Question

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

Answer 1

Let $P$ and $Q$ be such points that trisect the line segment $AB$ in such a way that $AP:PQ:QB=1:1:1$.

Consider point $P$ which divides the line segment $AB$ such that $AP:PB=1:2$

Therefore, by Sectional Formula,

$P(x,y)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$ where $m:n=1:2$

$=(\frac{1\times 2+2\times (-4)}{3},\frac{1\times (-1)+2\times \left(3\right)}{3})$

$=(\frac{-6}{3},\frac{5}{3})$

$\therefore P(x,y)=(-2,\frac{5}{3})$.

Now, $Q$ is the mid point of $PB$.

$\therefore Q(x,y)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

$=(\frac{-2+2}{2},\frac{\frac{5}{3}-1}{2})$

$=(0,\frac{1}{3})$

$\therefore Q(x,y)=(0,\frac{1}{3})$