The vertices of the triangle A (6; 2), B (6; 3), C (7; 1) are given. Equate the height of a triangle from vertex A.

1. Find the coordinates of the BC vector:

BC (7-6; 1-3) = BC (1; -2)

2. The equation of a straight line perpendicular to a vector with coordinates (K, N) has the form: Kx + Ny + C = 0

In this example: 1 * x -2 * y + C = 0

3. Find S. Since. the desired straight line passes through point A (6; 2), then substituting the coordinates of the point into the equation, we find the parameter C:

1 * 6 – 2 * 2 + C = 0

2 + C = 0

C = -2

4. Substitute the value of C into the equation of the straight line:

1 * x – 2 * y – 2 = 0

x – 2y -2 = 0

Let us express y through x:

y = x / 2 -1 – the desired equation