40
Find a relation between $x$ and $y$ such that the point P$(x, y)$ is equidistant from the points A$(5, 3)$ and B$(1, 7)$.
Show SolutionHide Solution↓
Since P $(x, y)$ is equidistant from A$(5, 3)$ and B$(1, 7)$
$\therefore$ PA = PB $\Rightarrow$ PA$^2$ = PB$^2$ ($\frac{1}{2}$ Mark)
$\Rightarrow (x-5)^2 + (y - 3)^2 = (x - 1)^2 + (y - 7)^2$ (1 Mark)
$\Rightarrow x^2 + 25 - 10x + y^2 + 9 - 6y = x^2 + 1 - 2x + y^2 + 49 - 14y$ (1 Mark)
$\Rightarrow x - y = -2$ or $x - y + 2 = 0$ ($\frac{1}{2}$ Mark)
$\therefore$ PA = PB $\Rightarrow$ PA$^2$ = PB$^2$ ($\frac{1}{2}$ Mark)
$\Rightarrow (x-5)^2 + (y - 3)^2 = (x - 1)^2 + (y - 7)^2$ (1 Mark)
$\Rightarrow x^2 + 25 - 10x + y^2 + 9 - 6y = x^2 + 1 - 2x + y^2 + 49 - 14y$ (1 Mark)
$\Rightarrow x - y = -2$ or $x - y + 2 = 0$ ($\frac{1}{2}$ Mark)