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Which point on y-axis is equidistant from $(2, 3)$ and $(-4, 1)$?
Given:
Given points are $(2, 3)$ and $(-4, 1)$.
To do:
We have to find the point on y-axis which is equidistant from $(2, 3)$ and $(-4, 1)$.
Solution:
Let the co-ordinates of the two points be $A (2, 3)$ and $B (-4, 1)$.
We know that,
The $x$ co-ordinate of a point on y-axis is $0$.
Let the co-ordinates of the point which is equidistant from the points $A$ and $B$ be $C(0, y)$.
This implies,
$AC = CB$
The distance between two points \( \mathrm{A}\left(x_{1}, y_{1}\right) \) and \( \mathrm{B}\left(x_{2}, y_{2}\right) \) is \( \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \).
Therefore,
\( \mathrm{AC}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \)
\( =\sqrt{(0-2)^{2}+(y-3)^{2}} \)
\( =\sqrt{(-2)^{2}+(y-3)^{2}} \)
\( =\sqrt{4+(y-3)^{2}} \)
\( \mathrm{BC}=\sqrt{(0-4)^{2}+(y-1)^{2}} \)
\( =\sqrt{(4)^{2}+(y-1)^{2}} \)
\( =\sqrt{16+(y-1)^{2}} \)
Here,
\( \mathrm{AC}=\mathrm{BC} \)
\( \therefore \sqrt{4+(y-3)^{2}}=\sqrt{16+(y-1)^{2}} \)
Squaring on both sides, we get,
\( 4+(y-3)^{2}=16+(y-1)^{2} \)
\( 4+y^{2}+9-6 y=16+y^{2}+1-2 y \)
\( -6 y+13=-2 y+17 \)
\( -6 y+2 y=17-13 \)
\( -4 y=4 \)
\( y=\frac{4}{-4} \)
\( y=-1 \)
Therefore, the required point is $(0,-1)$.
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