28
Prove that abscissa of the point $P$ which is equidistant from points with coordinates $A(7, 1)$ and $B(3, 5)$ is 2 more than its ordinate.
Show SolutionHide Solution↓
Let $P(x, y)$ be equidistant from $A(7, 1)$ and $B(3, 5)$. $PA = PB \Rightarrow PA^2 = PB^2$ ($\frac{1}{2}$ mark). $(x-7)^2 + (y-1)^2 = (x-3)^2 + (y-5)^2$ ($\frac{1}{2}$ mark). $x^2 + 49 - 14x + y^2 + 1 - 2y = x^2 + 9 - 6x + y^2 + 25 - 10y$ ($\frac{1}{2}$ mark). $x = 2 + y$. Thus, abscissa of the point $P$ is 2 more than its ordinate ($\frac{1}{2}$ mark).