Linear Equations with Two Variables - Part 2


Objectives

After completing this lesson, you will be able to:

  • Identify if two lines are parallel.
  • Identify if two lines are perpendicular.
  • Identify and find the formula of vertical lines.
  • Recognize and use different formulas for a linear relationship.


Introduction

In the lesson on Linear Equations — Part 1, you saw y = mx + b is a line with slope m and y-intercept b. Next, you will explore several relations between lines and alternatives to representing lines by y = mx + b.



Parallel Lines

Two parallel lines can be described in a variety of ways:

  • Two lines in a plane that never intersect.
  • Two lines which maintain a constant distance between them.
  • Two lines with the same slopes.

To solve mathematical problems, generally it is simpler to characterize parallel lines as two lines with the same slopes.

Examples:

Example 1: The lines y = 4 3 x + 2 and y = 4 3 x + 5 are parallel.


Example 2: Follow the instructions in the application below and observe the effects of different changes in the formulas of the line graphs:

  1. Move the slider m1 until the two lines are parallel.
  2. Now, move the slider b1 from b1 = -5 until b1 = 5. Changing the y-intercept b1 does not change the slope of the line. So the value of the y-intercept does not have any affect on whether the lines are or are not parallel. If the slopes of the two lines are the same, then the lines are parallel.
  3. Move the red sliders until you change the formula of line 1 to y = 2x + 1. Then move the blue sliders to find the formulas of four lines parallel to y = 2x + 1.
  4. Move the red sliders until you change the formula of line 1 to y = -3x + 4. Then move the blue sliders to find the formulas of four lines parallel to y = -3x + 4.

Example 3: Find the equation of the line that passes through point (3,2) and is parallel to line y = 2 x + 3 .

Solution:

  1. The slope of y = 2 x + 3 is equal to 2.

  2. The slope of a line parallel to y = 2 x + 3 has to be 2 also. So, the parallel line has to have a formula y = 2 x + b .

  3. If the formula of the line is y = 2x + b, what's left is to find the value of y-intercept b. If the line passes through point (3,2), by substituting 3 for x in the formula, the result (the value of y) has to be 2. So:
    • Substituting 3 for x and 2 for y in the formula y = 2x + b, gives 2 = 2 x 3 + b

    • Now you can solve for the value of b which will make the equation true:

    2 = 6 + b
    2 – 6 = b
    –4 = b


  4. So the formula of the line which passes through point (3,2) and is parallel to line y = 2x + 3 is y = 2 x 4

  5. Verification
    • y = 2x + 2 and y = 2x - 4 have the same slope 2. Therefore, they are parallel.

    • Point (3,2) satisfies formula y = 2x - 4 because substituting 3 for x and 2 for y results in:
      2 = 2 × 3 - 4, which is true.

    So line y = 2x - 4 is parallel to line y = 2x + 3 and passes through point (3,2), which can be seen with the graphs of the two lines.




Perpendicular Lines

Two perpendicular lines are frequently described in one of two ways:

  • Two lines that form 90 degree angles where they intersect.
  • Two lines of a plane where the product of their slopes is -1.

To solve mathematical problems, the characterization which is most used is: two lines where the product of its slopes is -1 are perpendicular.

Examples:

Example 1: If the slope of a line is 2 3 , then for each 3 units the coordinate of x moves to right on this line, the coordinate of y is increases by 2 units.

According to the characterization, a line which is perpendicular to this line will have a slope of 3 2 . In the perpendicular line, for every 2 units the coordinate of x moves to right, the y coordinate decreases by 3. (Or for every 3 units that the coordinate y moves up, the coordinate x decreases by 2.)



Example 2: Follow the instructions on the application below and observe the effects of changes to the formula on the line graphs:  

  1. Move the slider m1 until both lines are perpendicular.
  2. Now, move the slider b1 from b1 = -5 until b1 = 5. Changing the y-intercept b1 does not change the slope of the line. So the value of the y-intercept has no effect on whether the lines are or are not perpendicular. If the product of the two slopes is -1, then the lines are perpendicular.
  3. Move the red sliders until you change the formula of line 1 to y = 2x + 1. Then move the blue sliders to find the formulas of four lines perpendicular to y = 2x + 1.
  4. Move the red sliders until you change the formula of line 1 to y = -3x + 4. Then move the blue sliders to find the formulas of four lines perpendicular to y = -3x + 4.



Vertical lines

In the following diagram, the two points (0,1) and (0,5) are labeled and there is a line drawn which passes through the two points. In Linear Equations - Part 1, we saw that slopes can be expressed in different ways:

  • Slope m equals the amount that y changes when x increases by 1.
  • m = rise/run = Δy/Δx
  • rise = m × run or Δy = m × Δy

In the case of two points (0,1) and (0,5), the value of x is constant. Therefore, x never increases by 1, never advances and, as a consequence, Δx is zero in all cases. So the slope is not defined for the line that passes through these points. The usual intercept-slope formula y = mx + b does not represent the set of points that belong to this or any other straight vertical line. By examining various points which belong to this line, such as

(0,-1), (0,0), (0,1) and (0,2),

we observe that the x coordinate is zero at any and all points. So a formula for the line is x = 0.

In general, with vertical lines:

  • The value of x does not change and the equation is expressed in the form of x = a , where a is a constant.
  • Δx is equal to zero. The slope of a vertical line is not defined and we say that a vertical line has no slope.

Examples:

Example 1: Determine the equation of the line which passes through points (-1,2) and (-1,5).

Given the first coordinates of both points is the same, the only possibility is that the line which passes through two points is a vertical line, in other words, for any value for y, the value of x has to be -1. So the equation of the line would be: x = -1.



Example 2: Follow the instructions on the application below and observe the effects of changes in the formulas on the line graphs:

  1. Move point J until it is on top of point (2,0). You will observe the equation of line, x = 2 written in red.
  2. Move the blue slider until the value is 4 to see four points labeled on the line. Verify that the four points satisfy the equation x = 2.
  3. Move J until the four vertical lines correspond to x = -1, x = -3, x = 0 and x = 3. For each line, verify that the four points satisfy the equation of the line labeled in red.




Other Formulas for Linear Relationships

Until now, we have used y = mx + b to represent non-vertical lines. This way of representing a line is called slope-intercept. There are circumstances where other forms of expressing the equation of a line are more useful:

  1. Point-Slope: Given that a line which passes through (2,3) with a slope equal to 4. Any other point (x,y) on the line has to serve to calculate the slope of the line. Since the slope is 4, y − 3 / x − 2 has to be 4, in other words:

    y 3 x 2 = 4

    When you multiply the two sides of the equation by (x - 3) we get the form (equation) known as point-slope:

    (y - 3) = 4(x - 2)

    Generally, you can represent the line with slope m which passes through point (a,b) with the point-slope equation:

    (y - b) = m (x - a)

    By habit, we write the equation to calculate the value of y. In this case, the point-slope equation will be:

    y = m(x - a) + b

  2. Standard Formula: When a line is written as ax + by = c, we say the equation is in the standard form. The standard form is very useful to find the x and y-intercept. For example, in the case of equation 2x + 4y = 8, when x = 0, the equation becomes 4y = 8. Therefore y = 2 is the y-intercept. When y = 0 (substituting 0 for y) the equation becomes 2x = 8. Therefore x = 4 and the intercept of the x axis is 4. With the two intercepts (0,2) and (4,0), you can plot the graph of the line efficiently. See the following graph and confirm that all the points on the line satisfy the equation 2x + 4y = 8.

    Generally, if ax + by = c, the intercepts are ( c a , 0 ) and ( 0 , c b ) .

To practice representing lines in various forms, follow the instructions on the application below.

  1. With the blue sliders, select m = -1 and b = 20. The graph which results should be y = -1x + 20 with slope -1 and y-intercept (0,20).
  2. Click on the green text labeled Point-Slope. Move the green point on the line labeled P to the following points. Write down the point-slope formula shown for each point.
  3. P Point-Slope Formula Point-Slope Formula
    (20,0) (y – 0) = –1(x – 20) y = –1(x – 20) + 0
    (0,20) (y – 20) = –1(x – 0) y = –1(x – 0) + 20
    (-20,40) (y – 40) = –1(x + 20) y = –1(x + 20) + 40
  4. Verify that all the resulting formulas are equivalent to the slope-intercept formula y = - x + 20.
  5. Click on the red text labeled Standard Form. With the red sliders, select intercepts L = 20 and M = 20. Verify that the line which has intercepts (0,20) and (20,0) is identical to the line seen on the previous steps and that its standard form is x + y = 20 is equivalent to the slope-intercept formula:

 y = –x  + 20

  1. To practice, use the application to express the following lines with formulas in the point-slope form and in the standard form:
    1. y = -2x + 40
    2. y = 2x - 30
    3. y = -5x + 30
    4. y = 6x - 20





Summary

Now that you have completed this lesson, you will be able to:

  • Identify if two lines are parallel.
  • Identify if two lines are perpendicular.
  • Identify and find the formulas of vertical lines.
  • Recognize and use different formulas to represent a linear relationship.
Kentucky Center for Mathematics
Northern Kentucky University, MEP 475
Highland Heights, KY 41076
ph: 859-572-7690, fax: 859-572-7677
kcm@nku.edu