| The coordinates of the point which is equidistant from the three vertices of the \[\Delta AOB\] as shown in the figure is a |
|
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 :
| 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 |
| \[\Rightarrow \,\,\,\,\,P{{O}^{2}}=P{{A}^{2}}=P{{B}^{2}}\] ...(i) |
| By distance formula, |
| \[{{\left[ \sqrt{{{\left( h-0 \right)}^{2}}+\left( k-0 \right)} \right]}^{2}}\] |
| \[={{[\sqrt{{{\left( h-0 \right)}^{2}}+{{\left( k-2y \right)}^{2}}}]}^{2}}\] |
| \[={{[\sqrt{{{\left( h-2x \right)}^{2}}+{{\left( k-0 \right)}^{2}}}]}^{2}}\] |
| \[\Rightarrow \,\,\,{{h}^{2}}+{{k}^{2}}={{h}^{2}}+{{\left( k-2y \right)}^{2}}\] |
| \[={{\left( h-2x \right)}^{2}}+{{k}^{2}}\] ...(ii) |
| Taking first two equations, we get |
| \[{{h}^{2}}+{{k}^{2}}={{h}^{2}}+{{\left( k-2y \right)}^{2}}\] |
| \[\Rightarrow \,\,\,{{k}^{2}}={{k}^{2}}+4{{y}^{2}}-4yk\Rightarrow 4y\left( y-k \right)=0\] |
| \[\Rightarrow \,\,\,y=k\] \[\left[ \because \,\,\,\,y\ne 0 \right]\] |
| Taking first and third equations, we get |
| \[{{h}^{2}}+{{k}^{2}}={{\left( h-2x \right)}^{2}}+{{k}^{2}}\] |
| \[\Rightarrow \,\,\,{{h}^{2}}={{h}^{2}}+4{{x}^{2}}-4xh\] |
| \[\Rightarrow \,\,\,\,\,4x\left( x-h \right)=0\] |
| \[\Rightarrow \,\,\,\,\,x=h\] \[\left[ \because \,\,\,x\ne 0 \right]\] |
| \[\therefore \] Required points = (h, k) = (x, y) |
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