If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# Slope of a line secant to a curve

What is the slope of a line between two points on a curve?  This is called a "secant" line. Created by Sal Khan.

## Want to join the conversation?

• What are the main differences between a tangent line and a secant line in a curve? •  A tangent line touches the curve at one point and has the same slope as the curve does at that point. A secant line intersects at 2 or more points and has a slope equal to the average rate of change between those points.
• I'm having difficulty understanding the concept of a secant line as it pertains to a sine graph. Specifically, there is a problem in the "Slope of secant lines" exercise, where there are four questions. In each you are asked to evaluate the rates of change between secant lines for four different points.

The second question asks whether [Sin 5/2 Pi - Sin 1/2 Pi / 5/2 Pi - 1/2 Pi] is greater than, less than or equal to [Sin 2/3 Pi - Sin 1/3 Pi / 2/3 Pi - 1/3 Pi]. Evaluating the first part, I see that 5/2 Pi is the same as 1/2 Pi, and therefore the change in Y is zero. However, when evaluating the second part, it seems to me that Delta Y/Delta X is Sin 1/3 Pi / 1/3 Pi.

But the lesson tells me that the secant line is horizontal, which I do not understand. Can anyone help me with understanding secant lines on a sine graph? I did all of the trig lessons and most of the videos already, but if anyone can clarify this concept or point me in the direction of some valuable review materials, that would be legit. ( : • You're correct that sin(5π/2) - sin(π/2) = 0, so the slope of the first line is 0 making it horizontal.

Let's look at the second slope. In the numerator it has sin(2π/3) - sin(π/3).
Can we simplify that?

Remember that sin x = sin(π - x), so
sin(2π/3) = sin(π - 2π/3) = sin(π/3).

The numerator then becomes
sin(2π/3) - sin(π/3) = sin(π/3) - sin(π/3) = 0.

So, both slopes are horizontal and thus equal.

Maybe the mistake you made was that you thought that
sin(2π/3) - sin(π-3) = sin(2π/3 - π/3)?
The two aren't equal.
• Does a secant line always only touch 2 points on the graph? (is this not doable for some periodic functions where it would hit it 3 times + for example?) • As the term is typically used in calculus, a secant line intersects the curve in two places locally -- it may or may not intersect the curve somewhere else. So the requirement of just two intersections applies just to the small region of interest and is not a strict requirement for regions you are not concerned with at the moment.

Note: in some fields of geometry the requirement of exactly two points of intersection is much more strict than what we usually have in calculus.
• what is the difference between a chord and a secant? • Why does he use x not and sub 1 in the first graph but sub 1 and sub 2 in the second graph? Is there any special situation that those are reserved for? • Can't we calculate slope in the same way as we do for lines?We have coordinates in the curve. • At Sal mentions interval. Can someone explain to me what this 'interval' means? I keep on hearing this term more and more often. • Can't we call that secant line is a chord? Or is it both? Or one in specific?
(1 vote) • In the end, he ends up with finding the slope of a line with points (X0, Y0), (X1, Y1). This way, when I know two points on the line, I can find out the equation of that line. Now, that line, suppose line m, was secant to the curve, suppose L. So Line m is secant to the curve L. Now, let's imagine a line n, which is parallel to line m and tangent to the curve L. So i know the slope of the line m is equal to the slope of the line n, because they are parallel. Hence, can I find out the point at which the tangent n intersects the curve L using just algebra? I think I can, using some concepts in co-ordinate geometry. Suppose that point is (X3, Y3). So without actually using Differential calculus, we have the value of the instantaneous point. So if this was the example of a car running on a road and a graph or it's speed on y and time on x axis, then i guess that at X3, dy/dx = Y3....i.e. instantaneous velocity at X3 = Y3. So we solved it without using calculus. Someone please tell me if i'm correct or wrong!  