Question Number 34803 by Tinkutara last updated on 11/May/18
Commented by ajfour last updated on 11/May/18
Commented by ajfour last updated on 11/May/18
$$\mathrm{2}{x}+{y}=\mathrm{0} \\ $$
Answered by MJS last updated on 11/May/18
$${y}=\frac{\mathrm{1}}{{x}} \\ $$$${y}=\mathrm{2}{x}+{d} \\ $$$$\mathrm{2}{x}+{d}=\frac{\mathrm{1}}{{x}} \\ $$$${x}^{\mathrm{2}} +\frac{{d}}{\mathrm{2}}{x}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =−\frac{{d}}{\mathrm{4}}−\frac{\sqrt{{d}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{4}};\:{y}_{\mathrm{1}} =\frac{{d}}{\mathrm{2}}−\frac{\sqrt{{d}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}} \\ $$$${x}_{\mathrm{2}} =−\frac{{d}}{\mathrm{4}}+\frac{\sqrt{{d}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{4}};\:{y}_{\mathrm{2}} =\frac{{d}}{\mathrm{2}}+\frac{\sqrt{{d}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}} \\ $$$${M}=\begin{pmatrix}{\frac{{x}_{\mathrm{1}} +{x}_{\mathrm{2}} }{\mathrm{2}}}\\{\frac{{y}_{\mathrm{1}} +{y}_{\mathrm{2}} }{\mathrm{2}}}\end{pmatrix}=\begin{pmatrix}{−\frac{{d}}{\mathrm{4}}}\\{\frac{{d}}{\mathrm{2}}}\end{pmatrix} \\ $$
Commented by Tinkutara last updated on 12/May/18
Thank you very much Sir! I got the answer. ��������