Slope of the Secant Line Formula

A secant line is a straight line that connects two points on the curve of a function f(x). A secant line, also known as a secant, is basically a line that passes through two points on a curve. It tends to a tangent line when one of the two points is brought towards the other one. It is used to evaluate the equation of tangent line to a curve at a point only and only if it exists for a value (a, f(a)). 

Slope of the Secant Line Formula

The slope of a line is defined as the ratio of change in y coordinate to the change in x coordinate. If there are two points (x₁, y₁) and (x₂, y₂) connected by a secant line on a curve y = f(x) then the slope is equal to the ratio of differences between the y-coordinates to that of the x-coordinates. The slope value is represented by the symbol m.

m = (y₂ – y₁)/(x₂ – x₁)

If the secant line is passing through two points (a, f(a)) and (b, f(b)) for a function f(x), then the slope is given by the formula:

m = (f(b) – f(a))/(b – a)

Sample Problems

Problem 1. Calculate the slope of a secant line that joins the two points (4, 11) and (2, 5).

Solution:

We have, (x₁, y₁) = (4, 11) and (x₂, y₂) = (2, 5)

Using the formula, we have

m = (y₂ – y₁)/(x₂ – x₁)

= (5 – 11)/(2 – 4)

= -6/(-2)

= 3

Problem 2. The slope of a secant line that joins the two points (x, 3) and (1, 6) is 7. Find the value of x.

Solution:

We have, (x₁, y₁) = (x, 3), (x₂, y₂) = (1, 6) and m = 7

Using the formula, we have

m = (y₂ – y₁)/(x₂ – x₁)

=> 7 = (6 – 3)/(1 – x)

=> 7 = 3/(1 – x)

=> 7 – 7x = 3

=> 7x = 4

=> x = 4/7

Problem 3. The slope of a secant line that joins the two points (5, 4) and (3, y) is 4. Find the value of y.

Solution:

We have, (x₁, y₁) = (5, 4), (x₂, y₂) = (3, y) and m = 4

Using the formula, we have

m = (y₂ – y₁)/(x₂ – x₁)

=> 4 = (y – 4)/(3 – 5)

=> 4 = (y – 4)/(-2)

=> -8 = y – 4

=> y = -4

Problem 4. Calculate the slope of a secant line for the function f(x) = x² that joins the two points (3, f(3)) and (5, f(5)).

Solution:

We have, f(x) = x²

Calculate the value of f(3) and f(5).

f(3) = 3² = 9

f(5) = 5² = 25

Using the formula, we have

m = (f(b) – f(a))/(b – a)

= (f(5) – f(3))/ (5 – 3)

= (25 – 9)/2

= 16/2

= 8

Problem 5. Calculate the slope of a secant line for the function f(x) = 4 – 3x³ that joins the two points (1, f(1)) and (2, f(2)).

Solution:

We have, f(x) = 4 – 3x³

Calculate the value of f(1) and f(2).

f(3) = 4 – 3(1)³ = 4 – 3 = 1

f(5) = 4 – 3(2)³ = 4 – 24 = -20

Using the formula, we have

m = (f(b) – f(a))/(b – a)

= (f(2) – f(1))/ (2 – 1)

= -20 – 1

= -21

Problem 6. The slope of a secant line that joins the two points (x, 7) and (9, 2) is 5. Find the value of x.

Solution:

We have, (x₁, y₁) = (x, 7), (x₂, y₂) = (9, 2) and m = 5.

Using the formula, we have

m = (y₂ – y₁)/(x₂ – x₁)

=> 5 = (2 – 7)/(9 – x)

=> 5 = -5/(9 – x)

=> 45 – 5x = -5

=> 5x = 50

=> x = 10

Problem 7. The slope of a secant line that joins the two points (1, 5) and (8, y) is 9. Find the value of y.

Solution:

We have, (x₁, y₁) = (1, 5), (x₂, y₂) = (8, y) and m = 9

Using the formula, we have

m = (y₂ – y₁)/(x₂ – x₁)

=> 9 = (y – 5)/(8 – 1)

=> 9 = (y – 5)/7

=> y – 5 = 63

=> y = 68