| The coordinates of the point which is equidistant from the three vertices of the \[\Delta AOB\] as shown in the figure is: (NCERT EXEMPIAR) |
 |
A) \[(x,y)\]
B) \[(y,x)\]
C) \[\left( \frac{x}{2},\frac{y}{2} \right)\]
D) \[\left( \frac{y}{2},\frac{x}{2} \right)\]
Correct Answer: A
Solution :
| [a] Let the coordinates 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, \[PO=PA=PB\Rightarrow {{(PO)}^{2}}={{(PA)}^{2}}={{(PB)}^{2}}\] ...(1) |
| By distance formula, |
| \[[\sqrt{{{(h-0)}^{2}}+{{(k-0)}^{2}}{{]}^{2}}}=\,[\sqrt{{{(h-0)}^{2}}+{{(k-2y)}^{2}}{{]}^{2}}}\] \[=[\sqrt{{{(h-2x)}^{2}}+{{(k-0)}^{2}}{{]}^{2}}}\] |
| \[\Rightarrow \,\,\,{{h}^{2}}+{{k}^{2}}={{h}^{2}}+{{(k-2y)}^{2}}={{(h-2x)}^{2}}+{{k}^{2}}\] (2) |
| Taking first two terms, we get |
| \[{{h}^{2}}+{{k}^{2}}={{h}^{2}}+{{(k-2y)}^{2}}\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{k}^{2}}={{k}^{2}}+4{{y}^{2}}-4yk\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4y(y-k)=0\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,k=y\] \[[y\ne 0]\] |
| Taking first and third terms, we get |
| \[{{h}^{2}}+{{k}^{2}}={{(h-2x)}^{2}}+{{k}^{2}}\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,{{h}^{2}}={{h}^{2}}+4{{x}^{2}}-4xh\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,4x(x-h)=0\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,h=x\] \[[x\ne 0]\] |
| \[\therefore \] Required coordinates of the point are \[(x,y)\]. |
You need to login to perform this action.
You will be redirected in
3 sec
