Triangle ABC A (5; 3), B (-11; -9), C (-4; 15) find the equation of the AC side.

In order to find the equation of the AC side, it is necessary to substitute the coordinates of the points A (5; 3) and C (-4; 15) into the equation of the straight line y = k * x + b, and find k and b from the resulting system:

{3 = k * 5 + b;

{15 = k * (-4) + c.

{in = 3 – 5 * k;

{15 = -4 * k + 3 – 5 * k.

{in = 3 – 5 * k;

{15 = -9 * k + 3.

{in = 3 – 5 * k;

{k = (15 – 3) / (- 9).

{in = 3 – 5 * (-4/3) = 3 + (5 * 4) / 3 = 3 + 6 + 2/3 = 9 + 2/3;

{k = -12/9 = -4/3.

Substituting these values into the equation of the straight line, we obtain the equation of the AC side:

y = -4/3 * x + 9 + 2/3.