Given two points A and B, any point P on the line can be located by the signed ratio t = AP/AB. This ratio t defines an affine coordinate system on the line; for any point P, the coordinate of P is the ratio t(P) = AP/AB. . The point A has coordinate t(A) = 0 and the point B has coordinate t(B) = 1. There is a different affine coordinate system on the line for each choice of A and B.
Also if we start with any real number t, we can find a unique P on the line with this signed ratio t. The map taking t to the point P(t) with ratio AP(t)/AB = t, is a parametrization of the line. This is called the affine parametrization of the line with origin A and unit point B.
In this Java Sketchpad figure, you can move A and B and also P to see the corresponding values of t. The ratio s = BP/BA is also shown. This s is also an affine coordinate system with origin B and unit A.
Observe that for P on the line, 1 = (AP + PB)/AB = t + s, so s = 1-t for all P on the line, including when t or s is negative.
Origin and Vector Parametrization of a Line
In this figure is also a button that will show a point O. If we think of O as an origin point in the plane, we see that the ratios t and s are also vector coordinates x and y, so that OP = xOB + yOA. The significant point is that we can see from the geometry of parallel lines and similar triangles that for P on the line, no matter what point O is, the vector equation OP = uOB + vOA is true for the same u = t and v = s, with u + v = 1!
This means that in any (x,y) coordinate system in the plane or any (x,y,z) coordinate system in 3-space, the coordinates of P can be computed from the coordinates of A and B by viewing them as position vectors.
For example, if t = 1/3 and A = (1, 30, -13) and B = (100, -18, 32), then P = (2/3)A + (1/3)B =
((1/3)*(2+100), (1/3)*(60-18), (1/3)*(-26+32)) = (34, 14, 2).
Note: There is no reason in general for the numbers to be "nice" as in this example. Also, t need not even be a rational number. It could be the square root of 2, or e, or any other real number.
Exercise. For what point P on line AB, does P lie on the plane with equation x+y+z = 1? Hint: Let P(t) = (1-t)A+tB = (x,y,z). Substitute into the equation for the plane and solve for t = t0. Then compute P(t0).
For any triangle OAB, one can define affine coordinates (u,v) of any point P in the plane as was done above. Construct the lines through P parallel to OA and AB and intersect these new lines with OA and OB to form a parallelogram OP'PP''. Then the ratios u = OP'/OB and v = OP''/OA are the affine coordinates of P in the affine coordinate system with origin O, first unit point B and second unit point A.
Important! Notice that in this case, the coordinates u and v of P will change if you move the origin. The coordinates will always change unless u+v = 1, which only occurs when P is on line AB.
Exercise. If O = (1, 0, 0), B = (0, 1, 0), C = (0, 0, 1) and for what P is u = 2 and v = 3?
Exercise: In the plane, let O = (1, 1), B = (2, 0), A = (-1, -1). If P = (2, 3) what are u and v?
James King, 1/10/2003