The locus of a point which moves so that it i | vedclass

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(B) Let the moving point be $P(h, k)$.Given that the distance $PA = PB$.Using the distance formula,we have:$\sqrt{(h - a)^2 + (k - 0)^2} = \sqrt{(h -

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Perpendicular bisector of the line segment $AB$

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(B) Let the moving point be $P(h, k)$.Given that the distance $PA = PB$.Using the distance formula,we have:$\sqrt{(h - a)^2 + (k - 0)^2} = \sqrt{(h - (-a))^2 + (k - 0)^2}$Squaring both sides:$(h - a)^2 + k^2 = (h + a)^2 + k^2$$h^2 - 2ah + a^2 + k^2 = h^2 + 2ah + a^2 + k^2$$-2ah = 2ah$$4ah = 0$Since $a 
eq 0$,we get $h = 0$.Replacing $(h, k)$ with $(x, y)$,the locus is $x = 0$,which is the $y$-axis.The $y$-axis is the perpendicular bisector of the line segment joining $A(a, 0)$ and $B(-a, 0)$.

The locus of a point which moves so that it is always equidistant from the points $A(a, 0)$ and $B(-a, 0)$ is

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