5. Points of trisection of a line segment

We know that a mid-point divides the line segment into two halves. If the same line segment is divided into $3$ equal parts, the points obtained are called points of trisection.

 

Let us look at the figure given below.

 

 

Here, $M$ and $N$ are the end-points of the line segment.

 

$A$ and $B$ are the points of trisection.

 

Distance between $MA$ $=$ Distance between $AB$ $=$ Distance between $BN$

 

Let us look carefully at the graph given below.

 

 

To find the points of trisection, we should find the values of $P$ and $Q$.

 

To find the coordinates of $P$:

 

From the graph, we know that $P$ is $(a$, $b)$.

 

To find the co-ordinates of $P$, we should have the values of '$a$' and '$b$'.

 

Since '$a$' is a $x$-coordinate, we should consider the distance $OP'$.

 

$a = OP' = OA' + A'P'$

 

$a = x_1 +$ $(\frac{x_2 - x_1}{3})$

 

$= \frac{3x_1 + x_2 - x_1}{3}$

 

$= \frac{2x_1 + x_2}{3}$

 

Therefore, $a = \frac{2x_1 + x_2}{3}$.

 

Since $b$ is a $y$-coordinate, we should consider the distance $PP'$.

 

$b = PP' = PA'' + A''P'$

 

$=$ $(\frac{y_2 - y_1}{3})$ $+ y_1$

 

$= \frac{y_2 - y_1 + 3y_1}{3}$

 

$= \frac{2y_1 + y_2}{3}$

 

Therefore, the coordinates of $P(a$, $b)$ is $(\frac{2x_1 + x_2}{3}$, $\frac{2y_1 + y_2}{3})$.

 

Similarly, the coordinate $Q(c$, $d)$ is $(\frac{2x_2 + x_1}{3}$, $\frac{2y_2 + y_1}{3})$.