The triangle ABC on the plane has the coordinates of the vertices: A (-13; 3); B (-1; -2); C (2; 2).

The triangle ABC on the plane has the coordinates of the vertices: A (-13; 3); B (-1; -2); C (2; 2). find the equations of the straight lines: for the side AB-in the form y = kx + b.

We will use the fact that the equation of a straight line that passes on the coordinate plane through two points with coordinates (x1; y1) and (x2; y2) for x1 ≠ x2 and y1 ≠ y2 can be written in the following form:

(x – x1) / (x2 – x1) = (y – y1) / (y2 – y1).

Substituting into this equation the values x1 = -13, y1 = 3, x2 = -1 and y2 = -2, we write the equation of the straight line containing the AB side:

(x – (-13)) / (- 1 – (- 13)) = (y – 3) / (- 2 – 3).

Simplifying this ratio, we get:

(x + 13) / (- 1 + 13) = (y – 3) / (- 5);

(x + 13) / 12 = (y – 3) / (- 5);

-5 * (x + 13) = 12 * (y – 3);

-5x – 65 = 12y – 36;

12y = -5x – 65 + 36;

12y = -5x – 29;

y = – (5/12) x – 29/12.

Answer: y = – (5/12) x – 29/12.