Given 5x−3y=6 . find min value of (x−1)^2 +(y+1)^2 ?


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Question Number 99003 by bramlex last updated on 18/Jun/20

$G i v e n 5 x - 3 y = 6 . f i n d m i n v a l u e$ $o f {(x - 1)}^{2} + {(y + 1)}^{2} ?$
	Answered by bobhans last updated on 18/Jun/20
	$let (x−1)^2 +(y+1)^2 =R^2 min {(x−1)^2 +(y+1)^2 } if 5x−3y = 6 is tangent line of a circle . min value = R^2 =((∣5.1−3.(−1)−6∣^2 )/(25+9)) = (4/(34)) = (2/(17)) ■$
	$let {(x - 1)}^{2} + {(y + 1)}^{2} = R^{2}$ $\min {{(x - 1)}^{2} + {(y + 1)}^{2}} if 5 x - 3 y = 6 is$ $tangent line of a circle .$ $\min value = R^{2} = \frac{∣ 5.1 - 3 . (- 1) - 6 ∣^{2}}{25 + 9}$ $= \frac{4}{34} = \frac{2}{17} ◼$
	Commented by bramlex last updated on 18/Jun/20

	$t h a n k y o u$
	Answered by 1549442205 last updated on 18/Jun/20
	$5(x−1)−3(y+1)=−2.ApplyingAM−GM we get (−2)^2 ≤(5^2 +(−3)^2 )[(x−1)^2 +(y+1)^2 ] ⇒(x−1)^2 +(y+1)^2 ≥(4/(34))=(2/(17)).The equality occurs when { (((5/(x−1))=((−3)/(y+1)))),((5x−3y=6)) :}⇔ { ((3x+5y=−2)),((5x−3y=6)) :}⇔ { ((x=((12)/(17)))),((y=((−14)/(17)))) :} Hence,S=(x−1)^2 +(y+1)^2 has the least value equal to (2/(34)) when { ((x=((12)/(17)))),((y=((−14)/(17)))) :}$
	$5 (x - 1) - 3 (y + 1) = - 2 . ApplyingAM - GM$ $we get {(- 2)}^{2} ⩽ (5^{2} + {(- 3)}^{2}) [{(x - 1)}^{2} + {(y + 1)}^{2}]$ $\Rightarrow {(x - 1)}^{2} + {(y + 1)}^{2} ⩾ \frac{4}{34} = \frac{2}{17} . The equality occurs$ $when {\begin{cases} \frac{5}{x - 1} = \frac{- 3}{y + 1} \\ 5 x - 3 y = 6 \end{cases} \Leftrightarrow {\begin{cases} 3 x + 5 y = - 2 \\ 5 x - 3 y = 6 \end{cases} \Leftrightarrow {\begin{cases} x = \frac{12}{17} \\ y = \frac{- 14}{17} \end{cases}$ $Hence, S = {(x - 1)}^{2} + {(y + 1)}^{2} has the least value equal to$ $\frac{2}{34} when {\begin{cases} x = \frac{12}{17} \\ y = \frac{- 14}{17} \end{cases}$
	Commented by bemath last updated on 18/Jun/20

	$the last line \frac{2}{34} or \frac{2}{17} or \frac{4}{34} ?$
	Commented by 1549442205 last updated on 18/Jun/20

	$Excuse me, thank sir you, I wrote a mistake$ $it is \frac{2}{17}$
	Answered by MJS last updated on 18/Jun/20

	$why you folks do such complicated things ?$ $5 x - 3 y = 6 \Rightarrow y = \frac{5}{3} x - 2$ ${(x - 1)}^{2} + {(y + 1)}^{2} = \frac{34}{9} x^{2} - \frac{16}{3} x + 2$ $\frac{d}{d x} [\frac{34}{9} x^{2} - \frac{16}{3} x + 2] = 0$ $\frac{68}{9} x - \frac{16}{3} = 0$ $x = \frac{12}{17}$ $\frac{d}{d x} [\frac{68}{9} x - \frac{16}{3}] > 0 \Rightarrow minimum$ $\frac{34}{9} x^{2} - \frac{16}{3} x + 2 = \frac{2}{17}$
	Commented by bramlex last updated on 18/Jun/20

	$t h a n k y o u$
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