In triangle ABC, the vertices have coordinates of point A (4; 2), point B (1; 5), point C (-2; 6)

In triangle ABC, the vertices have coordinates of point A (4; 2), point B (1; 5), point C (-2; 6). Write the equations for side AB, height BK and median CM

The coordinates of the vertices of the triangle are given: A (4.2), B (1.5), C (-2.6).

Equation of straight line AB.

The canonical equation of the straight line:

(x – 4) / (1 – 4) = (y – 2) / (5 – 2)

(x – 4) / (- 3) = (y – 2) / (3)

y = -x + 6 or y + x – 6 = 0

Triangle median equation

Let us denote the midpoint of the side AB by the letter M. Then the coordinates of the point M can be found by the formulas for dividing the segment in half.

xm = (xA + xB) / 2 = (4 + 1) / 2 = 5/2

ym = (yA + yB) / 2 = (2 + 5) / 2 = 7/2

M (5/2; 7/2)

We find the median equation CM using the formula for the equation of a straight line passing through two given points. The median CM passes through the points C (-2; 6) and M (5/2; 7/2), therefore:

The canonical equation of the straight line:

(x + 2) / (5/2 – (-2)) = (y – 6) / (7/2 – 6)

(x + 2) / (9/2) = (y – 6) / (-5 / 2)

y = -5 / 9 x + 44/9 or 9y + 5x – 44 = 0

Height equation through vertex B

The straight line passing through the point N0 (x0; y0) and perpendicular to the straight line Ax + By + C = 0 has a direction vector (A; B) and, therefore, is represented by the equations:

(x – x0) / A = (y – y0) / B

Canonical equation of line AC:

(x – 4) / (-2 – 4) = (y – 2) / (6 – 2)

(x – 4) / (- 6) = (y – 2) / 4

y = -x + 6 or y + x – 6 = 0

y = (-2 / 3) x + 14/3 or 3y + 2x – 14 = 0

Let’s find the equation of height through the vertex B:

(x – 1) / 2 = (y – 5) / 3

y = (3/2) x + 7/2 or 2y -3x – 7 = 0