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# Find a point on y-axis which is equidistant from the points $\left(5,-2\right)$ and $\left(-3,2\right)$.

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## Find a point on y-axis which is equidistant from the points $\left(5,-2\right)$ and $\left(-3,2\right)$:Let the point be $P\left(0,a\right)$ be equidistant from the points $A\left(5,-2\right)$ and $B\left(-3,2\right)$.$⇒PA=PB$.According to the distance formula, the distance between two points $A\left({x}_{1},{y}_{1}\right)$ and $B\left({x}_{2},{y}_{2}\right)$ is given by:$AB=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$Let's use the distance formula to substitute the given coordinates in $PA=PB$: $⇒$ $\sqrt{{\left(5-0\right)}^{2}+{\left(-2-a\right)}^{2}}$ $=\sqrt{{\left(-3-0\right)}^{2}+{\left(2-a\right)}^{2}}$$⇒$ ${\left(5-0\right)}^{2}+{\left(-2-a\right)}^{2}$ $={\left(-3-0\right)}^{2}+{\left(2-a\right)}^{2}$ [ squaring both the sides ]$⇒$ $25+{4+4a+a}^{2}$ $=9+{4-4a+a}^{2}$$⇒$ $29+4a$ $=13-4a$$⇒$ $29-13$ $=-4a-4a$ $⇒$ $16$ $=-8a$$⇒$ $a$ $=-2$ Hence, $\left(0,-2\right)$ is a point on y-axis which is equidistant from the points $\left(5,-2\right)$ and $\left(-3,2\right)$.

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