11.02 Parallel and perpendicular lines

We will use the slopes of parallel lines theorem to prove the lines are parallel by showing the slopes are equal.

If we find nice points, we can count the rise and run to determine the slope.

Begin with the definition of parallel lines: "Parallel lines are lines in the same plane that do not intersect."

We can prove this informally using definitions and diagrams. We will start with one line on the coordinate plane and try to find a line which never intersects it.

For a line to never intersect the one that we started with, it must be a translation of our starting line without any reflection or rotation. A translation is defined as "a transformation in which every point in a figure is moved in the same direction and by the same distance."

This diagram shows the possible vertical translations.

Whether we translate the line up or down, the slope triangle is translated along with the line. Translations preserve distance, so the distances a and b are preserved.

This means the rise and run remain the same, so the slope of the line remains the same.

This diagram shows the possible horizontal translations. Whether we translate the line left or right, the slope triangle is translated with the line.

Using the same reasoning as we did above, we know the slope of the line remains the same.

Since translating a line does not change its slope, we know that two lines are parallel if and only if their slopes are the same.

Another strategy for proving this is to make a system of two non-vertical linear equations which has no solution. We can start with:

\begin{cases} y=m_1x+b_1 \\ y=m_2x+b_2 \end{cases}

If b_1=b_2, then system will have a solution when x=0, so we must have b_1\neq b_2.

Now, we can consider what a solution to the equation will look like. If we solve the equation by letting the y-values be equal, we get:

We can see that there will be no viable solutions to the system only when m_1=m_2 because that would force us to divide by zero which is not possible. So two lines are parallel if and only if their slopes are the same.

Begin with the definition of perpendicular lines: "Perpendicular lines are two lines that are at right angles to each other."

One way to show they are at right angles is to start with one line and consider its slope using a right triangle. For example:

To find a line that is at a right angle to our starting line, we can just rotate it by 90\degree.

Notice that rotating the line by 90\degree also rotated the right triangle by 90\degree.

This shows us that the values of the rise and run have switched, so the slope of the new line will be the reciprocal of the starting line.

We can also see that the slope has changed sign, since rotating any line by 90\degree will change its slope from positive to negative or vice versa. The sign of the slope of the new line will be the negative of the starting line.

Therefore, two lines are perpendicular if their slopes are opposite reciprocals, which is the same as their product being -1.

To show that two non-vertical lines whose slopes have a product of -1 are perpendicular, we would need to begin with a line that has a slope of \dfrac{a}{b} and a line that has a slope of -\dfrac{b}{a} and show they meet at a 90\degree angle.

We can use a compass and straightedge or technology to construct parallel lines.

One approach for constructing parallel lines follows:

1. Use the line segment tool to construct line \overleftrightarrow{AB} and an arbitrary point C not on the line.

   

2. Use the point tool to add point D anywhere on line \overleftrightarrow{AB} and use the line tool to draw line \overleftrightarrow{CD}.

   

3. Use the point tool to place point E on \overleftrightarrow{CD} closer to C and use the compass tool to construct a circle with radius equal to ED centered at D.

   

4. With the compass tool again, construct a circle with radius equal to ED centered at C.

   

5. Use the point tool to label the intersection of circle C and \overleftrightarrow{CD} with F and use the point tool to label the intersection of circle D and \overleftrightarrow{AB} with G.

   

6. Select the compass tool and create a circle with radius GE centered at F and use the point tool to label the intersection of circle F and circle C with H.

   

7. Use the line tool to draw a line through points C and H. This line is parallel to \overleftrightarrow{AB}.

   

To prove this theorem, construct a segment \overline{PM} \perp \overline{AB} such that M is the midpoint of \overline{AB}.

Since both \triangle AMP and \triangle BMP are right angled triangles with AM=BM (given) and \overline{PM} common to both we can use the Pythagorean theorem and substitution to find the relationship between PA and PB.

We can use a flow chart proof, two column proof, bulleted list, or paragraph proof to prove that l \perp m. For this proof, we will use a two column proof.

One approach Ludek can use to construct perpendicular lines is by using dynamic technology software.

1. Use the 'Line' tool to construct a line through two points.

   

2. Use the 'Circle with Center' tool to construct a circle about each point on the line, such that the circles intersect at two distinct points.

   

3. Use the 'Line' tool once more to construct a line through the points of intersection of the circles. This line will be perpendicular to the first line.

   

Notice that while we have constructed a perpendicular line, it is not a perpendicular bisector because the radii of the two circles used in the construction are not the same. Therefore, the line does not pass through the midpoint of \overline{AB}.

One approach to constructing the perpendicular bisector through \overline{AB} is using a compass and straightedge.

1. Construct an arc centered at point A so that the arc is longer than half of \overline{AB}.

2.Construct an arc centered at point B using the same compass setting, such that this arc intersects the arc constructed in Step 1 at two distinct points. Label the two points C and D.

3. Draw \overleftrightarrow{CD}. Label the point M where \overleftrightarrow{CD} intersects \overline{AB}. M is the midpoint of \overline{AB}.

\overleftrightarrow{CD}\perp \overline{AB} at the midpoint of \overline{AB}, so \overleftrightarrow{CD} is the perpendicular bisector of \overline{AB}.