Find the coordinates of the points of trisection of the line segment joining

General
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find the coordinates of the points of trisection of the line segment joining

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Find the coordinates of the points of trisection of the line segment joining

Answer: To find the coordinates of the points of trisection of the line segment joining two points ( A(x_1, y_1) ) and ( B(x_2, y_2) ), we need to find the points that divide the line segment into three equal parts. These points are known as the points of trisection.

Let’s denote the points of trisection as ( P ) and ( Q ), such that ( P ) divides the segment ( AB ) in the ratio ( 1:2 ) and ( Q ) divides the segment ( AB ) in the ratio ( 2:1 ).

Step-by-Step Solution:

  1. Coordinates of Point ( P ) (Ratio ( 1:2 )):

    The formula for the coordinates of a point dividing a line segment in a given ratio ( m:n ) is:

    \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right)

    For point ( P ) dividing ( AB ) in the ratio ( 1:2 ):

    • ( m = 1 )
    • ( n = 2 )

    Therefore, the coordinates of ( P ) are:

    P \left( \frac{1 \cdot x_2 + 2 \cdot x_1}{1+2}, \frac{1 \cdot y_2 + 2 \cdot y_1}{1+2} \right)

    Simplifying this:

    P \left( \frac{x_2 + 2x_1}{3}, \frac{y_2 + 2y_1}{3} \right)

  2. Coordinates of Point ( Q ) (Ratio ( 2:1 )):

    For point ( Q ) dividing ( AB ) in the ratio ( 2:1 ):

    • ( m = 2 )
    • ( n = 1 )

    Therefore, the coordinates of ( Q ) are:

    Q \left( \frac{2 \cdot x_2 + 1 \cdot x_1}{2+1}, \frac{2 \cdot y_2 + 1 \cdot y_1}{2+1} \right)

    Simplifying this:

    Q \left( \frac{2x_2 + x_1}{3}, \frac{2y_2 + y_1}{3} \right)

Summary:

  • The coordinates of the first point of trisection ( P ) are \left( \frac{x_2 + 2x_1}{3}, \frac{y_2 + 2y_1}{3} \right)
  • The coordinates of the second point of trisection ( Q ) are \left( \frac{2x_2 + x_1}{3}, \frac{2y_2 + y_1}{3} \right)

These points ( P ) and ( Q ) divide the line segment joining A(x_1, y_1) and B(x_2, y_2) into three equal parts.

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