Final answer:
The parametric equation of the line is x = 2 + 4t and y = 5 + 8t. The general equation of the line can be obtained by eliminating the parameter t from the parametric equation. The ordered slope at the origin can be found by substituting x = 0 and y = 0 into the general equation and simplifying.
Step-by-step explanation:
The given vector equation of a line is L = {(2-2k, 5-2k): k ∈ R}. To find the parametric equation of the line, we can express it in the form of x = x0 + at and y = y0 + bt, where (x0, y0) is a point on the line and (a, b) is the direction vector. In this case, x0 = 2 and y0 = 5. The direction vector can be obtained by subtracting the coordinates of two points on the line, which gives (2, 5) - (-2, -3) = (4, 8). Therefore, the parametric equation of the line is x = 2 + 4t and y = 5 + 8t.
The general equation of a line can be obtained by eliminating the parameter t from the parametric equation. By solving the two equations x = 2 + 4t and y = 5 + 8t simultaneously, we can express t in terms of x and y. Then substitute the value of t into one of the equations (x = 2 + 4t or y = 5 + 8t) to obtain the general equation of the line.
The ordered slope at the origin is the slope of the line when it passes through the origin (0, 0). We can find the slope by substituting x = 0 and y = 0 into the general equation of the line obtained previously, and then simplify.