Question

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

• A. $\frac{1}{1}$
• B. $\frac{2}{1}$
• C. $\frac{1}{2}$
• D. $\frac{2}{3}$

Answer 1

The correct option is A

$\frac{1}{1}$

For finding the ratio in which the midpoint of AB divides points of trisection of AB. First, the midpoint and points of trisection have to be found. Then find the ratio in which the midpoint divides points of trisection.
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 divide 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

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

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

$\Rightarrow k=1$