Given the coordinates of the vertices of the triangle abc: A (4; 1) B (-2; 4) C (-2; 5) Equate the median AD.

Since point D is the middle of the BC, we can find its coordinates:
D ((x1 + x2) / 2; (y1 + y2) / 2)), where B (x1; y1), C (x2; y2).
Then the point D ((- 4/2); (9/2)) = D (-2; 9/2).
Let’s compose the equation of the median, for this we use the equation of the straight line:
(x – x1) / (x2 – x1) = (y – y1) / (y2 – y1).
In our case, A (x1; y1), D (x2; y2).
Then the median equation:
(x – 4) / (-6) = (y – 1) / (7/2);
(7/2) (x – 4) = -6 (y – 1);
7x – 28 = -12y + 12;
7x + 12y – 40 = 0.
Answer: 7x + 12y – 40 = 0.