# Let P be a point on the perpendicular bisector of the segment joining (2, 3) and (6, 5). If the abscissa and the ordinate of P are equal, find the coordinates of P.

Let the point =       $P(x, y)$ is equal to coordinate then we have $y= x$

therefore, the coordinate of P =$(x, x )$

Now, let A and B denote the point (2,3) and (6, 5) Since P is equidistant from A and B

We get,   $PA^{2} = PB^{2}$

i.e.

$(x-2)^{2} + (x-3)^{2} - (x - 6)^{2} + (x -5)^{2} \\ x^{2} -4x + 4 + x^{2} - 6x + 9 = x^{2} - 12 x + 36 + x^{2} - 10 x + 25\\ 2x^{2} -10x + 13 = 2x^{2} - 22x +61\\ 22x - 10x = 61 -13\\ 12x = 48\\ x = \frac{48}{12} = 4 \\$

Hence, the coordinate of P = $(4,4)$

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