We can establish the general equation of a line from the three-point alignment condition.
Given a straight r, being THE(xTHE, yTHE) and B(xB, yB) known and distinct points of r and P(x, y) a generic point, also of rbeing THE, B and P aligned, we can write:
Doing yTHE - yB = a, xB - xTHE = b and xTHEyB - xByTHE= c, as a and b are not simultaneously null we have:
ax + by + c = 0
(general equation of line r)
This equation relates x and y to any point P generic straight line. So given the point P(m, n):
if m + bn + c = 0, P is the point of the line;
if m + bn + c 0, P It is not the point of the line.
Follow the examples:
Let's consider the general equation of the line r that goes through THE(1, 3) and B(2, 4).
Considering one point P(x, y) of the line, we have:
Let's check if the points P(-3, -1) and Q(1, 2) belong to line r of the previous example. Overriding the coordinates of P at x - y + 2 = 0, we have:
-3 - (-1) + 2 = 0 -3 + 1 + 2 = 0
Since equality is true, so P r.
Overriding the coordinates of Q at x - y + 2 = 0, we get:
1 - 2 + 2 0
Since equality is not true, so Q r.Next: Segmental Equation