A) \[(-4,2)\]
B) \[(-4,5)\]
C) \[(2,\,5)\]
D) \[(10,\,8)\]
Correct Answer: A
Solution :
Let the coordinates of A, B, C are \[({{x}_{1}},{{y}_{1}}),\] \[({{x}_{2}}{{y}_{2}}),\,({{x}_{3}}{{y}_{3}})\] respectively. Given mid points \[D(6,1),\,\,\,\,\,E(3,5)\] and \[F(-1,-2)\] of the sides AB, BC and CA of the triangle. \[\therefore \] \[\frac{{{x}_{1}}+{{x}_{2}}}{2}=6,\] \[\frac{{{y}_{1}}+{{y}_{2}}}{2}=1\] \[\Rightarrow \] \[\left. \begin{matrix} {{x}_{1}}+{{x}_{2}}=12, \\ {{y}_{1}}+{{y}_{2}}=2 \\ \end{matrix} \right\}\] ?..(i) Similarly, \[\left. \begin{matrix} {{x}_{2}}+{{x}_{3}}=6, \\ {{y}_{2}}+{{y}_{3}}=10 \\ \end{matrix} \right\}\] ?..(ii) and \[\left. \begin{matrix} {{x}_{1}}+{{x}_{3}}=-2 \\ {{y}_{1}}+{{y}_{3}}=-4 \\ \end{matrix} \right\}\] ?.(iii) On solving Eqs. (i), (ii) and (iii), we get \[{{x}_{1}}=2,\,{{x}_{2}}=10,\,{{x}_{3}}=-4\] and \[{{y}_{1}}=-6,\,{{y}_{2}}=8,\,{{y}_{3}}=2\] Now, the vertex opposite to D is C ie, \[(-4,2)\]You need to login to perform this action.
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