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