1. Lesson 5-2 Direct Variation
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Lesson 5-2 Direct Variation
Direct Variation (an equation): y = kx where k does not equal Constant of Variation (a number): The constant of variation is k,
k is a number.
**k is the slope of the line!

2. An equation is a direct variation if:
its graph is a line that passes through (0,0)

4. Is an Equation a Direct Variation
Is an Equation a Direct Variation? If it is, find the constant of variation.
5x + 2y = 0 Solve for y!!!!
1. Subtract 5x.
2. Divide by 2.
Yes, it’s a direct variation.
Constant of variation, k, is

5. Is an Equation a Direct Variation
Is an Equation a Direct Variation? If it is, find the constant of variation.
5x + 2y = 9 Solve for y!!!!
1. Subtract 5x.
2. Divide by 2.
No, it’s not a direct variation.
It’s not in the form y = kx.

6. Is an Equation a Direct Variation
Is an Equation a Direct Variation? If it is, find the constant of variation.
7y = 2x
Yes, it is a direct variation.
The constant of variation, k, is

7. Using Direct Variation to find unknowns (y = kx)
Given that y varies directly with x, and y = 6 when x=-5,
Find y when x = -8.
1. Set-up a proportion!!
2. Cross Multiply
-5y = -8*6
-5y = -48
y= 48/5 or 9.6
Therefore:
x = -8 when y = 9.6

9. Real-World Problem Solving
Your distance from lightning varies directly with the time it takes you to hear thunder. If you hear thunder 10 seconds after you see the lightning, you are about 2 miles from the lightning. Write an equation for the relationship between time and distance.

10. Real-World Problem Solving
Relate:
The distance varies directly with the time. When x = 10, y = 2.
Define:
Let x = number of seconds between seeing lightning and hearing thunder.
Let y = distance in miles from lightning.

11. Real-World Problem Solving
y = kx Use general form of direct variation.
2 = k(10) Substitute 2 for y and 10 for x.
Solve for k.
Write an equation using the value for k.

12. Real-World Problem Solving
The force you must apply to lift an object varies directly with the object’s weight. You would need to apply lb of force to a windlass to lift a 28-lb weight. How much force would you need to lift 100 lb?
Relate: A force of lifts 28 lb. What lifts 100 lb?

13. You need about 2.2 lb of force to lift 100 lb.
Use a proportion.
Cross multiply n = 62.5
Solve for n. n ≈ 2.2
You need about 2.2 lb of force to lift 100 lb.