The vertices of triangle ABC are given: A (0; 2), B (12; -7) and C (16; 15). Write down the equations of its sides.

The equation is y = k * x + b. Substitute the values of the coordinates of the points in the function formula and get the equation of the straight line.

1) A (0; 2), B (12; -7).

2 = 0 * k + b;

-7 = 12 * k + b;

b = 2;

-7 = 12 * k + 2;

12 * k = -9;

k = -3/4;

We get the equation of the straight line:

y = -3/4 * x + 2.

2) A (0; 2), C (16; 15).

2 = 0 * k + b;

15 = 16 * k + b;

b = 2;

15 = 16 * k + 2;

16 * k = 13;

k = 13/16;

We get the equation of the straight line:

y = 13/16 * x + 2.

3) B (12; -7), C (16; 15).

-7 = 12 * k + b;

15 = 16 * k + b;

Subtract the first from the second equation by the Gauss method:

16 * k – 12 * k = 15 + 7;

4 * k = 22;

k = 11/2;

b = 15 – 16 * 11/2 = 15 – 88 = -73;

We get the equation of the straight line:

y = 11/2 * x – 73.