Equate the sides of a triangle ABC, given two vertices A (-3,3) B (5, -1) and the intersection point of heights M (4,3).

Side AB equation:

x – x1 / x2 – x1 = y – y1 / y2 – y1;

x + 3/8 = y – 3 / -4;

-4 * (x +3) = 8 * (y – 3);

x + 2y – 3 = 0.

The equation of the height lowered to the side of the AC passing through the points B (5; -1) and M (4; 3):

x – 4/1 = y – 3 / -4;

16 – 4x = y – 3;

4x + y – 19 = 0.

By the condition of perpendicularity of two lines:

k2 = -1 / k1; k1 = -4; k2 = ¼.

AC⊥ BM, then the AC equation:

y – 3 = ¼ * (x +3);

x – 4y + 15 = 0.

Height equation on the BC side:

x – 4 / -3 – 4 = y – 3/3 – 3;

x – 4/7 = y – 3/0;

-7 * (y – 3) = 0;

y = 3.

The slope of the line is 0.

Height lowered on the aircraft || X-axis.

Hence, the equation of the BC side: x – 5 = 0.