A) \[\left( \frac{13}{4},\frac{-9}{4} \right)\]
B) \[\left( \frac{-13}{4},\frac{9}{4} \right)\]
C) \[\left( \frac{-13}{4},\frac{-9}{4} \right)\]
D) \[\left( \frac{13}{4},\frac{9}{4} \right)\]
E) \[\left( \frac{13}{2},\frac{9}{4} \right)\]
Correct Answer: D
Solution :
Let the vertices of a triangle are\[P(2,1),\] Q (5, 2) and R (3, 4) and\[A(x,\text{ }y)\]be the circumcentre of\[\Delta PQR.\] \[\therefore \] \[A{{P}^{2}}=A{{Q}^{2}}\] \[\Rightarrow \]\[{{(2-x)}^{2}}+{{(1-y)}^{2}}={{(5-x)}^{2}}+{{(2-y)}^{2}}\] \[\Rightarrow \] \[4+{{x}^{2}}-4x+1+{{y}^{2}}-2y\] \[=25+{{x}^{2}}-10x+4+{{y}^{2}}-4y\] \[\Rightarrow \] \[6x+2y=24\] \[\Rightarrow \] \[3x+y=12\] ...(i) Similarly, \[A{{P}^{2}}=A{{R}^{2}}\] \[\Rightarrow \]\[{{(2-x)}^{2}}+{{(1-y)}^{2}}={{(3-x)}^{2}}+{{(4-y)}^{2}}\] \[\Rightarrow \]\[4+{{x}^{2}}-4x+1+{{y}^{2}}-2y\] \[=9+{{x}^{2}}-6x+16+{{y}^{2}}-8y\] \[\Rightarrow \] \[2x+6y=20\] \[\Rightarrow \] \[x+3y=10\] ...(ii) On solving Eqs. (i) and (ii), we get \[x=\frac{13}{4}\]and\[y=\frac{9}{4}\] \[\therefore \]Circumcentre is\[\left( \frac{13}{4},\frac{9}{4} \right)\]
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