1. MAT 1033C – INTERMEDIATE ALGEBRA /CRN 10682
Liudmila Kashirskaia
August 29,2018

2. Chapter 2.3 The Slope of a Line The slope m of the line passing through the points (x1, y1) and (x2, y2) is where x1 ≠ x2. That is, slope equals rise over run.

3. Example 1: Find the slope of the line passing through the points (0, -4) and (2, 2). Plot these points and graph the line. Interpret the slope.
Solution
Graph the line passing through these points. The slope indicates that the line rises 3 unit for every 1 units of run.

4. Slope: positive, negative, horizontal, undefined
A line with positive slope rises from left to right, m>0

5. A line with negative slope falls from left to right, m<0

6. A horizontal line has a zero slope, m=0

7. A line with undefined slope is a vertical line, m is undefined.

8. Example 2: Find the slope of the line passing through each pair of points, if possible. a. (-3, 2), (2, 2) b. (3, -1), (3,4 )
Solution a b

9. Example 3: Sketch a line passing through the point (1, 2) and having slope 3/4.
Solution:
Start by plotting (1, 2). The slope is ¾ which means a rise (increase) of 3 and a run (horizontal) of 4. The line passes through the point (1 + 4, 2 + 3) = (5, 5).

10. SLOPE-INTERCEPT FORM of a Line
The line with slope m and
y-intercept (0, b) is given by
y = mx + b,
the slope-intercept form of a line.

11. Example 4 : For the graph write the slope-intercept form of the line
Example 4 : For the graph write the slope-intercept form of the line. Solution The graph intersects the y-axis at (-3), so the y-intercept is (-3). The graph falls 1 units for each 1 unit increase in x, the slope is –1. The slope intercept-form of the line is y = mx + b y = –x -3.

12. Example 5: Identify the slope and y-intercept for the three lines y=2x-1, y=2x and y= 2x+1. Compare the lines.
The slopes are all 2,
the y- intercepts are
(0, -1), (0,0), (0,1).
The lines are parallel.

13. Example 6: The graph represents the gallons of water in a small swimming pool after x hours. Assume that a pump can either add water or remove water from the pool. 1. Estimate the slope of each line segment m1=125, m2=0, m3= - 125

14. Example 6(cont): 2. Interpret each slope as a rate of change
m1: The pump added water at the rate of 125 gallons per hour
m2: The pump neither added nor removed water
m3: The pump removed water at the rate of 125 gallons per hour

15. The pump was turned off for 4 hours
Example 6(cont): 3. Describe what happened to the amount of water in the pool.
The pool was empty
The pump added 500 gallons of water over the first 4 hours at the rate of 125 gallons per hour
The pump was turned off for 4 hours
The pump removed 500 gallons of water over the last 4 hours at a rate of 125 gallons per hour and the pool was empty

16. Chapter 2.4 Equations of Lines and Linear Models Working with the Point-Slope Form of a Line POINT-SLOPE FORM The line with slope m passing through the point (x1, y1) is given by y = m(x – x1)+ y1 or equivalently, y – y1 = m(x – x1), the point-slope form of a line.

17. Example7: Find an equation of the line passing through (-1, 2) and (5, -3). Solution
First find the slope of the line.
Substitute −5/6 for m and (-1, 2) for x and y in the
slope intercept form.
The point (5,- 3) could be used instead.
The slope-intercept form is
y = mx+ b

18. Writing Equations of Horizontal and Vertical Lines
The equation of a horizontal line with y-intercept (0, b) is
y = b.
The equation of a vertical line with x-intercept (h, 0) is x = h.
Example 8:
Find equations of the vertical and horizontal lines that pass through the point (−5, -2). Graph these two lines.
Solution
x = −5
y = -2

19. Writing Equations of Parallel and Perpendicular Lines Two lines with the same slope are parallel. Example 9: Find the slope-intercept form of a line parallel to y = 2x + 1 and passing through the point (4, 1). Solution. The line has a slope of 2 any parallel line also has slope 2. y=2(x-4)+1=2x-7

20. PERPENDICULAR LINES y= − x+5
Two lines with nonzero slopes m1 and m2 are perpendicular if m1m2 = −1.
Example 10: Find the slope-intercept form of the line perpendicular to y = x – 5 passing through the point
(2, 3).
Solution :
m1m2 = −1, m2 = −1, y= − 1(x − 2)+3,
y= − x+5