A) \( \Large \left(3,\ 6\right) \) |
B) \( \Large \left(1,\ 2\right) \) |
C) \( \Large \left(4,\ 8\right) \) |
D) \( \Large \left(-3,\ 6\right) \) |
B) \( \Large \left(1,\ 2\right) \) |
Let \( \Large \left(x_{1},\ y_{1}\right),\ \left(x_{2},\ y_{2}\right)\ and\ \left(x_{3},\ y_{3}\right) \) are co ordinates of the points D, E, F which divide each AB, BC and CA respectively in the ratio 3 : 1 (internally)
Therefore, \( \Large x_{1}=\frac{3 \times 6-1 \times 1}{4}=\frac{17}{4} \)
\( \Large y_{1}=\frac{-2 \times 3+4 \times 1}{4}=-\frac{2}{4}=-\frac{1}{2} \)
Similarly, \( \Large x_{2}= 0,\ y_{2}=\frac{5}{2}\ and\ x_{3}=-\frac{5}{4},\ y_{3}=4 \)
Let \( \Large \left(x,\ y\right) \) be the co-ordinates of centroid of \( \Large \triangle DEF \)
Therefore, \( \Large x=\frac{1}{3} \left(\frac{17}{4}+0-\frac{5}{4}\right)=1 \)
\( \Large y=\frac{1}{3} \left(-\frac{1}{2}+\frac{5}{2}+4\right)=2 \)
Hence, Co-ordinates of centroid are \( \Large 1,\ 2 \).