Question

Hey guys plz give answer of this question...

Find a relation between x and y such that the point ( x,y) is equidistant from the points ( 7, 1 ) and ( 3, 5).

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Answer 1

GivEn point (x,y) is equidistant from the points ( 7, 1 ) and ( 3, 5).

We have to find, relation between x and y.

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☯ Let it be P(x,y) , A(7,1) and B(3,5).

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⋆ Reference of image is shown in diagram

$\setlength{\unitlength}{1.6mm}\begin{picture}(50,20)\linethickness{0.1mm}\put(-3,-3){\line(1,1){20}}\put(36.6,-2.8){\line(-1,1){19.8}}\put(-3,-3){\line(1,0){39.5}}\put(15.5,17.5){P}\put(17.3,17.3){(x,y)}\put(-4,-5){A}\put(-8.4,-3.8){(7,1)}\put(35,-5){B}\put(37,-3){(3,5)}\end{picture}$

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Here,

★ PA = PB

$\therefore\sf PA^2 = PB^2$

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${\underline{\sf{\bigstar\;By\;Using\; Distance\;Formula\;:}}}$

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$\star\;{\boxed{\sf{\pink{ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}}}}$

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${\underline{\sf{\bigstar\;Putting\; values\;:}}}$

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$:\implies\sf \sqrt{(x - 7)^2 + (y - 1)^2} = \sqrt{x - 3)^2 + (y - 5)^2}$

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$:\implies\sf (x - 7)^2 + (y - 1)^2 = (x - 3)^2 + (y - 5)^2$

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$:\implies\sf x^2 - 14x + 49 + y^2 - 2y + 1 = x^2 - 6x + 9 + y^2 - 10y + 25$

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$\;\;\;\;:\implies\sf x^2 - 14x + y^2 - 2y + 50 = x^2 - 6x + y^2 - 10y + 34$

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$\;\;\;:\implies\sf \cancel{x^2} - 14x + \cancel{y^2} - 2y + 50 = \cancel{x^2} - 6x + \cancel{y^2} - 10y + 34$

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$:\implies\sf - 14x - 2y + 50 = - 6x - 10y + 34$

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$:\implies\sf - 14x + 6x + 50 - 34 = - 10y + 2y$

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$:\implies\sf - 8x + 16 = - 8y$

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$:\implies\sf - 8x + 8y + 16 = 0$

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$:\implies\sf 8( - x + y + 2) = 0$

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$:\implies\sf - x + y + 2 = 0$

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$:\implies{\underline{\boxed{\bf{\pink{x - y = 2}}}}}\;\bigstar$

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$\therefore$ The relation between x and y is (x - y = 2).