question_answer

The mid points of the sides of a triangle are \[D(6,1),\,\,E(3,\,5)\] and \[F(-1,-2),\] then the vertex opposite to D is

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, \[(-4,2)\]