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