# Draw the secants of a curve (line?)

1. Jul 28, 2009

### zeion

1. The problem statement, all variables and given/known data
An automobile's distance is given by d = 30t (t in seconds, d in metres):

"Draw the secants from the point (2, 60) to each of the points on the graph ending at t = 3, 2.5, 2.1, 1.01"

"Determine the equations of each of the secants above."

2. Relevant equations

3. The attempt at a solution

To my understanding, d = 30t is a line.
According to the text, "A line that intersects a curve in two places is called a secant".
How can a line have any secants?

Last edited: Jul 28, 2009
2. Jul 28, 2009

### Staff: Mentor

A curve in its most liberal interpretation can also include a straight line.

3. Jul 28, 2009

### tiny-tim

How can a point end?

4. Jul 28, 2009

### zeion

Uh I'm not sure, but I'm guessing it's asking me to draw the secant from (2, 60) to these points (3, d(3)), (2.5, d(2.5)), etc.

Though I still don't see the point since they're all gonna overlap anyway lol.

5. Jul 28, 2009

### ideasrule

Perhaps the point is to show that for a straight line, all secant lines have the same slope.