Question

Find the ratio in which the midpoint of A(12, 8) and B(4, 6) divides the line joining the points of trisection of the line AB.

• A. $1:1$
• B. $2:1$
• C. $\frac{1}{2}:1$
• D. $4:1$

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 the points divide the line AB in the ratio of 1 : 2 and 2 : 1 respectively.
Let P divide 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 it divides the trisection in the ratio $1:1$.