Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
4. Point-Slope Form
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Exercise 29 Page 319

Use the information about altitude and boiling point to find two points.

Equation: b=- 0.0018a +212
Boiling Point of Water: 207.5^(∘)F

Practice makes perfect

To begin, let's make sense of the given information. It is given that the temperature at which water boils depends on altitude. This means that altitude a is the independent variable, and boiling point b is the dependent variable. We can write the given information as coordinates.

  • Altitude 8000 feet, boiling point 197.6^(∘) F: (8000, 197.6)
  • Altitude 4500 feet, boiling point 203.9^(∘) F: (4500, 203.9)
We can write an equation of b in terms of a in slope-intercept form. We will begin by finding the slope of the line that connects the points by using our points.
m=y_2-y_1/x_2-x_1
m=203.9- 197.6/4500- 8000
m=6.3/- 3500
m=- 6.3/3500
The slope - 6.33500 means that for every 3500 feet higher in altitude the boiling point decreases by 6.3^(∘). We can simplify the expression of the slope more.
m=- 6.3/3500
Simplify right-hand side
m=- 63/35000
m=- 63/35000
m=- 9/5000
m=- 18/10000
m=- 0.0018
At this point, we have the following equation. b=- 0.0018a +n To determine n we can substitute any of the given points into the equation for a and b and solve for n. We arbitrarily choose to use the point (4500,203.9).
b=- 0.0018a +n
203.9=- 0.0018* 4500 + n
Solve for b
203.9 = - 8.1 +n
212=n
n=212
The y-intercept of 212 means that at an altitude of 0 feet the boiling point is 212^(∘). Our complete equation is as follows. b=- 0.0018a +212 We can use the equation to determine the boiling point at an altitude of 2500 feet. To do this, we will substitute a=2500 into the equation and solve for b.
b=- 0.0018a +212
b=- 0.0018* 2500 +212
b = - 4.5 +212
b = 207.5
At an altitude of 2500 feet, the boiling point is 207.5^(∘) F.