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Question Number 57212 by Tinkutara last updated on 31/Mar/19

Answered by ajfour last updated on 31/Mar/19

$$\mathrm{a}-2\sqrt{\mathrm{bc}}=\mathrm{b}+\mathrm{c}$$$$\Rightarrow \mathrm{c}+2\sqrt{\mathrm{b}}\sqrt{\mathrm{c}}+(\mathrm{b}-\mathrm{a})=0$$$$\Rightarrow \sqrt{\mathrm{c}}=\frac{-2\sqrt{\mathrm{b}}\pm \sqrt{4\mathrm{b}-4(\mathrm{b}-\mathrm{a})}}{2}$$$$\Rightarrow \sqrt{\mathrm{c}}=-\sqrt{\mathrm{b}}\pm \sqrt{\mathrm{a}}$$$$\mathrm{but}\mathrm{since}\sqrt{\mathrm{c}}>0\mathrm{so}\mathrm{we}\mathrm{just}\mathrm{consider}$$$$\sqrt{\mathrm{c}}=-\sqrt{\mathrm{b}}+\sqrt{\mathrm{a}}$$$$\Rightarrow -\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}+\sqrt{\mathrm{c}}=0$$$$\mathrm{while}\sqrt{\mathrm{a}}\mathrm{x}+\sqrt{\mathrm{b}}\mathrm{y}+\sqrt{\mathrm{c}}=0$$$$\Rightarrow \mathrm{fixed}\mathrm{point}\mathrm{through}\mathrm{which}\mathrm{the}$$$$\mathrm{family}\mathrm{of}\mathrm{lines}\mathrm{pass}\mathrm{is}(-1,1).$$

Commented by Tinkutara last updated on 31/Mar/19

Isn't the root over x and y also?

Commented by ajfour last updated on 31/Mar/19

$$\mathrm{how}\mathrm{will}\mathrm{the}\mathrm{equation}\mathrm{then}\mathrm{be}\mathrm{for}$$$$\mathrm{family}\mathrm{of}\mathrm{lines}?!$$

Commented by Tinkutara last updated on 01/Apr/19

Yes Sir, Thanks a lot!

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