1. Basic Examples
2. Graphing Lines
3. Intersection of Lines
4. Measurement Conversions
5. Word Problems

Basic Examples

Example 1: Find the equation of a line with a slope of 2, passing through the point (3,-2). What is the y-intercept?

Solution: The point slope form of the equation gives:

y = 2x - 8,

so the y-intercept is -8.

Example 2: Find the equation of a line passing through the points (-2,1) and (3,-2).

Solution: The slope is

Recall the point slope equation of the line is

y - y₀ = m(x - x₀)

so

 

Parallel and Perpendicular Lines

Consider two lines given by the equations:

The two lines are parallel if the slopes are equal, so

and the y-intercepts are different. (If b₁ = b₂, then the lines are the same.)

The two lines are perpendicular if the slopes are negative reciprocals of each other, that is

Example 3: Find the equation of the line perpendicular to the line 5x + 3y = 6, passing through the point (-2,1). Sketch a graph of both lines.

Solution: Line #1 satisfies

The slope of the perpendicular line is the negative reciprocal of Line #1 or

The point slope equation of the line (Line #2) is

Graphing Lines

sourcecode for plotexample1c.java

Intersection of Lines

Example 4: Find the intersection of the line parallel to the line y = 2x passing through (1,-3) and the line given by the formula 3x + 2y = 5.

Solution: The first line is parallel to y = 2x, so has a slope m = 2.

The point slope equation gives:

Substituting into the formula for the second line gives

or

 Substituting the x value into the first line equation gives

 The point of intersection is

 source code for crosspoint.java

Measurement Conversions

Example 5: Find the weight of a 175 pound man in kilograms.

Solution: Since 1 kg = 2.2046 lb,

Example 6: Suppose a ball is thrown at a speed of 95 miles per hour. Find the speed of this ball in meters per second.

Solution: Since 1 mile = 1609.3445 m and since there are 3600 s in 1 hour,

Use the conversion javascript to check your answers.

Word Problems

Example 6: The growth curve in the lecture notes shows that over a short period of time, using a straight line to estimate growth is quite reasonable. Suppose that a population of white sea urchins (Lytechinus pictus) has a mean diameter of 28 mm at the beginning of June and 33 mm at the beginning of July. Estimate the mean diameter for the population of Lytechinus pictus on June 20, July 10, August 1, and August 15. Which estimates do you trust more and why?

Solution:

• There are 30 days in June, so assume two data points, (0, 28), and (30, 33), where t corresponds to the number of days, and d corresponds to the mean diameter in mm
• Assume a linear relationship between time and diameter
• The slope is

• From the point slope equation

• Note that b, the d-intercept, is essentially the initial diameter measurement
• The slope m must then be the growth rate

Linear Predictions

• For the dates June 20, July 10, August 1, and August 15 correspond to the points (19, 31.2), (39, 34.5), (61, 38.2), and (75, 40.5), respectively
• The estimate for June 20 is probably the most accurate because it falls between the two measurements actually taken
• Also, growth estimates are more accurate over shorter time intervals

 

Example 7: The pressure of air delivered by the regulator to a Scuba diver varies linearly with the depth of the water. When the diver is at 33 ft., the regulator delivers 29.4 psi (pounds/square inch), while at 66 ft., the regulator delivers 44.1 psi. Find the pressure of air delivered at the surface (0 ft.), at 50 ft., and at 130 ft. (the maximum depth for recreational diving).

Solution:

• The measurements are (33, 29.4), and (66, 44.1)
• Let d be the depth in feet, and p be the regulator pressure in psi
• The slope is

• The equation of the line is

• From the linear relationship, the surface (d = 0), the air pressure is 14.7 psi
• This calculation can be represented as a point, (0, 14.7)
• For depths of d = 50 ft and 130 ft, the model gives (50, 36.95), and (130, 72.55), respectively.