Which point on y-axis is equidistant from (2, 3) and (-4, 1)?

Question

Which point on $y$-axis is equidistant from $(2, 3)$ and $(-4, 1)$?

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Answer

$A(2, 3)$ and $B(-4, 1)$ are the given points.
Let C(0, y) be the points are y-axis.
$\text{AC}=\sqrt{(0-2)^2+(\text{y}-3)^2}$
$\Rightarrow\ \text{AC}=\sqrt{4+\text{y}^2+9-6\text{y}}$
$\Rightarrow\ \text{AC}=\sqrt{\text{y}^2-6\text{y}+13}$
$\text{BC}=\sqrt{(0+4)^2+(\text{y}-1)^2}$
$\Rightarrow\ \text{BC}=\sqrt{16+\text{y}^2+1-2\text{y}}$
$\Rightarrow\ \text{BC}=\sqrt{\text{y}^2-2\text{y}+17}$
Since $AC = BC$
$AC^2 = BC^2$
$y^2 - 6y + 13 = y^2 - 2y + 17$
$\Rightarrow -6y + 2y = 17 - 13$
$\Rightarrow -4y = 4$
$\Rightarrow y = -1$
$\therefore$ The points on $y$-axis is $(0, -1).$

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