Parallel And Perpendicular Linear Polynomials
At the end of chapter 6, you discovered a way to make a parabola by folding paper. By taking points on the bottom edge of the paper (the directrix), and matching them up with a given point (the focus), you were able to make enough crease lines to outline a parabola.
Knowing that all points of a parabola are the same distance to the focus as they are vertically to the directrix, it’s possible to determine the coordinates of the points on the parabola as well.
In other words, if we could determine the linear polynomial of a crease line, we could figure out the coordinates of the tangent point (the point where the parabola and the crease line meet).
However, to determine the linear polynomial of a crease line, we would need to determine its slope and its y-intercept. All we know though, are the coordinates for the focus and the point on the directrix, and the x-value of the tangent point.
As it turns out, that’s all we need, once we realize the focus and the point on the directrix are reflections of each other. In other words, the crease line is a perpendicular bisector (a line which goes through the middle of the line segment, and is perpendicular to the line segment).
So it comes down to, determining the coordinates of the line segment’s middle point, and
the slope of a perpendicular line.
We’ll determine the coordinates of the line segment’s middle point first.
Below is a snapshot of an applet, which will allow you to discover how to determine the coordinates of a line segment’s middle point, referred to as, the midpoint.
Run the applet “midpoint.html” , and move around the line segment’s endpoints, A and B, until you have discovered how to determine the coordinates of the line segment’s midpoint.
Use a copy of the pdf appletnotes2g, to record examples and your discoveries.
After discovering how to determine the midpoint of a line segment, and recording examples, go on to the next page.
As we discovered, the midpoint is the half way point, or average point of the two endpoints. All you have to do is find the average of the x-values of the endpoints, and the average of the y-values of the endpoints. This explains the midpoint formula, shown below, for any two given points, ( x1 , y1 ) and ( x2 , y2 ).
The above formula is a special case of a more general one. Open the pdf midpoint to read further.
We now know ( 2 , 1 ) is a point on the perpendicular bisecting crease line. 2 2.
We still need the slope, though, but unfortunately, you need two points to determine a slope. We could determine the slope of the line segment however, if we knew it would be helpful.
Below are snapshots of an applet, which will allow you to discover if there is a connection between the slopes of two perpendicular lines.
Run the applet “perpendicular.html” , to see if there is a connection between the slopes of perpendicular lines.
Use a copy of the pdf appletnotes to record examples and any discoveries.
After determining if there is a connection between the slopes, and defending why or why not, go on to the next page.
As we discovered, there is a connection between the slopes of perpendicular lines.
As it turns out, the slopes are negative (opposite signed) reciprocals of each other.
For example, the line shown in black has a slope of -3/2, and the perpendicular line shown in blue, has a slope of 2/3.
In other words, -3/2 and 2/3 are reciprocals of each other, but with opposite signs.
For some evidence the two lines are actually perpendicular, fit the corner of a sheet of paper snugly between the two lines, to show they form 90º angles.
Of course, fitting the corner of a sheet of paper between the two lines, doesn’t prove they’re perpendicular.
Open the pdf perpslopes, to explore a proof that perpendicular lines do have negative reciprocal slopes.
After you are convinced that perpendicular lines have negative reciprocal slopes, go on to the next page.
Now that we know perpendicular lines have negative reciprocal slopes, we can use the slope of the line segment to determine the slope of the perpendicular crease line.
The means the slope, or m, of the perpendicular crease line, is 2/1 or just 2.
We now have the slope of the crease line, but in order to determine the linear polynomial, we still need the y-intercept.
Since the values involved are integers, we can get an exact value of the y-intercept by using the definition of slope, ∆y over ∆x, and making a table, starting with ( 2 , 1 ).
Since the y-intercept will have an x-value of zero, we now know the y-intercept of the crease line is -3, and the linear polynomial of the crease line is y = 2x -3.
Now that we know the linear polynomial for the crease line (tangent line), we can determine the coordinates of the tangent point.
Since points on the directrix, and the resulting tangent points line up vertically, they have the same x-values.
Plugging in 4 for x into y = 2x – 3, gives us y = 2( 4 ) – 3 = 8 – 3 = 5, so the tangent point is ( 4 , 5 ).
As we will discover later. With enough points, it’s possible to determine the quadratic polynomial for the parabola as well.
Run the applet “creases.html” , to recap the steps taken to find the coordinates of the tangent point ( 4 , 5 ).
Determine the coordinates of the tangent point, for a different point on the directrix (without using the applet).
Use a copy of booknotes to record your work.
After checking your work with the applet, and redoing any of the steps if necessary, go on to the next page.
Now that we know perpendicular lines have negative reciprocal slopes, what about the slopes of parallel lines?
Below is a snapshot of an applet, which will allow you to refresh your memory.
Run the applet “parallel.html” , until you can describe the connection between the slopes of parallel lines.
After being able to describe the connection between the slopes of parallel lines, go on to the next page.
As we discovered back in chapter 1, when we subtracted two linear polynomials with the same slopes, we ended up with a horizontal line, which is described by a constant. In other words, the vertical difference between them, is always the same (which is why the lines are parallel to each other).
2x + 5
– ( 2x + 2 )
As we discovered back in chapters 3 and 5, and were just reminded, another way to look at it is, parallel lines have the same slopes because adding or subtracting a constant, just means the lines are vertical translations of each other (which doesn’t change their steepness, just where they cross the y-axis).
-.5x + 0
-.5x + 1
Below are snapshots of a program, which will randomly generate a line in slope-intercept form, and then ask for another line which is either parallel or perpendicular to it.
Run program 701 , until you determine an acceptable line, for at least 10 of the 12 problems.
After determining an acceptable line, for at least 10 of the 12 problems, go on to the following challenge.
Open the pdf parperchallenge and run program 702 , until you solve 4 of the 7 challenges.
Use a copy of the pdf parpergrids, to record and label your 4 solutions.
After solving and recording 4 of the 7 challenges, go on to the next page.
For a given line, there are an infinite number of lines which are parallel or perpendicular to it. Having so many choices available to us in the last couple of programs, made finding solutions a little nicer.
If we were given a line and a particular point for the parallel or perpendicular lines to go through, things would have been a little trickier, because there would have been only one possible solution for each problem.
Below are snapshots of an applet which will allow you to convince yourself, for a given line and a particular point, there is only one possible line which can be parallel or perpendicular to the given line, and also go through the given point.
Run the applet “paraperp.html” until you are convinced, given a line and a point, there is only one possible line
which can be parallel or perpendicular to the given line, and also go through the given point.
After you have convinced yourself of the uniqueness of such a parallel or perpendicular line, go on to the next page.
To find a line which is parallel or perpendicular to a given line, which also goes through a particular point, we need to work with a form of a line based on slope and any given point.
The slope-intercept form of a line comes close, but the given point would have to be the y-intercept. The standard and intercept forms of a line won’t work either, because neither of them are based on slope.
Again, if we go back to the definition of slope, ∆y over ∆x, we can create a new form of a line.
There isn’t any challenge in finding lines which are parallel or perpendicular to horizontal or vertical lines, so we only have to worry about lines without a slope of zero or an undefined slope. In other words, we don’t have to worry about numerators or denominators equalling zero. Since x2 – x1 can’t equal zero, we can multiply both sides of the equal sign by it, which gives us the following.
This isn’t a formula though, because ( x1 , y1 ) and ( x2 , y2 ) both represent specific points. For lines though, the slope is always the same, no matter what two points are used.
This means we can keep ( x1 , y1 ) to represent a particular given point, and use variable coordinates ( x , y ) to represent any other and every other point on the line.
The resulting formula, shown above, is called the point-slope form of a line.
Below are snapshots of an applet, which will allow you to explore lines written in point-slope form.
Run the applet “pointslope.html” , until you see a connection between the point-slope form, and the graph of a line.
After the point-slope form of a line makes sense to you, go on to the next page.
Given a slope and a particular point, the point-slope form of a line makes it easy to set up a formula, but it’s a little awkward to work with afterwards.
After the point-slope form of a line is used, it’s usually more helpful to convert it to slope-intercept form, or standard form.
Below are snapshots of a program, which will randomly generate a line and a point. The challenge is to determine the unique line, in slope-intercept form, which is either parallel or perpendicular to the given line, and goes through the point.
Run program 703 , and use the point-slope form of a line, to solve at least 4 of the 5 challenges.
Use a copy of the pdf programnotes and/or programnotes2g, to record the challenges and your work.
After being able to determine at least 4 of the 5 unique lines, and recording your work neatly, go on to the next page.
Since parallel and perpendicular lines come up quite often in math, and can often be found in human-made objects, the point-slope form of a line is quite useful. The point-slope form of a line can do much more than find parallel or perpendicular lines however. It can find any unique line, described by any two given points.
Below are snapshots of a program, which will randomly generate two given points, and challenge you to determine the unique line going through them (in standard form this time, to avoid messy fractions).
Run program 704 , and use the point-slope form of a line, to solve at least 4 of the 5 challenges.
Run the applet “ptslope2pts.html” , for hints and/or to see examples.
After finding at least 4 of the 5 unique lines, and recording your work neatly, go on to the next page.
The triangle shown below, has been sandwiched between two lines.
One line is an extension of a side, and goes through the two vertices ( -1 , 6 ) and ( 3 , 5 ).
The other line goes through the remaining vertex ( 1 , 3 ), and is parallel to the first line.
Use the point-slope formula, to determine the two lines pictured above, and convert them to slope-intercept form.
Use a copy of the pdf booknotes to sketch the above picture, and record your work neatly.
After determining the above two lines in slope-intercept form, go on to the next page.
Since the directions on the last page didn’t mention anything about using decimals or fractions, any one of the following would be considered acceptable ways for writing the two lines in slope-intercept form.
Mixed numbers and/or slashes aren’t usually considered acceptable, because they can sometimes be read incorrectly.
If the directions had asked for standard form, as seen below, we would have avoided decimals and fractions altogether,
Slope-intercept form was chosen though, to make it obvious the two lines had the same slopes, and were parallel.
The important thing though, isn’t in what form the lines are written. The fact that we can determine the lines at all, and therefore graph them, is the important thing to understand.
Below are snaphsots of an applet (shown at step 4), which will allow you to explore triangles sandwiched between parallel lines.
Run the applet “anglesum.html” , until you can answer the two questions, found at steps 7 and 9.
Use a copy of the pdf appletnotes to make a sketch and record both of your answers.
After defending both of your answers, and recording them, go on to the next page.
Because vertical angles and their corresponding angles all have the same measure, the angles on the inside of the parallel lines, but on opposite sides of the transversal, will also have the same measure. These types of angles are called alternate interior angles.
Lines have angle measures of 180º, and are called straight angles.
Since the three angles of any triangle form a straight angle, their measures will always add up to 180º.
Below are snapshots of a program, which will randomly generate a pair of parallel lines cut (crossed) by a transversal. Given the measure of one of the resulting 8 angles, the challenge is to determine the measures of 4 of the remaining angles, chosen at random.
Run the applet “transversals.html” , to first explore an example of two parallel lines being cut by a transversal.
Run program 705 , until you successfully determine the measures of at least 10 of the 12 angles.
After determining the measures of at least 10 of the 12 angles, read the next line, and go on to the next page,
For more on angles, run program 706 and/or open the pdf angles
A surprising number of properties about triangles, can be applied or discovered, using the point-slope form of a line. Given any three points, there exists a fourth point, which is the exact same distance from all three of the given points. This point is called the circumcenter.
To the right is a snapshot of an applet, which will give enough information to figure out how to determine the circumcenter of 3 points.
Run the applet “chords.html” , until the steps for finding the circumcenter of 3 points, make sense to you.
After you understand the steps for finding the circumcenter of 3 points, go on to the next page.
If the three points represented three small towns on a map, finding the circumcenter, would help determine where to build a new firehouse, a radio tower, or a local dump, for example. For safety reasons, the firehouse should be as equally close as possible to all three of the towns it serves. For quality of reception reasons, a radio tower should be close to the same distance to all three towns as well. Looking at it the other way, the three towns might want a dump to be equally far away
Below are the coordinates for three small towns in Maine. Determine the circumcenter, to help the three towns decide on a fair location for a radio tower, which will help reduce cost, radio interference, and land devaluation (assuming there aren’t any wildlife or environmental issues of course).
Determine the circumcenter of the three towns, represented by the three points pictured above.
Use a copy of the pdf booknotes, for sketches, and to record the work needed to determine the circumcenter.
After determining the circumcenter, and recording your work neatly, go on to the next page.
Below are the coordinates, slopes, and linear polynomials involved in finding the circumcenter of the three towns.
If you determined the correct circumcenter, and all of your work is neat and easy to follow, go on to the next page.
If not, redo whatever work is necessary to defend the correct circumcenter,
and/or make all of your work neat and easy to follow, before going on to the next page.
If you found the design at the beginning of the chapter interesting,
run program 713 to explore Poincaré disks, named after Henri Poincaré (on-REE pwawn-ka-RAY)
Open the pdf disk, to learn more about Poincaré disks, and non-Euclidean geometry.
Below is a snapshot of the chapter 7 programs menu program.
Run the program “7menu.bas” , or its alias to access every program used in chapter 7.
“A scientist worthy of his or her name, above all a mathematician, experiences in his or her work the same impression as an artist; his or her pleasure is as great and of the same nature” — Henri Poincare —
“The study of mathematics cannot be replaced by any other activity that will train and develop man’s purely logical faculties to the same level of rationality.” — C.O. Oakley —
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