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. 1 : 1
• B. 2 : 1
• C. 1 : 2
• D. 4 : 1

Answer 1

The correct option is A 1 : 1

$\because$ The midpoint of the points $(x_1,y_1)\text{and}(x_2,y_2)$ is $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

$\therefore$ The midpoint of AB $=(\frac{12+4}{2},\frac{8+6}{2})$
$=(8,7)$
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' 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.

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

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

Now, let us assume the point C (8, 7) divide PQ in the ratio of k : 1.

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

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

$\Rightarrow k=1$

$\therefore$ Midpoint of the line segment AB divides its point of trisection in the ratio 1 : 1.