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Question Number 164417 by HongKing last updated on 16/Jan/22

Answered by Rasheed.Sindhi last updated on 17/Jan/22

$$y=4x-3$$$$\sqrt{x^2+y^2+6y+9}+\sqrt{x^2+y^2-2x-2y+2}$$$$=\sqrt{x^2+{(y+3)}^2}+\sqrt{x^2-2x+1+y^2-2y+1}$$$$=\sqrt{x^2+{(y+3)}^2}+\sqrt{{(x-1)}^2+{(y-1)}^2}$$$$=\sqrt{x^2+{(4x-3+3)}^2}+\sqrt{{(x-1)}^2+{(4x-3-1)}^2}$$$$=\sqrt{x^2+{\left(4x\right)}^2}+\sqrt{{(x-1)}^2+4{(x-1)}^2}$$$$=\mid x\mid \sqrt{17}+\mid x-1\mid \sqrt{5}$$$$=x\sqrt{17}+(1-x)\sqrt{5}[\because x\in (0,1)]$$$$=(\sqrt{17}-\sqrt{5})x+\sqrt{5}$$$$=(\sqrt{17}-\sqrt{5})(0,1)+\sqrt{5}$$$$=(0,\sqrt{17}-\sqrt{5})+\sqrt{5}$$$$=(\sqrt{5},\sqrt{17})$$

Commented by Rasheed.Sindhi last updated on 17/Jan/22

$$\mathrm{Thanks}\mathrm{to}\mathrm{guide}\mathrm{me}\mathbf{mr}\mathbf{W}\mathbf{sir}!{\mathrm{I}}^{\prime }\mathrm{ve}$$$$\mathrm{corrected}\mathrm{now}!$$

Commented by HongKing last updated on 18/Jan/22

$$\mathrm{thank}\mathrm{you}\mathrm{dear}\mathrm{Sir}\mathrm{cool}$$

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