Parallel Lines


Explore This 1


Created with GeoGebra. Author: University of Waterloo. CC BY-NC-SA 4.0. https://ggbm.at/pmxdd5uc#


Explore This 1 Summary

  • In the Explore This, you may have noticed that if two lines are parallel, they will share the same slope.
  • We can also say, if two lines have the same slope, the lines are parallel.

Slide Notes

Glossary

All Slides

Navigation

Parallel

Objects are considered to be parallel if the distance between them remains constant along their entire length.

Sources: Telephone Wires - azndc/E+/Getty Images; Crops - ollo/E+/Getty Images; Lined Notepad - bonetta/E+/Getty Images

Parallel Lines

Lines are parallel if they share the same slope. If lines have the same slope, they are parallel.

 

 

Example 1

Line A is parallel to Line B, which is represented by the equation \(y=6x+1\). Line A passes through the point \((2,5)\). Determine the equation of Line A in the form \(y=mx+b\).

 

 

Example 1 Continued

Line A is parallel to Line B, which is represented by the equation \(y=6x+1\). Line A passes through the point \((2,5)\). Determine the equation of Line A in the form \(y=mx+b\).

 

 

Example 2

 

 

 

Example 2 Continued

Line A

\(y=-\dfrac{2}{3}x+5\)

Line B

\(y=-\dfrac{3}{2}x+3\)

Line C

\(4x+6y=11\)

Are any of the lines parallel? Justify your answer.

 

Example 2 Continued

Line A

\(y=-\dfrac{2}{3}x+5\)

Line B

\(y=-\dfrac{3}{2}x+3\)

Line C

\(4x+6y=11\)

Are any of the lines parallel? Justify your answer.

Solution

\(m_A=-\dfrac{2}{3}\)

\(m_B=-\dfrac{3}{2}\)

\(m_C=-\dfrac{2}{3}\)

 

Paused Finished
Slide /

Check Your Understanding 1


Created with GeoGebra. Author: University of Waterloo. CC BY-NC-SA 4.0. https://ggbm.at/urzdehnf#


Example 3

A picture of railway tracks is placed on a grid. Lines are drawn along the picture as shown. Show that the railway tracks are parallel. 

A railway superimposed on a graph shows that points EF lie on one rail and GH lie on the other rail. The two rails define two parallel lines. The points are E(1.35, 5.00), F(6.32, 2.84), G(0.54, 3.22), and H(5.51, 1.06).

Source: Simple Rail Tracks - ziggy1/iStock/Getty Images

Solution 

  • We are given the points on railway tracks: \(E(1.35,5.00)\), \(F(6.32,2.84)\), \(G(0.54,3.22)\), and \(H(5.51,1.06)\). 
  • To determine if the lines are parallel, we need to calculate the slope of line segment \(EF\) and the slope of line segment \(GH\).

    Line Segment \(EF\)

    \[\begin{align*} m_{EF}&=\dfrac{\Delta y}{\Delta x}\\ &=\dfrac{y_2-y_1}{x_2-x_1}\\ &=\dfrac{2.84-5.00}{6.32-1.35}\\ &=\dfrac{-2.16}{4.97}\\ &\approx-0.435 \end{align*}\]

    Line Segment \(GH\)

    \[\begin{align*} m_{GH}&=\dfrac{\Delta y}{\Delta x}\\ &=\dfrac{y_2-y_1}{x_2-x_1}\\ &=\dfrac{1.06-3.22}{5.51-0.54}\\ &=\dfrac{-2.16}{4.97}\\ &\approx-0.435 \end{align*}\]

Therefore, \(m_{EF}= m_{GH}\) and the railway tracks are parallel.

We can state that the line segments \(EF\) and \(GH\) are parallel by writing \({EF}\parallel {GH}\).

When we draw line segments, we can also indicate that they are parallel by showing matching arrows in the middle of the line segments.

Example 4

In the given diagram, \(AB\) is parallel to \(CD\). Determine algebraically the missing \(y\)-coordinate for point \(D\).

A graph depicts two parallel line segments, AB and CD. The points are A(negative 6, 1), B(negative 2, 3), C(4, 2), and D(8, v).

Solution

  • For the two line segments to be parallel, their slopes must be equal. 
  • Therefore, \(m_{AB}\) must equal \(m_{CD}\).
    • Because we have two complete points for line segment \(AB\), we will first calculate \(m_{AB}\).
      • Point 1: \(A(-6,1)\rightarrow (x_1,y_1)\)
      • Point 2: \(B(-2,3)\rightarrow (x_2,y_2)\)
    \[\begin{align*} m_{AB}&=\dfrac{\Delta y}{\Delta x}\\ &=\dfrac{y_2-y_1}{x_2-x_1}\\ &=\dfrac{3-1}{-2-(-6)}\\ &=\dfrac{2}{4}\\ &=\dfrac{1}{2} \end{align*}\]
  • Knowing \(m_{AB}\) means we also know that \(m_{CD}=\dfrac{1}{2}\). 
  • Knowing \(m_{CD}=\dfrac{1}{2}\), we can substitute the known values into the slope formula and solve for \(v\).
    • Point 1: \(C(4,2)\rightarrow (x_1,y_1)\)
    • Point 2: \(D(8,v)\rightarrow (x_2,y_2)\)
\[\begin{align*} m_{CD}&=\dfrac{\Delta y}{\Delta x}\\ m_{CD}&=\dfrac{y_2-y_1}{x_2-x_1}\\ \dfrac{1}{2}&=\dfrac{ v-2}{8-4}\\ \dfrac{1}{2}&=\dfrac{v-2}{4}\\ 4(1)&=2(v-2)\\ 4&=2v-4\\ 4+4&=2v\\ \dfrac{8}{2}&=v\\ 4&=v \end{align*}\]

Therefore, the \(y\)-coordinate of point \(D\) is equal to \(4\).
You may have thought that \(v=4\) was the missing coordinate solely by using the diagram. In this case that is true; however, calculating the slope is the only way we would know for sure. This is especially true if the coordinates are not integers.


Check Your Understanding 2


Part 1: null
How Did I Do?Try Another