Question Number 164417 by HongKing last updated on 16/Jan/22
Answered by Rasheed.Sindhi last updated on 17/Jan/22
![y=4x−3 (√(x^2 +y^2 +6y+9)) +(√(x^2 +y^2 −2x−2y+2)) =(√(x^2 +(y+3)^2 ))+(√(x^2 −2x+1+y^2 −2y+1)) =(√(x^2 +(y+3)^2 ))+(√((x−1)^2 +(y−1)^2 )) =(√(x^2 +(4x−3+3)^2 ))+(√((x−1)^2 +(4x−3−1)^2 )) =(√(x^2 +(4x)^2 ))+(√((x−1)^2 +4(x−1)^2 )) =∣x∣(√(17)) +∣x−1∣(√5) =x(√(17)) +(1−x)(√5) [∵ x∈(0,1)] =((√(17)) −(√5) )x+(√5) =((√(17)) −(√5) )(0,1)+(√5) =(0,(√(17)) −(√5) )+(√5) =((√5) ,(√(17)) )](https://www.tinkutara.com/question/Q164435.png)
$${y}=\mathrm{4}{x}−\mathrm{3} \\ $$$$\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{6}{y}+\mathrm{9}}\:+\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{2}{y}+\mathrm{2}}\: \\ $$$$=\sqrt{{x}^{\mathrm{2}} +\left({y}+\mathrm{3}\right)^{\mathrm{2}} }+\sqrt{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}+{y}^{\mathrm{2}} −\mathrm{2}{y}+\mathrm{1}} \\ $$$$=\sqrt{{x}^{\mathrm{2}} +\left({y}+\mathrm{3}\right)^{\mathrm{2}} }+\sqrt{\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\sqrt{{x}^{\mathrm{2}} +\left(\mathrm{4}{x}−\mathrm{3}+\mathrm{3}\right)^{\mathrm{2}} }+\sqrt{\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{4}{x}−\mathrm{3}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\sqrt{{x}^{\mathrm{2}} +\left(\mathrm{4}{x}\right)^{\mathrm{2}} }+\sqrt{\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{4}\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\: \\ $$$$=\mid{x}\mid\sqrt{\mathrm{17}}\:+\mid{x}−\mathrm{1}\mid\sqrt{\mathrm{5}}\: \\ $$$$={x}\sqrt{\mathrm{17}}\:+\left(\mathrm{1}−{x}\right)\sqrt{\mathrm{5}}\:\:\left[\because\:{x}\in\left(\mathrm{0},\mathrm{1}\right)\right] \\ $$$$=\left(\sqrt{\mathrm{17}}\:−\sqrt{\mathrm{5}}\:\right){x}+\sqrt{\mathrm{5}}\: \\ $$$$=\left(\sqrt{\mathrm{17}}\:−\sqrt{\mathrm{5}}\:\right)\left(\mathrm{0},\mathrm{1}\right)+\sqrt{\mathrm{5}}\: \\ $$$$=\left(\mathrm{0},\sqrt{\mathrm{17}}\:−\sqrt{\mathrm{5}}\:\right)+\sqrt{\mathrm{5}}\: \\ $$$$=\left(\sqrt{\mathrm{5}}\:,\sqrt{\mathrm{17}}\:\right) \\ $$
Commented by Rasheed.Sindhi last updated on 17/Jan/22
$$\mathrm{Thanks}\:\mathrm{to}\:\mathrm{guide}\:\mathrm{me}\:\boldsymbol{\mathrm{mr}}\:\boldsymbol{\mathrm{W}}\:\boldsymbol{\mathrm{sir}}!\:\mathrm{I}'\mathrm{ve} \\ $$$$\mathrm{corrected}\:\mathrm{now}! \\ $$
Commented by HongKing last updated on 18/Jan/22
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{cool} \\ $$