Question

Find the points on the y-axis which is equidistant from points 6,5and-4,3

Answer 1

$Let \: the \: Coordinates \: of \: the \: points$

$on \:the \:Y-axis \: which \:is \: equidistant \\from \: A(6,5) = (x_{1},y_{1}) \\ and \: B(-4,3) = (x_{2},y_{2}) \:is \: P( 0, y )$

$PA = PB$

$\implies PA^{2} = PB^{2}$

$\implies ( x_{1} - 0 )^{2} + ( y_{1} - y)^{2} = ( x_{2} - 0 )^{2} + ( y_{2} - y)^{2}$

$\implies 6^{2}+(5-y)^{2} =(-4)^{2} + (3-y)^{2}$

$\implies 36 + 5^{2} - 2\times 5\times y = 16+3^{2}-2\times 3\times y$

$\implies 36 + 25- 10 y = 16+9-6 y$

$\implies 61 - 10 y = 25-6 y$

$\implies - 10 y +6y= 25-61$

$\implies - 4y = -36$

$\implies y = \frac{-36}{-4}$

$\implies y = 9$

Therefore.,

$\red{ Coordinates \: of\: required \:point } \green { = ( 0 , 9 ) }$

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