Question

Find the ratio in which the midpoint of A(14, 10) and B(2, 4) divides the line joining the points of trisection of the line AB.

• A. 1 : 1
• B. 1 : 2
• C. 2 : 1
• D. 3 : 2

Answer 1

The correct option is A 1 : 1

Let the points of trisection of the line segment AB be P(x,y) and Q(h,k)
$\therefore AP=PQ=QB$

$\Rightarrow AP:PB=1:2$

We know, by section formula, that the coordinates of the point that divides a line in the ratio m : n is,

$(\frac{(n\times x_1+m\times x_2)}{m+n},\frac{n\times y_1+m\times y_2}{m+n})$

where $(x_1,y_1)and(x_2,y_2)$ are the coordinates of the endpoints of the line segment.

$\Rightarrow x=\frac{1\left(2\right)+2\left(14\right)}{1+2}=\frac{2+28}{3}=\frac{30}{3}=10$
$\Rightarrow y=\frac{1\left(4\right)+2\left(10\right)}{1+2}=\frac{4+20}{3}=\frac{24}{3}=8$
$\therefore P=(10,8)$

Now, $AQ:QB=2:1$
$\Rightarrow h=\frac{2\left(2\right)+1\left(14\right)}{1+2}=\frac{4+14}{3}=\frac{18}{3}=6$
$\Rightarrow k=\frac{2\left(4\right)+1\left(10\right)}{1+2}=\frac{8+10}{3}=\frac{18}{3}=6$
$\therefore Q=(6,6)$

$\text{Let the midpoint of AB be}C(a,b)$
$\Rightarrow a=\frac{2+14}{2}=\frac{16}{2}=8$
$\Rightarrow b=\frac{4+10}{2}=\frac{14}{2}=7$
$\therefore C=(8,7)$

$\text{Let, 'C' divides PQ in the ratio}k:1$

$\Rightarrow 8=\frac{6k+10}{k+1}$
$\Rightarrow 8(k+1)=(6k+10)$
$\Rightarrow 8k-6k=10-8$
$\Rightarrow 2k=2$
$\Rightarrow k=1$
$\therefore$ The required ratio is 1 : 1 i.e., the mid point of AB is also the midpoint of PQ.