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Question: The points of trisection of the line segment joining the points (3, –2) and (–3, –4) are....

The points of trisection of the line segment joining the points (3, –2) and (–3, –4) are.

A

(32,−52),(−32,−134)\left( \frac{3}{2}, - \frac{5}{2} \right),\left( - \frac{3}{2}, - \frac{13}{4} \right)

B

(−32,52),(32,134)\left( - \frac{3}{2},\frac{5}{2} \right),\left( \frac{3}{2},\frac{13}{4} \right)

C

(1,−83),(−1,−103)\left( 1, - \frac{8}{3} \right),\left( - 1, - \frac{10}{3} \right)

D

None of these

Answer

(1,−83),(−1,−103)\left( 1, - \frac{8}{3} \right),\left( - 1, - \frac{10}{3} \right)

Explanation

Solution

Let A(3,−2)A ( 3 , - 2 ) , B(−3,−4)B ( - 3 , - 4 ) and two points C and D are the trisection points of AB. Obviously C divides the line AB in 1 : 2, therefore coordinates of C are

(1(−3)+2(3)1+2,1(−4)+2(−2)1+2)=(1,−83)\left( \frac { 1 ( - 3 ) + 2 ( 3 ) } { 1 + 2 } , \frac { 1 ( - 4 ) + 2 ( - 2 ) } { 1 + 2 } \right) = \left( 1 , - \frac { 8 } { 3 } \right).

Similarly D divides the line AB in 2 : 1, hence

coordinates of D are

(2(−3)+1(3)2+1,2(−4)+1(−2)2+1)=(−1,−103)\left( \frac { 2 ( - 3 ) + 1 ( 3 ) } { 2 + 1 } , \frac { 2 ( - 4 ) + 1 ( - 2 ) } { 2 + 1 } \right) = \left( - 1 , - \frac { 10 } { 3 } \right).