In a triangle with vertices A (- 2; 0), B (2; 6) and C (4; 2), the median BE is drawn. Write the equation for the median BE.

Since BE is the median, point E is the midpoint of the AC side. Let’s calculate the coordinates of point E:

A (-2; 0), C (4; 2).

E ((- 2 + 4) / 2; (0 + 2) / 2) = E (1; 1).

BE is straight. The straight line equation has the form y = kx + b. This line must pass through points B and E. Substitute the coordinates of points B and E into the equation of the line and calculate the value of the coefficients k and b.

B (2; 6) x = 2, y = 6: 6 = 2k + b.

E (1; 1) x = 1, y = 1: 1 = k + b.

The result is a system of equations: 2k + b = 6; k + b = 1.

Let us express k from the second equation and substitute it into the first equation:

k = 1 – b.

2 (1 – b) + b = 6.

2 – 2b + b = 6.

-b = 4.

b = -4.

Find the coefficient k: k = 1 – b = 1 – (-4) = 5.

The equation of the straight line BE has the form: y = 5x – 4.

Move everything to the left side: 5x – y – 4 = 0.

Answer: 5x – y – 4 = 0.