question_answer

The coordinates of the point of trisection of the line segment joining (2, - 3) and (4, - 1). (when the point is near the point (2, - 3)) is

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]

\[=P\left( \frac{1\times 4+2\times 2}{1+2},\,\frac{1\times \left( -1 \right)+2\times \left( -2 \right)}{1+2} \right)\]

\[=P\left( \frac{4+4}{3},\,\frac{-1-6}{3} \right)=P\left( \frac{8}{3},\,\frac{-7}{3} \right)\]