We’re s--o excited about this!!! As in, E–X–C–I–T–E–D!!! Indeed, this is expected, considering what is to be expounded: “any” geometric sequence can be encapsulated as slopes of a simple triangle like the one below:
And the procedure is easy: first find a triangle with slopes , then the remaining terms
can be obtained as slopes of line segments drawn from a fixed vertex
to a fixed side
.
Positive common ratio
that is, (positive) powers of .
Example
Find coordinates for the vertices of with slopes
.
There are different ways to obtain a triangle whose slopes are . One can begin with an arbitrary point
. Then the two other points can be chosen as
and
. With these, the slopes of sides
will be
, respectively.
In the present case, let’s first choose as
, for simplicity. Since the sequence is
, we have
and
. So
and
.
The entire geometric sequence can be obtained from the above triangle.
Example
For each , PROVE that
lies within
, where
is the point
and
is the point
.
Note that is a point on
. Let’s use distances to show that it’s indeed internal. We prove that
.
This proves that for each , the point
lies internally on
.
Example
For each , let
. PROVE that the slope of
is
, where
is the point
.
This follows by direct calculation:
Thus, when , the corresponding slopes will be
; that is,
.
By defining and using
with vertices at
, we’ve seen that the entire geometric progression
can be obtained as slopes of the line segments
, together with
with slope
. Notice that when
,
becomes
and when
,
becomes
. What happens to
when
?





as expected.
Put , then
has coordinates
. Together with
, the slope of line segment
is
as expected.
Put , then
has coordinates
. Together with
, the slope of line segment
is
as expected.
Three things to note from the above calculation:
- the points
are defined relative to the chosen coordinates for
;
- the line segments
that describe the geometric progression do not get to the midpoint of side
;
- the roles of vertex
and side
cannot be altered; doing so will distort the orderly arrangement of the line segments. See the next two examples.
Example
Given with vertices
, find coordinates for a point
on
such that the slope of the line segment
is
.
Let’s find the equations of and
and then solve the resulting linear system:
Thus, is the point
. It is external to side
— we don’t want that.
Example
Given with vertices
, find coordinates for a point
on
such that the slope of the line segment
is
.
Let’s find the equations of and
and then solve the resulting linear system:
Thus, is the point
. It is external to side
— we don’t want that.
Example
Find coordinates for the vertices of whose side slopes are
.
Take . Then the slope of
is
, the slope of
is
, and the slope of
is
.
Example
Let . For each
, PROVE that the slope of the line segment
is
, where
is the point
.
Observe that each lies within
(
). By direct calculation:
Thus, we obtain the geometric progression by putting
. Together with the first three terms that are the slopes of the original triangle, we obtain the entire geometric progression
.

Example
Explain how to obtain the entire geometric progression as slopes.
Easy-peasy.
Begin with a in which the slopes of the sides are
. To this end, take
. Then the slope of
is
, the slope of
is
, and the slope of
is
.
Next, let . Then, by direct calculation (as in and ), the slope of
is
, for
. Therefore, letting
successively, we obtain the terms
. If we now append the slopes of the sides of
(which are
), we obtain the entire geometric progression
.
Reciprocal terms
In addition to obtaining , we also get
, once we go past a certain point; let’s call it a terminal point (see the exercises at the end
).
Example
Given with vertices
, find coordinates for a point
on
such that the slope of
is
.
In Example we had . Put
, then
. Take
. Then
is an internal point on
; further, the slope of
is
as desired. Similarly, by putting , we obtain the terms
. Thus, inside a simple, single triangle lies the infinite geometric progression
disguised as slopes.
Example
Given with vertices
, let
. PROVE that
.
Put , then
. Put
, then
, as also found in a previous example. By the distance formula:
Observe that the slopes of the sides of are
; they form a geometric progression with a common ratio of
, which is the square of the common ratio of the geometric progression
formed by the slopes of the sides of the parent
.
Takeaway
Slopes are usually introduced sometime in grade (or earlier), while geometric sequences are usually studied in grade
(or earlier). Here, we’ve managed to show a close connection between the two concepts, using the medium of a triangle.
Throughout we worked with positive integer common ratios. But the common ratios can also be negative, and in such cases things become much more exciting.
Tasks
- (Depressed cubic) Consider
with vertices
. PROVE that:
- the foot of the altitude from vertex
is
;
- the length of the altitude from vertex
is
;
- the area of
is
(this becomes a depressed cubic if written in the form
. Moreover, if
is an integer (
of course), then this area is always a positive integer).
- the foot of the altitude from vertex
- Let
. PROVE that
as
.
- Let
. PROVE that
as
.
- (Special points) Given
with vertices
, PROVE that there are (internal) points
and
on
such that:
- the slope of
is undefined and the slope of
is zero (
for vertical,
for horizontal);
and
trisect side
(
);
- the areas of
are
sq. unit each (making the overall area of
to be
sq. units).
- the slope of
- (Higher powers) Given
with vertices at
, the slopes of its sides form a geometric progression with common ratio
. Find two points
and
on
such that the slopes of the sides of
form a geometric progression with common ratio
.
- (Negative region) Given
with vertices
, find two points
and
on
such that the slope of any line segment
, where
is a point between
and
, is negative.
- (Base five) Explain how to generate the geometric progression
using slopes.
- (Base six) Explain how to generate the geometric progression
using slopes.
- (Base seven) Explain how to generate the geometric progression
using slopes.
- (Terminal point) Given
with vertices
, determine the coordinates of the terminal point
on
(a point on
such that the slope of the line segment
is
; this is also where the slopes of the line segments from vertex
change from positive powers of
to negative powers of
).