question_answer

ABC   is   a   triangle   with   vertices\[A(-1,4),\] \[B(6,-2)\]and\[C(-2,4)\]. D, E and F are the points which divide each AB, BC and CA respectively in the ratio\[3:1\]internally. Then, the centroid of me triangle DEF is:

A)  (3, 6)                    

B)         (1, 2)

C)  (4, 8)                    

D)        \[(-3,\text{ }6)\]

E)  \[(-1,2)\]

Correct Answer: B

Solution :

Let \[({{x}_{1}},{{y}_{1}}),({{x}_{2}},{{y}_{2}})\]and\[({{x}_{3}},{{y}_{3}})\]are co-ordinates of the points D, E and F which divide each A3, BC and CA respectively in the ratio \[3:1\]internally \[\therefore \]  \[{{x}_{1}}=\frac{3\times 6-1\times 1}{4}=\frac{17}{4}\]                 \[{{y}_{1}}=\frac{-2\times 3+4}{4}=-\frac{2}{4}=-\frac{1}{2}\] Similarly,              \[{{x}_{2}}=0,{{y}_{2}}=\frac{5}{2}\]                                 \[{{x}_{3}}=-\frac{5}{4},{{y}_{3}}=4\] Let\[(x,\text{ }y)\]be the co-ordinate of centroid of\[\Delta DEF\]                 \[x=\frac{1}{3}\left( \frac{17}{4}+0-\frac{5}{4} \right)=1\]and                 \[y=\left( -\frac{1}{2}+\frac{5}{2}+4 \right)=\frac{1}{3}=2\] \[\therefore \]Co-ordinate of centroid is (1, 2).