# ASSIGNMENT 10

Write the parametric equations of a line segment through (7,5) with a slope of 3. Graph the line segment using your equations.

Using algebra, we can determine the y - intercept needed to write the cartesian equation of the line containing the point (7,5) with a slope of 3

5 = 3(7) + b

5 = 21 + b

b = -16

Since the y- intercept is -16 and the slope is 3, the cartesian equation is y = 3x - 16.

To write this equation parametrically, we need another point on the line segment. If I let x = 8, then y = 8 using our cartesian equation. We can determine parametric equations as follows:

(x, y) = (1 - t) (7, 5) + t (8, 8)

(x, y) = (7 - 7t + 8t, 5 - 5t + 8t)

(x, y) = (t + 7, 3t + 5)

The graph yields a line segment for values of 0 < t < 2.

Suppose we chose the point (5, -1). How would that change the parametric equations? How would that change the graph?

(x, y) = (1-t) (7, 5) + t (5, -1)

(x, y) = (7 - 2t, 5 - 6t)

The segment produced has the point (7, 5) as the endpoint, but the segment still lies on the line y = 3x - 16.