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Find the point on $y-$axis which is equidistant from the points $( 5,\ - 2)$ and $( -3,\ 2)$.
Given: There is a point on $y-$axis which is equidistant from the points $( 5,\ - 2)$ and $( -3,\ 2)$.
To do: To find the point.
Solution:
Since the point is on $y-$axis so,
$X-$coordinate is zero Let the point be $P( 0,\ y)$
It's distance from $A( 5,\ -2)$ and $B( - 3,\ 2)$ are equal
We know, If there two points $( x_{1}, y_{1})$ and $( x_{2}, y_{2})$,
Distance between the two points$=\sqrt{( x_{2}-x_{1})^{2}+( y_{2}-y_{1})^{2}}$
$PA=\sqrt{( 5-0)^{2}+( -2-y)^{2}}$
$\Rightarrow PA=\sqrt{(5)^{2}+(-(2+y))^{2}}$
$\Rightarrow PA=\sqrt{25+(y+2)^{2}}$
$\Rightarrow PA=\sqrt{25+y^{2}+4+4y}$
$\Rightarrow PA=\sqrt{y^{2}+4y+29}$
$\Rightarrow ( PA)^{2}=y^{2}+4y+29$ ..................$( 1)$
Similarly,
$PB=\sqrt{( -3-0)^{2}+( 2-y)^{2}}$
$\Rightarrow PB=\sqrt{( -3)^{2}+4+y^{2}-4y}$
$\Rightarrow ( PB)^{2}=9+4+y^{2}-4y$
$\Rightarrow ( PB)^{2}=y^{2}-4y+13$ .....................$(2)$
As given $PA=PB$
$\therefore ( PA)^{2}=( PB)^{2}$
$\Rightarrow y^{2}+4y+29=y^{2}-4y+13$
$\Rightarrow 4y+4y=13-29$
$\Rightarrow 8y=-16$
$\Rightarrow y=-\frac{16}{8}$
$\Rightarrow y=-2$
Thus, The point is $( 0,\ -2)$
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