Question

Question 15
The coordinates of the point which is equivalent from the three vertices of the $△AOB$ shown in the figure, is

(A) $(x,y)$
(B) $(y,x)$
(C) $(\frac{x}{2},\frac{y}{2})$
(D) $(\frac{y}{2},\frac{x}{2})$

Answer 1

Let the coordinate of the point which is equidistant from the three vertices O(0,0), A(0,2y) and B(2x,0) is P(h,k)
Then, PO = PA = PB
$\Rightarrow (PO{)}^2=(PA{)}^2=(PB{)}^2\text{By distance formula,}{\left[\sqrt{(h-0{)}^2+(k-0{)}^2}\right]}^2$
$={\left[\sqrt{(h-0{)}^2+(k-2y{)}^2}\right]}^2$
$={\left[\sqrt{(h-2x{)}^2+(k-0{)}^2}\right]}^2\Rightarrow h^2+k^2=h^2+(k-2y{)}^2=(h-2x{)}^2+k^2...\left(i\right)\text{Taking first two terms of equation}\left(i\right)\text{, we get}{h}^2+k^2=h^2+(k-2y{)}^2\therefore \text{Required points}=(h,k)=(x,y)\Rightarrow k^2=k^2+4y^2-4yk\Rightarrow 4y(y-k)=0\Rightarrow y=k[\because y\ne 0]\text{Taking first and third term of equation}\left(i\right)\text{, we get}{h}^2+k^2=(h-2x{)}^2+k^2\Rightarrow h^2=h^2+4x^2-4xh\Rightarrow 4x(x-h)=0\Rightarrow x=h[\because x\ne 0]$
Therefore, coordinates of the point P is (x,y).