Prove that abscissa of a point P which is equidistant from points with coordinates A$\displaystyle (7, 1)$ and B$\displaystyle (3, 5)$ is $\displaystyle 2$ more than its ordinate.

Let \(\displaystyle \mathrm{P}(x, y)\) be equidistant from \(\displaystyle \mathrm{A}(7,1)\) and \(\displaystyle \mathrm{B}(3,5)\) \[P A=P B \Rightarrow P A^{2}=P B^{2} \] \(\displaystyle (x-7)^{2}+(y-1)^{2}=(x-3)^{2}+(y-5)^{2}\) \(\displaystyle x^{2}+49-14 x+y^{2}+1-2 y=x^{2}+9-6 x+y^{2}+25-10 y\) \[x=2+y \] Thus, abscissa of the point P is $\displaystyle 2$ more than its ordinate.