question_answer

If the middle points of the sides of a triangle be (-2, 3),          (4, -3) and (4, 5), then the centroid of the triangle is

A) (5/3, 2)

B) (5/6, 1)

C) (2, 5/3)

D) (1, 5/6)

Correct Answer: C

Solution :

Let the vertices of the triangle are \[A\,({{x}_{1}},\,{{y}_{1}}),\,B\,({{x}_{2}},\,\,{{y}_{2}})\] and \[C\,({{x}_{3}},\,\,{{y}_{3}}),\] then   \[{{x}_{1}}+{{x}_{2}}=8\]     .....(i)   \[{{y}_{1}}+{{y}_{2}}=10\]        .....(ii)   \[{{x}_{2}}+{{x}_{3}}=-4\]   ?..(iii)   \[{{y}_{2}}+{{y}_{3}}=6\]     ?..(iv)   \[{{x}_{3}}+{{x}_{1}}=8\]      ?..(v)  and     \[{{y}_{3}}+{{y}_{1}}=-6\]      ?..(vi) On solving these equations, we get \[{{x}_{1}}=10,\,{{x}_{2}}=-2,\,\,{{x}_{3}}=-2,\,\,{{y}_{1}}=-1,\,\,{{y}_{2}}=11\] and \[{{y}_{3}}=-5\] Hence the centroid is \[\left( 2,\,\frac{5}{3} \right)\]. Aliter : As we know that the centroid of the triangle ABC and that of the triangle formed by joining the middle points of the sides of triangle ABC is same. Therefore, the required centroid is  \[\left( \frac{4+4-2}{3},\,\,\frac{5-3+3}{3} \right)\,\equiv \,\left( 2,\,\frac{5}{3} \right)\]