Question

Find the co-ordinates of the points which divide the line segment joining $A(-7,5)$ and $B(5,-1)$ into three congruent segments (Such points are called the points of trisection of segment).

Answer 1

$A(-7,5)$ and $B(5,-1)$ are the given points. There are two more points on $AB$ such that they divide $\overline{AB}$ into three congruent segments.
Suppose $M$ and $N$ are the trisection points of $\overline{AB}$
$M$ divides $\overline{AB}$ from $A$ in the ratio $\frac{AM}{MB}=\frac{k}{2k}=\frac{1}{2}=\frac{m}{n}$; where $k>0$
$\therefore$ $p=1$ and $q=2$ and $(x_1,y_1)=A(-7,5),(x_2,y_2)=B(5,-1)$ and $M(x,y)$
Now x co-ordinate of $M$
$x=\frac{m{x}_2+n{x}_1}{m+n}=\frac{1\left(5\right)+2(-7)}{1+2}=-3$
y co-ordinate of $M$
$y=\frac{m{y}_2+n{y}_1}{m+n}=\frac{1(-1)+2\left(5\right)}{1+2}=3$
Therefore $M(x,y)=(-3,3)........\left(1\right)$
Next, $M-N-B$ and $MN=NB=k$
So, $N$ is the midpoint of $MB$
$\therefore N(x,y)=(\frac{-3+5}{2},\frac{3-1}{2})$
$N(x,y)=(1,1).........\left(2\right)$
From (1) and (2) the trisection points of $AB$ are $(-3,3)$ and $(1,1)$