In a triangle with vertices A (-2; 0), B (2; 6) and C (4; 2), the height VD and median BE are drawn.

In a triangle with vertices A (-2; 0), B (2; 6) and C (4; 2), the height BD and median BE are drawn. Write the equations for the AC side, BE median and BD height.

The equation of a straight line passing through two points on a plane is:

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

In order to use this formula, we find the coordinates of the midpoint of the AC side, the point E. The coordinates of the point E are found by the formulas for dividing the segment in half:

XE = (XA + XB) / 2, YE = (YA + YB) / 2, XE = (-2 + 4) / 2 = 1, YE = (0 + 2) / 2 = 1. E (1; 1 ).

Let’s make the equation of the AC side:

(x + 2) / (4 + 2) = (y – 0) / (0 + 2), (x + 2) / 6 = y / 2 → x – 3y + 2 = 0 → y = 1 / 3x + 2/3.

Let’s compose the equation for the median BE:

(x – 2) / (1 – 2) = (y – 6) / (1 – 6) → -5x + 10 = – y + 6 → -5x – y – 4 = 0.

Let’s make an equation for the height of the air pressure. It is perpendicular to the AC side, so we use the formula of a straight line passing through a given point B, perpendicular to this AC straight line:

(y – y1) = -1 / k (x – x1) → (y – 6) = -1 / (1/3) (x – 2) → y – 6 = -3x + 6 → y + 3x – 12 = 0.