MATH 140 -- Intermediate Algebra -- Exercise Solutions
Section: 3.5

2. Write the equation of the line with slope and y-intercept (0, - 6).
Solution: The slope-intercept form of the equation of a line is y = mx + b where m is the slope and b is the y-intercept.
In this problem we have y = x + (- 6) = x - 6.
The equation is y = x - 6.

4. Write the equation of the line with slope - 3 and y-intercept (0, - ).
Solution: The slope-intercept form of the equation of a line is y = mx + b where m is the slope and b is the y-intercept.
In this problem we have y = - 3x + (- ) .
The equation is y = - 3x - .

13. Write the equation of the line that passes through the point (1, 2) with slope 3 .
Solution: The point-slope form of the equation of a line is y - y1 = m(x - x1) where the coordinates of the known point are (x1, y1) and m is the known slope.
 In this problem we have y - 2= 3(x - 1) = 3x - 3.   The equation is y - 2 = 3x - 3
Add 2 to both sides of the equation to obtain the more desirable slope-intercept form of the equation of this line
y = 3x - 1.

14. Write the equation of the line that passes through the point (5, 1) with slope 4 .
Solution: The point-slope form of the equation of a line is y - y1 = m(x - x1) where the coordinates of the known point are (x1, y1) and m is the known slope.
 In this problem we have y - 1= 4(x - 5) = 4x - 20.   The equation is y - 1 = 4x - 20
Add 1 to both sides of the equation to obtain the more desirable slope-intercept form of the equation of this line
y = 4x - 19.

16. Write the equation of the line that passes through the point (2, - 4) with slope - 4 .
Solution: The point-slope form of the equation of a line is y - y1 = m(x - x1) where the coordinates of the known point are (x1, y1) and m is the known slope.
 In this problem we have y - (- 4)= - 4(x - 2) = - 4x + 8.   The equation is y + 4 = - 4x + 8
Add - 4 to both sides of the equation to obtain the more desirable slope-intercept form of the equation of this line
y = - 4x + 4.

18. Write the equation of the line that passes through the point (- 9, 4) with slope .
Solution: The point-slope form of the equation of a line is y - y1 = m(x - x1) where the coordinates of the known point are (x1, y1) and m is the known slope.
 In this problem we have y - 4 = (x - [- 9]) = x + ()(9) =x + 6.   The equation is y - 4 = x + 6
Add 4 to both sides of the equation to obtain the more desirable slope-intercept form of the equation of this line
y = x + 10.

26. Write the equation of the line that passes through (3, 0) and (7, 8)
Solution: Begin by computing the slope
                   
Now use the slope-intercept form of the equation of a line y - y1 = m(x - x1) to find the equation of the line through (3, 0) with slope 2.   This gives the equation y - 0 = 2(x - 3). Simplification yields
y = 2x - 6 which is f(x) = 2x - 6 when functional notation is used.

28. Write the equation of the line that passes through (7, - 4) and (2, 6)
Solution: Begin by computing the slope
                   
Now use the slope-intercept form of the equation of a line y - y1 = m(x - x1) to find the equation of the line through (2, 6) with slope - 2.   This gives the equation y - 6 = - 2(x - 2). Simplification yields
y = - 2x + 10 which is f(x) = - 2x + 10 when functional notation is used.

30. Write the equation of the line that passes through (- 9, - 2) and (- 3, 10)
Solution: Begin by computing the slope
                   
Now use the slope-intercept form of the equation of a line y - y1 = m(x - x1) to find the equation of the line through (- 3, 10) with slope 2.   This gives the equation y - 10 = 2(x - (- 3)). Simplification yields
y - 10 = 2x + 6 which simplifies to y = 2x + 16 which is f(x) = 2x + 16 when functional notation is used.

48. Find the equation of the line through (1, 5) parallel to the graph of f(x) = 3x - 4.
Solution: Two lines are parallel if they have the same slopes so we are looking for the line through (1, 5) with slope 3 (the same slope as f(x) = 3x - 4). Use the point slope form of the equation of a line y - y1 = m(x - x1) to obtain
y - 5 = 3(x - 1) which can be written as y = 3x + 2 or as f(x) = 3x + 2 if functional notation is used.

50. Find the equation of the line through (- 4, 8) perpendicular to the graph of 2x - 3y = 1.
Solution: The slope of the desired line must be the negative reciprocal of the slope of 2x - 3y = 1.
The slope of 2x - 3y = 1 is found by solving 2x - 3y = 1 for y to obtain

The slope of 2x - 3y = 1 is
The desired equation is the equation of the line that passes through (- 4, 8) with slope
Use the point slope form of the equation of a line y - y1 = m(x - x1) to obtain
y - 8 = (x - (- 4))
y - 8 = (x + 4)

Y - 8 = x - 6
y = x + 2

Therefore the desired equation is y = x + 2 or f(x) = x + 2 if written using functional notation