Question

The ratio in which the midpoint of A (12 , 8) and B (4 , 6) divides the points of trisection of the line segment AB is

• A. 1 : 1
• B. 2 : 1
• C. 3 : 1

Answer 1

The correct option is A 1 : 1

The midpoint of AB is $(\frac{12+4}{2},\frac{8+6}{2})=(8,7)$.
Let this point be C.
The points of trisection of the line AB are the points which divides the line into three equal line segments. Hence these points divide the line AB in the ratio of 1 : 2 and 2 : 1 respectively.

Let P be the point which divides AB in the ratio of 1 : 2 and Q be the point which divides AB in the ratio
2 : 1.

$P=(\frac{2\times 12+1\times 4}{1+2},\frac{2\times 8+1\times 6}{1+2})=(\frac{28}{3},\frac{22}{3})$

$Q=(\frac{(1\times 12+2\times 4)}{1+2},\frac{1\times 8+2\times 6}{1+2})=(\frac{20}{3},\frac{20}{3})$

Let the point C (8, 7) divide PQ in the ratio of k : 1. Then

$\frac{(k\times \frac{20}{3}+\frac{28}{3})}{(k+1)}=8$

$\Rightarrow k\times \frac{20}{3}+\frac{28}{3}=8k+8$

$\Rightarrow k=1$
Hence, the midpoint of A and B divides the points of trisection in the ratio $1:1$.