Given:- ABC is a triangle and L, M and N are the mid-points of the AB, BC, and AC.
To find:- The coordinates of L, M, and N.
Solution:-
To calculate the coordinate of L, M and N. We are going to use the mid-point theorem.
The formula is:
[tex]Mid-point\text{ }coordinate=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})[/tex](a) First, calculate the coordinate of L.
Here coordinate of A is (0,0) and the coordinate of B is (6q, 6r).
So, the coordinate of L can be calculated as:
[tex]\begin{gathered} Coordinate\text{ }of\text{ }L=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}) \\ Coordinate\text{ }of\text{ }L=(\frac{0+6q}{2},\frac{0+6r}{2}) \\ Coordinate\text{ }of\text{ }L=(\frac{6q}{2},\frac{6r}{2}) \\ Coordinate\text{ }of\text{ }L=(3q,3r) \end{gathered}[/tex](b) Now calculating the coordinate of M.
Here the coordinate of B is (6q, 6r) and the coordinate of C is (6p, 0).
So, the coordinate of M can be calculated as:
[tex]\begin{gathered} Coordinate\text{ }of\text{ }M=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}) \\ Coordinate\text{ }of\text{ }M=(\frac{6q+6p}{2},\frac{6r+0}{2}) \\ Coordinate\text{ }of\text{ }M=(3q+3p,3r) \end{gathered}[/tex](c) Now calculating the coordinate of N.
Here the coordinate of A is (0,0) and the coordinate of C is (6p, 0).
So, the coordinate of N can be calculated as:
[tex]\begin{gathered} Coordinate\text{ }of\text{ }N=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}) \\ Coordinate\text{ }of\text{ }N=(\frac{0+6p}{2},\frac{0+0}{2}) \\ Coordinate\text{ }of\text{ }N=(3p,0) \end{gathered}[/tex]Final answer:-
Therefore, the answer is:
[tex]\begin{gathered} Coordinate\text{ }of\text{ }L=(3q,3r) \\ Coordinate\text{ }of\text{ }M=(3q+3p,3r) \\ Coordinate\text{ }of\text{ }N=(3p,0) \end{gathered}[/tex]