Question

In a line segment AB, P(2,1) and Q(6,2) are the points of trisection. The coordinates of A and B (P is closer to A and Q is closer to B) .

• A. A(-2,0) and B(10,3)
• B. A(10,3) and B(-2,0)
• C. A(-2,3) and B(10,-3)
• D. A(-2,0) and B(10,0)

Answer 1

The correct option is A A(-2,0) and B(10,3)

Let the coordinates of A and B be A(x,y) and B(a,b).

Considering the line segment AQ, P becomes the mid-point of AQ.

Therefore applying mid-point formula, we get
$2=\frac{x+6}{2}$ and $1=\frac{y+2}{2}$
$\Rightarrow x=-2$ and $y=0$

Now, considering line segment PB, Q becomes the mid-point.

Therefore applying mid-point formula, we get
$6=\frac{a+2}{2}$ and $2=\frac{b+1}{2}$
$\Rightarrow a=10$ and $b=3$

Therefore, coordinates are $A(-2,0)$ and $B(10,3)$.