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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

2x+y=0

$$\mathrm{2}{x}+{y}=\mathrm{0} \\ $$

Answered by MJS last updated on 11/May/18

y=(1/x)  y=2x+d  2x+d=(1/x)  x^2 +(d/2)x−(1/2)=0  x_1 =−(d/4)−((√(d^2 +4))/4); y_1 =(d/2)−((√(d^2 +4))/2)  x_2 =−(d/4)+((√(d^2 +4))/4); y_2 =(d/2)+((√(d^2 +4))/2)  M= ((((x_1 +x_2 )/2)),(((y_1 +y_2 )/2)) )= (((−(d/4))),((d/2)) )

$${y}=\frac{\mathrm{1}}{{x}} \\ $$$${y}=\mathrm{2}{x}+{d} \\ $$$$\mathrm{2}{x}+{d}=\frac{\mathrm{1}}{{x}} \\ $$$${x}^{\mathrm{2}} +\frac{{d}}{\mathrm{2}}{x}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =−\frac{{d}}{\mathrm{4}}−\frac{\sqrt{{d}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{4}};\:{y}_{\mathrm{1}} =\frac{{d}}{\mathrm{2}}−\frac{\sqrt{{d}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}} \\ $$$${x}_{\mathrm{2}} =−\frac{{d}}{\mathrm{4}}+\frac{\sqrt{{d}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{4}};\:{y}_{\mathrm{2}} =\frac{{d}}{\mathrm{2}}+\frac{\sqrt{{d}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}} \\ $$$${M}=\begin{pmatrix}{\frac{{x}_{\mathrm{1}} +{x}_{\mathrm{2}} }{\mathrm{2}}}\\{\frac{{y}_{\mathrm{1}} +{y}_{\mathrm{2}} }{\mathrm{2}}}\end{pmatrix}=\begin{pmatrix}{−\frac{{d}}{\mathrm{4}}}\\{\frac{{d}}{\mathrm{2}}}\end{pmatrix} \\ $$

Commented by Tinkutara last updated on 12/May/18

Thank you very much Sir! I got the answer. ��������

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