# Direct Variation: Equation, Graph, Formula, and Examples

*Ray is a licensed engineer in the Philippines. He loves to write about mathematics and civil engineering.*

## What Is Direct Variation?

Direct variation is a variation in which the quotient of two variables is a constant. If the ratio between two quantities is constant, it means that one quantity varies directly as the other quantity, or the two quantities are in direct variation. In symbols, the direct variation formula is y=kx or k=y/x where k is the constant of variation or proportionality constant.

A simple problem situation will show clearly the idea of a direct relationship between quantities. Say, a motorist drives along the North Expressway at a constant speed of 80 km per hour. How far will he go in 2 hours? In 5 hours? In 3.5 hours? In t hours? The table below shows the relation between the distance traveled by the motorist and the time spent.

Time in Hours | Distance in Kilometers |
---|---|

1 | 80 |

2 | 160 |

3 | 240 |

4 | 320 |

5 | 400 |

t | 80t |

The longer the time, the farther the distance traveled, that is, the distance varies directly as the time (t) or distance and time are in direct variation.

distance/time = 80/1 = 160/2 = 240/3 = 80t / t

This means that the ratio of any specific pair of corresponding values of distance to time is constant. Ratios of successive pairs of values may be equated to form a proportion. Thus, two variables in direct variation are sometimes said to be in direct proportion or to vary proportionally.

## Guidelines for Solving a Direct Variation Equation

- Write a general formula for direct variation that involves the variables and a constant of variation. Write the direct variation formula in the form y = kx, where k ≠ 0.
- Find the value of k in guideline 1 by using the initial data given in the statement of the problem. Before that, make sure that the given problem is a direct variation. If the ratio y/x is constant for all ordered pairs, then it is a direct variation.
- Substitute the value of k found in guideline 2 into the formula of guideline 1, obtaining a specific formula that involves the variables.
- Use the new data to solve the problem.

## Direct Variation Graph

Using the concept of linear equations, the variables in direct variation are positive and so the graph of the variation can be seen on the first quadrant of a coordinate plane. A direct variation may also relate four quantities in proportions as x_{1}/x_{2} = y_{1}/y_{2} or x_{1}y_{2} = x_{2}y_{1}. This equality of ratios is called a proportion. That is, y is directly proportional to x or x is directly proportional to y. Recall that the numbers x_{2} and y_{2} are the means, and y_{1} and x_{1} are the extremes of the proportion. Proportions can be used to solve problems that involve direct variation.

Using the motorist problem given earlier, find the distance (d) traveled in 12 hours. Therefore, the distance traveled in 12 hours is 960 kilometers.

t_{1}/d_{1} = t_{2}/d_{2}

1/80 = 2/160 = 4/320 = 12/d

1/80 = 12/d

d = 960 km

The graph of a direct variation equation y = kx is a line with the following properties.

- The line passes through (0,0)
- The slope of the line is k.

## Example 1: Direct Variation and Ordered Pairs

Which relation is a direct variation that contains the ordered pair (2,7)?

- y = 3.5x
- y = 0.286x
- y = 27x

**Solution**

The given ordered pair (2, 7) represents the generally ordered pair (x, y). This means that x = 2 and y = 7. Recall that the direct variation formula is k = y/x. Substitute the values of x and y to the formula to obtain the variation model.

k = y/x

k = 7/2

k = 3.5

3.5 = y/x

y = 3.5x

**Final Answer**

Therefore, the relation that contains the ordered pair (2,7) is y = 3.5x.

## Example 2: Direct Variation Through a Point

What is the constant of variation, k, of the direct variation, y = kx, through (5, 8)? Through (-3, 2)?

**Solution**

The ordered pair (5, 8) represents an ordered pair of the format (x, y). It means that the value of x is 5 while that of y is 8. Substitute these values to the direct variation formula y = kx in order to obtain the constant of variation, k.

y = kx

8 = k(5)

k = 8/5

The ordered pair (-3, 2) represents an ordered pair of the format (x, y). It means that the value of x is -3 while that of y is 2. Substitute these values to the direct variation formula y = kx in order to obtain the constant of variation, k.

y = kx

2 = k(-3)

k = -2/3

**Final Answer**

The constant of variations k is k = 8/5 and k = -⅔.

## Example 3: Direct Variation Equation Example

**Solution**

The number of kilograms of rice (*r*) that can feed a family varies directly as the number of days (d). So, we have the following proportions.

r_{1}/d_{1} = r_{2}/d_{2}

51/45 = r2/60

Use cross multiplication and solve for r_{2}.

45r_{2} = 51 (60)

r_{2} = 51(60) / 45

r_{2} = 68

**Final Answer**

Thus, 68 kilograms of rice are needed to feed the family for 60 days.

## Example 4: Direct Variation Formula Problem

If 12 workers can make 80 native baskets a day, how many workers are needed to complete a daily order of 100 baskets?

**Solution**

The number of workers (w) varies directly as the number of native baskets (b) that can be made.

w_{1}/b_{1} = w_{2}/b_{2}

12/80 = w2/100

Use cross multiplication and solve for w2.

80w_{2} = 12(100)

w_{2} = 12 (100) / 80

w_{2} = 15

**Final Answer**

Therefore, 15 workers are needed to complete a daily order of 100 baskets.

## Example 5: Finding the Constant of Proportionality

Suppose a variable q is directly proportional to a variable z.

- If q=12 when z=5, determine the constant of proportionality.
- Find the value of q when z=7 and sketch a graph of proportionality.

**Solution**

Since q is directly proportional to a variable z., q=kz where k is a constant of proportionality. Substituting q = 12 and z = 5 gives us the following computation.

12 = 5k

k = 12/5

Since k = 12/5, the formula q = kz has the specific form.

q=12z/5

Thus when z=7, the value of q will be as shown below.

q = 12(7) / 5

q = 84/5

q = 16.8

**Final Answer**

The value of k is equal to 12/5 and q is equal to 16.8 when z = 7.

## Example 6: Writing a Direct Variation Equation

If it is known that y varies directly as x and that y = 32 and x = 4, find the variation constant and the equation of variation.

**Solution**

To express the statement y varies directly as x, we write y=kx. Therefore, by substituting the given values in the derived equation we come up with the following equation.

32 = 4k

k = 8

**Final Answer**

The variation constant is 8 and the equation of variation is given by y = 8x.

## Example 7: Direct Variation Word Problem

The circumference of a circle varies directly with its diameter. If the diameter of 7 centimeters in diameter circle is 7π, what is the circumference of the circle whose diameter is 10 centimeters? 15 centimeters 18 centimeters? 20 centimeters?

The circumference of a circle c varies directly as its diameter d can be expressed as the following equation.

c/d = k

c = kd

Substitute the given value of c and d in the equation.

7π/7 = k

π = k

Express c=kd in terms of {\pi by substituting {\pi} in place of k.

c = πd

Substitute the value of d to solve for x when d = 10 centimeters, d = 15 centimeters, d = 18 centimeters, and d = 20 centimeters.

c = π (10)

c = 10π

c = π (15)

c = 15π

c = π (18)

c = 18π

c = π (20)

c = 20π

**Final Answer**

Therefore, the circumferences are 10π, 15π, 18π, and 20π.

## Example 8: Direct Variation Application

In a downtown office building, the monthly rent for an office is directly proportional to the size of the office. If a 450-square-foot office rents for $1350 per month, then what is the rent for an 800-square-foot office?

**Solution**

Given that the monthly rent is directly proportional to the office size, then the rent, R, varies directly with the area of the office, A. Since 450-square-foot office rents for $1350 per month, substitute to the direct variation formula and find k.

R = kA

1350 = k(450)

k = 3

Now that we have the value of k, which is 3, then write the new equation incorporating the value of k. Then, to get the rent for an 800-square-foot office, insert 800 into the new formula.

R = 3A

R = 3(800)

R = $2400

**Final Answer**

Therefore, an 800-square-foot office rents for $ 2400 per month.

## Example 9: Mortgage Payments as Direct Variation Example

The monthly payment P on a mortgage varies directly with the amount borrowed B. If the monthly payment on a 30-year mortgage is $6 for every $1000 borrowed, find a formula that relates the monthly payment P to the amount borrowed B for a mortgage with these terms. Then, find the monthly payment P when the amount borrowed B is $120,000.

**Solution**

Take note that P varies directly with B. Write the equation translating this mathematical statement. Then, find the variation constant k given the first condition where the monthly payment on a 30-year mortgage is P = $6 for every B = $1000 borrowed.

P = kB

6 = k (1000)

k = 0.006

Solve the value of P given the new B = $120,000 and obtained value of k = 0.006.

P = kB

P = 0.006 (120000)

P = $720

The figure below illustrates the relationship between the monthly payment P and the amount borrowed B.

**Final Answer**

Therefore, the monthly payment P when the amount borrowed B is $120,000 is $720.

## Example 10: Direct Variation as an Nth Power

The distance a ball rolls down an inclined plane is directly proportional to the square of the time it rolls. During the first second, the ball rolls 10 feet. See the figure below.

- Write an equation relating the distance traveled to the time.
- How far will the ball roll during the first three-second?

**Solution**

Let d be the distance in feet the ball rolls and I the time in seconds. Since this is an example of a direct variation as an nth power, write the general formula in the format A = kB^{2}.

d = ki^{2}

Now, since the value of d is equal to 10 when times is equal to 1 second, solve the value of constant of variation k.

d = ki^{2}

10 = k(1)^{2}

k = 10

So, the equation relating distance to time is given by the equation below putting the value of k = 10. When t = 3 seconds, find the distance traveled.

d = 10(3)^{2}

d = 90 feet

**Final Answer**

The ball will roll 90 feet during the first three seconds.

## Example 11: Application Involving Direct Variation

The speed of a racing canoe in still water varies directly as the square root of the length of the canoe.

- If a 20 feet canoe can travel 4 mph in still water, find a variation model that relates the speed of a canoe to its length.
- Find the speed of a 24-feet canoe.

**Solution**

Let s be the speed of the canoe and L be its length. Translate the given mathematical statement to come up with the general variation model.

s = k√L

Then, solve the value of variation constant k, given that s = 4 mph and L = 20 feet.

s = k√L

4 = k√20

k = 0.894

Create a general formula incorporating the value of constant of variation k = 0.894. Then, substitute the length of the 24-feet canoe in finding the new speed.

s = 0.894√L

s = 0.894√24

s = 4.38 mph

**Final Answer**

Therefore, given the variation model s = 0.894√L, the value of the 24-feet canoe is 4.38 mph.

## Example 12: Solving a Direct Variation Problem

The volume of blood, B, in a person's body varies directly as body weight, W. A person who weighs 160 pounds has approximately 5 quarts of blood. Estimate the number of quarts of blood in a person who weighs 200 pounds.

**Solution**

We know that y varies directly as x where y = kx. By changing letters, we can write an equation that describes the given mathematical statement.

B = kW

Use the given values to find k. A person who weighs 160 pounds has approximately 5 quarts of blood. Substitute 160 for W and 5 for B in the direct variation equation. Then solve for k. After obtaining the value of k, substitute the value of k into the general equation.

B = kW

5 = k(160)

k = 0.03125

B = kW

B = 0.03125W

Answer the problem’s question. We are interested in estimating the number of quarts of blood in a person who weighs 200 pounds. Substitute 200 for W in B = 0.03125W and solve for B.

B = 0.03125W

B = 0.03125(200)

B = 6.25

**Final Answer**

A person who weighs 200 pounds has approximately 6.25 quarts of blood.

## Example 13: Direct Variation and Inverse Variation

Does the relation in the table represent direct variation, inverse variation, or neither?

x | y |
---|---|

5 | 2 |

10 | 1 |

15 | 2/3 |

20 | 1/2 |

**Solution**

The main technique in solving this kind of problem is to find y/x for each ordered pair. Always remember that a direct variation suggests that the value of y/x is the same for all ordered pairs. See the table shown below. By observation, the values of y/x are not the same. Therefore, it does not represent a direct variation.

x | y | k=y/x |
---|---|---|

5 | 2 | 2/5 |

10 | 1 | 1/10 |

15 | 2/3 | 2/45 |

20 | 1/2 | 1/40 |

Let us try finding out if it represents an inverse variation. Find the value of xy. By observation, the values of k or xy are the same for all ordered pairs.

x | y | k=yx |
---|---|---|

5 | 2 | 10 |

10 | 1 | 10 |

15 | 2/3 | 10 |

20 | 1/2 | 10 |

**Final Answer**

Therefore, since the given set of ordered pairs are not the same for y/x, they do not represent a direct variation. Since the ordered pairs are the same for xy, then they represent an inverse variation.

## Example 14: Direct Variation Graph

Which graphs show functions with direct variation? Check all that apply.

**Solution**

A direct variation suggests that the value of k or the constant of proportionality is the same for every ordered pair. Look at the highlighted points on every direct variation graph and find the value of y/x. If all y/x values for all ordered pairs are the same, then they represent a direct variation. Tabulate the ordered pairs and results for a more organized solution.

x (Time in Hours) | y (Total Cost in $) | k=y/x |
---|---|---|

1 | 2.4 | 2.4 |

5 | 4 | 0.8 |

x (Quantity in Ounces) | y (Total Cost in $) | k=y/x |
---|---|---|

1 | 0.3 | 0.3 |

5 | 1.5 | 0.3 |

x (Number of Meals) | y (Total Cost in $) | k=y/x |
---|---|---|

1 | 1.2 | 1.2 |

5 | 6 | 1.2 |

x (Number of Persons) | y (Total Cost in $) | k=y/x |
---|---|---|

1 | 2 | 2 |

5 | 8 | 1.6 |

**Final Answer**

Therefore, tables 2 and 3 show functions with a direct variation.

## Example 15: Identifying Which Represents a Direct Variation Function

Which table represents a direct variation function?

x | y |
---|---|

-3 | -4.5 |

-1 | -3 |

2 | -1.5 |

5 | 0 |

10 | 1.5 |

x | y |
---|---|

-3 | -7.5 |

-1 | -2.5 |

2 | 5 |

5 | 12.5 |

10 | 25 |

**Solution**

Recall that a direct variation suggests that the value of y/x is the same for all ordered pairs. See the table shown below showing the solution for y/x for each ordered pair of Table 1. As you can observe, the values of y/x or k are different for every ordered pair. It means that Table 1 does not represent a direct variation.

x | y | k=y/x |
---|---|---|

-3 | -4.5 | 1.5 |

-1 | -3 | 3 |

2 | -1.5 | 0.75 |

5 | 0 | 0 |

10 | 1.5 | 0.15 |

Now, see the table shown below showing the solution for y/x for each ordered pair of Table 2. As you can observe, the values of y/x or k are the same for every ordered pair. It means that Table 2 represents a direct variation.

x | y | k=y/x |
---|---|---|

-3 | -7.5 | 2.5 |

-1 | -2.5 | 2.5 |

2 | 5 | 2.5 |

5 | 12.5 | 2.5 |

10 | 25 | 2.5 |

**Final Answer**

Therefore, only Table 2 represents a direct variation.

## Example 16: Identifying the Graph of a Direct Variation Equation

What is the graph of the direct variation equation y=5/2x?

**Solution and Answer**

The graph of the direct variation equation y = 5/2x is a linear graph with a slope of 5/2. To graph the given direct variation equation, determine two points on the line. For instance, solve for the value of y for x = 0, and x = 2. You could pick other points for your solution, whichever you like.

For x = 0,

y = 5/2 (0)

y = 0

Point = (0, 0)

For x = 2,

y = 5/2 (2)

y = 5

Point = (2, 5)

Plot these points and connect them in a straight line.

## Example 17: Direct Variation Relationship

A professional sports team notes that the ambient crowd noise at home games is directly proportional to the attendance at those games. During one game, the cheering of a maximum capacity crowd of 75,000 fans averages 90 decibels. Identify the constant of proportionality k.

**Solution**

Let a represent attendance and n represent the noise (in decibels) generated by that population. If n varies directly with a, then n = ka. Substitute n = 90 and a = 75000 into the equation and solve for k.

n = ka

90 = k (75000)

k = 90/75000

k = 0.0012

**Final Answer**

The constant of proportionality k of the given problem is k = 0.0012.

## Explore More Topics on Variation

- Inverse Variation: Definition, Formula, Graph and Examples

Learn how to solve inverse variation problems in algebra. This article also includes definitions, formulas, graphs, and examples explaining how to translate mathematical statements of inverse variations to equations and vice versa. - Joint Variation: Solving Joint Variation Problems in Algebra

Learn how to solve joint variation problems in algebra. This article includes definitions and various examples about joint variation and combined variation that will help gauge understanding of the topic.

*This content is accurate and true to the best of the author’s knowledge and is not meant to substitute for formal and individualized advice from a qualified professional.*

**© 2021 Ray**

## Comments

**Miebakagh Fiberesima** from Port Harcourt, Rivers State, NIGERIA. on June 12, 2021:

Ray, this article contain many examples that greatly help the reader. But its a pity that you did not include a book or two for exploring the subject further. Thanks for the interesting read.