Question

find the point only on y axis which is equidistant from (1,2) and (3,4)

Answer 1

Answer:

Let the coordinates be 0,y

(0-1)^2+(y-2)^2=(0-3)^2+(y-4)^2

1+y^2-4y+4=9+y^2-8y+16

4y=20

y=5

So coordinates are 0,5

Answer 2

Answer:

hey here is your solution

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Step-by-step explanation:

$so \: we \: know \: that \\ whenever \: there \: is \: a \: point \: on \: y \: axis \: is \: x \: coordinate \: is \: 0 \\ so \: let \: the \: point \: be \: P \: such \: that \\ P = (x,y) = (0,y) \\ \\ moreover \: let \: here \: \\ A = (1,2) = (x1,y1) \\ B = (3 ,4) = (x2,y2) \\ now \: here \: \\ AP = PB \\ squaring \: both \: sides \\ we \: get \\ AP {}^{2} = PB {}^{2}$

$now \: using \: distance \: formula \\ we \: get \\ (x - x1) {}^{2} + (y - y1) {}^{2} = (x - x2) {}^{2} + (y - y2) {}^{2} \\ (0 - 1) {}^{2} + (y - 2) {}^{2} = (0 - 3) {}^{2} + (y - 4) {}^{2} \\ ( - 1) {}^{2} + (y - 2) {}^{2} = ( - 3) {}^{2} + (y - 4) {}^{2} \\ 1 + y {}^{2} + 4 - 4y = 9 + y {}^{2} - 8y + 16 \\ 5 - 4y = 25 - 8y \\ - 4y + 8y = 25 - 5 \\ 4y = 20 \\ y = 5$

$hence \: the \: point \: only \: on \: y \: axis \: which \: is \: equidistant \: from \: (1,2) \: and \: (3,4) \: is \: 5$