Question

In a line segment AB, P(2,1) and Q(6,2) are the points of trisection. Find the coordinates of A and B (P is closer to A and Q is closer to B).
$\left[2Marks\right]$

Answer 1

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 $y=0$
$\left[1Mark\right]$

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 $b=3$
$\left[1Mark\right]$

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