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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 $$(\frac{2x_2 + x_1}{3}$$, $$\frac{2y_2 + y_1}{3})$$.