Question

The ratio in which the midpoint of A(0, 6) and B(12, 12) divides the points of trisection of the line AB is .

• A. 1:1
• B. 1:2
• C. 2:1

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.

$\Rightarrow x=\frac{2\left(0\right)+1\left(12\right)}{1+2}=\frac{0+12}{3}=\frac{12}{3}=4$
$\Rightarrow y=\frac{2\left(6\right)+1\left(12\right)}{1+2}=\frac{12+12}{3}=\frac{24}{3}=8$
$\therefore P\equiv (4,8)$

$\Rightarrow AQ:QB=2:1$
$\Rightarrow h=\frac{1\left(0\right)+2\left(12\right)}{1+2}=\frac{0+24}{3}=\frac{24}{3}=8$
$\Rightarrow k=\frac{1\left(6\right)+2\left(12\right)}{1+2}=\frac{6+24}{3}=\frac{30}{3}=10$
$\therefore Q\equiv (8,10)$

Let the midpoint of AB be C(a,b)
$\therefore AC=CB$
$\Rightarrow a=\frac{1\left(0\right)+1\left(12\right)}{1+1}=\frac{0+12}{2}=\frac{12}{2}=6$
$\Rightarrow b=\frac{1\left(6\right)+1\left(12\right)}{1+1}=\frac{6+12}{2}=\frac{18}{2}=9$
$\therefore C\equiv (6,9)$

Let PC : CQ = k : 1

$6=\frac{8k+4}{k+1}$
$6(k+1)=(8k+4)$
$6k-8k=4-6$
$-2k=-2$
$k=1$
$\therefore$ The required ratio is 1 : 1 i.e., the mid point of AB is also the midpoint of PQ.