Question

Find the points of trisection of the line segment joining $(4,-1)$ and $(-2,-3)$.

Answer 1

Using the section formula, if a point $(x,y)$ divides the line joining the points $(x_1,y_1)$ and $(x_2,y_2)$ in the ratio $m:n$, then

$(x,y)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$

Let $A(4,-1)$ and $B(-2,-3)$ be the given points.
Let $P(x,y)$ and $Q(a,b)$ be the points of trisection of $AB$ so that $AP=PQ=QB$
Hence $P$ divides $AB$ internally in the ratio $1:2$ and $Q$ divides $AB$ internally in the ratio $2:1$
$\therefore$ By the section formula, the required points are
$P(\frac{1(-2)+2\left(4\right)}{1+2},\frac{1(-3)+2(-1)}{1+2})$ and
$Q(\frac{2(-2)+1\left(4\right)}{2+1},\frac{2(-3)+1(-1)}{2+1})$
$\Rightarrow P(x,y)=P(\frac{-2+8}{3},\frac{-3-2}{3})$ and
$Q(a,b)=Q(\frac{-4+4}{3},\frac{-6-1}{3})$
$=P(2,-\frac{5}{3})$ and $Q(0,-\frac{7}{3})$
Note that $Q$ is the midpoint of $PB$ and $P$ is the midpoint of $AQ$.