The triangle is given by the vertices A (7; -6) B (-2; -2) and C (1; 2) find the equation of the straight line AM

The triangle is given by the vertices A (7; -6) B (-2; -2) and C (1; 2) find the equation of the straight line AM parallel to the side BC.

We have points:

A (7; -6), B (-2; -2), C (1; 2).

First, let’s write the equation of the line BC.

The equation has the formula y = k * x + b. Substitute the coordinates of points B and C and solve the system:

-2 = -2 * k + b;

2 = k + b;

Subtract the first from the second equation:

k + 2 * k = 2 + 2;

3 * k = 4;

k = 4/3;

Since line AM is parallel to line BC, their slopes are equal.

Linear AM equations:

y = 4/3 * x + b;

Substitute the coordinates of point A:

-6 = 4/3 * 7 + b;

b = -6-28/3;

b = -46/3;

y = 4/3 * x – 46/3 is the equation of the line AM.