}} =\mathrm{0} \\ $$
Answered by MJS last updated on 13/Apr/18
$${y}_{\mathrm{1}} =−\frac{\mathrm{2}+\sqrt{\mathrm{19}}}{\mathrm{5}}{x} \\ $$$${y}_{\mathrm{2}} =\frac{\mathrm{2}+\sqrt{\mathrm{19}}}{\mathrm{5}}{x} \\ $$$$\mathrm{bisectors}\:\mathrm{are}\:{y}=\mathrm{0}\:\left(={x}−\mathrm{axis}\right)\:\mathrm{and} \\ $$$${x}=\mathrm{0}\:\left(={y}−\mathrm{axis}\right) \\ $$