A) (10/3,-5/3)
B) (8/3,-7/3)
C) (3,-2)
D) None of the above
Correct Answer: B
Solution :
Let P and Q be the points of trisection as shown below |
Then, \[AP:\,PB=1:2\]and \[AQ:QB+2:1\] |
(i) When P divides AB in the ratio 1:2. |
Then, \[\frac{{{m}_{1}}}{{{m}_{2}}}=\frac{1}{2}\] |
Here, \[A\left( {{x}_{1}},\,{{y}_{1}} \right)=\left( 2,-3 \right)\] and \[B\left( {{x}_{2}},\,{{y}_{2}} \right)=\left( 4,\,-1 \right)\] |
Now, \[P\left( x,\,y \right)=P\left( \frac{{{m}_{1}}{{x}_{2}}+{{m}_{2}}{{x}_{1}}}{{{m}_{1}}+{{m}_{2}}},\,\frac{{{m}_{1}}{{y}_{2}}+{{m}_{2}}{{y}_{1}}}{{{m}_{1}}+{{m}_{2}}} \right)\] |
[by section formula] |
\[=P\left( \frac{1\times 4+2\times 2}{1+2},\,\frac{1\times \left( -1 \right)+2\times \left( -2 \right)}{1+2} \right)\] |
\[=P\left( \frac{4+4}{3},\,\frac{-1-6}{3} \right)=P\left( \frac{8}{3},\,\frac{-7}{3} \right)\] |
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