Two posts ago, we promised to discuss the general theory behind triangles whose sides have slopes of — and here we go!!! We also considered, in the previous post, triangles whose sides have slopes of the form
. If you loved those two, and you read this one through, you’ll love this one too. Or even “befriend” it as we do.
That’s true.
Slopes in arithmetic progression
Example 1
Find coordinates for the vertices of a triangle whose sides have slopes of .
Let have vertices as shown below:
Since the slopes of sides are
, respectively, we have the following linear equations:
which we can recast into a “homogeneous” system:
(1)
(2)
(3)
Notice that there are six unknowns in three equations, so the system is underdetermined. Being also homogeneous, it has an infinite number of solutions, as our friend from L.A. (a.k.a Linear Algebra, not Los Angeles), just informed us. By row reduction, our system is found to have a rank of as well as three degrees of freedom — the latter meaning that the solution can be expressed in terms of three parameters (or “free” variables). In fact, we have:
where are our three “free” variables. Observe that the coefficients of these variables add up to one in the above equations, so in a sense they’re weighted.
To sum up, the required coordinates are:
As you may have thought, the formula breaks down when , but this leads to an important special case that deserves a separate treatment.
Example 2
In the notation of our first diagram, PROVE that if, and only if,
.
This is immediate; just another way of saying that the slope of side is zero. However, if we want to be fancy, we can use equation (3). First assume that
. Then (3) reduces to
. Conversely, if
, then we’re left with
which in turn leaves us with two choices: or
. However,
is impossible because otherwise it would mean that
. Together with the fact that
(by the assumption in the converse), the points
and
will coincide, meaning that we don’t have a triangle. So we choose
and this completes the proof.
As a by-product, we also obtain that is equivalent to
. We’ll have more to say about this case later.
Example 3
Find the -coordinate of the centroid of a triangle whose sides have slopes of
.
The -coordinate of the centroid is given by
. Using the coordinates we found in Example 1, we have:
Example 4
PROVE that if the slopes of the sides of a triangle follow an arithmetic progression, then the triangle contains one vertical median.
Let have vertices
and let sides
have slopes of
, respectively. In Example 3, we saw that the
-coordinate of the centroid is precisely
. Since the centroid shares an
-coordinate with a triangle’s vertex (in this case
), we know — from our previous post — that the median through that vertex is vertical.
Example 5
PROVE that if the slopes of the sides of a triangle follow an arithmetic progression , then so do the x-coordinates.
Note that the converse is not true.
To prove the above statement, we can use a result from our previous post, but let’s calculate it directly here. Enumerate the -coordinates as follows:
So the -coordinates differ consecutively by
.
The fact that the -coordinates follow an arithmetic progression whereas the
-coordinates do not (necessarily) follow an arithmetic progression makes the resulting triangle diagram quite weird, which was what we called “ugly-looking” two posts ago.
Slopes in geometric progression
Example 6
Find coordinates for the vertices of a triangle whose sides have slopes of .
Of course, we want and
. If there be any other restriction, it will become obvious when we’re done with the computations.
To find the coordinates, suppose we work with whose vertices are as shown below:
As before, it boils down to solving a linear system:
where, after row reduction and back-substitution, the solution is:
In each of the above expressions, the coefficients of the “free variables” add up to one, so in a sense they’re weighted.
Example 7
Find the centroid of the triangle with vertices given in Example 6 above.
The -coordinate of the centroid is
:
If , then the
-coordinate coincides with
. (Note that we can’t have
since otherwise one of the slopes will be zero, and we’re working with slopes in geometric progression.)
For the -coordinate of the centroid, we use
:
If , the
-coordinate of the centroid coincides with
.
COOL.
Example 8
Let be such that sides
have slopes
, respectively. PROVE that the slope of the median from vertex
is
.
The fact that the slope of the median from vertex to side
and the slope of side
are negatives of each other has an important implication and application (see Example 9 below).
To prove the present statement, let’s refer to an earlier diagram:
We found that
The midpoint of is
and so the slope of the median from vertex
is:
Example 9
Let be such that sides
have slopes
, respectively. PROVE that
if, and only if,
.
Suppose that (a similar argument holds for
). By Example 8 above, the slope of the median from vertex
is
. But then this means that the median from vertex
is perpendicular to side
. Therefore,
and
is isosceles.
Conversely, suppose that . Then the median from vertex
is perpendicular to side
, which in turn means that the product of their slopes is
. Since their slopes are
and
, it follows that
So we obtain a characterization of isosceles triangles, using slopes in geometric progression.
Example 10
Let be a right-triangle in which sides
have slopes
, respectively. PROVE that
is necessarily negative, and that side
cannot be the hypotenuse.
Since the location of the right-angle was not specified, we have to consider all possibilities. We use the fact that the product of the slopes of two perpendicular lines is , and take the product of two slopes at a time:
From the first equation, , which is always negative because the denominator
is always positive whatever (the real number)
is. From the third equation, we have
, and again
is negative since it’s only the cube of a negative (real) number that can be negative.
Now from the second equation, we have . Since the left side of this equation is always positive whereas the right side is always negative, the equation doesn’t have real solution. In turn, this means that side
(with slope
) and side
(with slope
) do not form a right-angle. Therefore, side
cannot be the hypotenuse.
Numerical examples
We now move from the abstract angle to actual examples.
- Let
and
. Then the arithmetic sequence
is
and the corresponding coordinates for the vertices of a triangle with slopes
are:
are “free” variables. We obtain an assemblage of triangles by specifying values for
, so long as
.
- Put
and
. Then
. The resulting triangle has coordinates at
:
Notice the presence of a vertical median
.
- Put
and
. Then
. We obtain a triangle with coordinates at
:
Notice the presence of a vertical median
.
- Put
- Let
and
. Then the corresponding arithmetic sequence of slopes is
; moreover:
Depending on what we choose for the “free variables”, we obtain different sets of triangles. Let
. Then
. This results in a triangle with coordinates at
:
Notice the presence of a vertical median
. Also this is a right triangle.
- An example of a triangle with slopes in geometric progression. Let
and
; the corresponding geometric sequence of slopes is
. If we put
and
in our formula, we obtain
Next stop: choose values for the “free variables”
, bearing in mind that
. Let
and
. We obtain
and the resulting triangle has vertices located at
:
Notice that this is a right triangle.
- Let
and
; we obtain the geometric sequence of slopes
. In turn:
For convenience, let’s choose our “free variables”
and
. We then obtain
and a triangle with coordinates at
results:
- Here’s an example in which some of the coordinates are irrational numbers. Let
and
. Let’s also choose our “free variables” right away:
and
. We obtain
and a triangle with coordinates at
results:
There’s something fascinating about this example that is worth mentioning. Observe that the given triangle doesn’t contain
. Now the midpoints of sides
are
Consequently, the lengths of the medians from
are
, respectively. How are these related?
COOL.
Takeaway
If the slopes of the sides of a triangle are (a geometric progression), and the slopes of the sides of another triangle are
(an arithmetic progression), then properties that correspond to
have analogues in
. These properties are mainly lateral properties — isosceles or equilateral.
We encourage you to investigate different cases of slopes in arithmetic progression and slopes in geometric progression. You’ll encounter exciting surprises.
Tasks
- PROVE that if the slopes of the sides of a triangle form a geometric progression
, then the slopes of its three medians also form a geometric progression (with a different common ratio), provided
.
- PROVE that if the slopes of the sides of a triangle form an arithmetic progression, then the slopes of its three medians do not form an arithmetic progression.
- Let
(
) be the slopes of the sides of a
. PROVE that the product of the slopes of the three medians is
.
- PROVE that if the slopes of the sides of a triangle are
, then the triangle can never be a right triangle.
Of course
.
- PROVE that if the slopes of the sides of a triangle are
, where
and
are rational numbers, then the triangle can never be equilateral.
Actually if either
or
is a rational number, same conclusion.
- PROVE that if the slopes of the sides of a triangle are
, where
and
are rational numbers, then the triangle can never be equilateral.
It appears that equilateral triangles don’t like rational numbers?
- Let
be such that sides
have slopes
, respectively. PROVE that the median from
and the median from
can never intersect at
, except when
.
Any triangle whose side slopes form a geometric progression with common ratio
is a peach.
- PROVE that if the
-coordinates of the vertices of a triangle form an arithmetic progression, then the slopes of the sides do not necessarily form an arithmetic progression.
A counter-example will do.
- If the slopes of the sides of a triangle form a geometric sequence with a common ratio of
, PROVE that both the
-coordinates and the
-coordinates form arithmetic progressions.
Cool.
- Let
be such that sides
have slopes of
, respectively. PROVE that
if, and only if
.
Pretend that you’re unaware of the Pythagorean theorem and perpendicular slopes. Solve the problem without recourse to these two.
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