# The coordinates of the point

Question:

The coordinates of the point which is equidistant from the three vertices of the ΔAOB as shown in the figure is

(a) $(x, y)$.

(b) $(y, x)$

(c) $\left(\frac{x}{2}, \frac{y}{2}\right)$

(d) $\left(\frac{y}{2}, \frac{x}{2}\right)$

Solution:

(a) Let the coordinate of the point which is equidistant from the three vertices 0(0, 0), A(0,2y) and B(2x, 0) is P(h,k).

Then,                                     PO = PA = PB

⇒                                           (PO)² = (PA)²= (PB)2                                                        … (i)

By distance formula,

$\left[\sqrt{(h-0)^{2}+(k-0)^{2}}\right]^{2}=\left[\sqrt{(h-0)^{2}+(k-2 y)^{2}}\right]^{2}=\left[\sqrt{(h-2 x)^{2}+(k-0)^{2}}\right]^{2}$

$\Rightarrow \quad h^{2}+k^{2}=h^{2}+(k-2 y)^{2}=(h-2 x)^{2}+k^{2}$ ....(ii)

Taking first two equations, we get

$h^{2}+k^{2}=h^{2}+(k-2 y)^{2}$

$\Rightarrow \quad k^{2}=k^{2}+4 y^{2}-4 y k \Rightarrow 4 y(y-k)=0$

$\Rightarrow \quad y=k \quad[\because y \neq 0]$

Taking first and third equations, we get

$h^{2}+k^{2}=(h-2 x)^{2}+k^{2}$

$\Rightarrow \quad h^{2}=h^{2}+4 x^{2}-4 x h$

$\Rightarrow \quad 4 x(x-h)=0$

$\Rightarrow \quad x=h \quad[\because x \neq 0]$

$\therefore \quad$ Required points $=(h, k)=(x, y)$