The orthocenter is defined as the point at which the three altitudes of a triangle meet. Let's prove that the three altitudes of any triangle really do meet in a single point.
First, let's find equations for the altitudes of a triangle.
is perpendicular to 
, so their slopes must be negative reciprocals.
is horizontal, so is a vertical line.
And goes through the point (2b, 2c), so is the line described by
x = 2b.
is perpendicular to 
, so their slopes must be negative reciprocals.
If is not vertical (i.e., if 2b ≠ 0), its slope is 
so the slope of is 
goes through the point (2a,0), so is the line described by
If is vertical (i.e., if 2b = 0), then the slope of is 0.
goes through the point (2a,0), so is the line described by
.
So, whether is vertical or not, is the line described by
is perpendicular to 
, so their slopes must be negative reciprocals.
If is not vertical (i.e., if 2b ≠ 2a), its slope is 
so the slope of is 
goes through the point (0,0), so is the line described by
or, more simply,
If is vertical (i.e., if 2b = 2a), then the slope of is 0.
goes through the point (0,0), so is the line described by
.
So, whether is vertical or not, is the line described by
Let's find the point 
at which and intersect, and then we can determine whether lies on as well. If it does, we will have shown that all three altitudes intersect in a single point.
We need to find a point that is on both and
The y-coordinate of both lines must be the same at x = 
.
So we must have:
And
Does our point lie on ? Let's find out.
The x-coordinate of any point on is 2b.
= 2b, so does indeed lie on .
|
So for any triangle, all three altitudes intersect at a single point, the orthocenter.
For a triangle whose vertices are (0,0), (2a,0), and (2b,2c), the coordinates of the orthocenter are
|
Click here to continue.
<< Previous (Centroid) Next (Circumcenter) >>
Return to my page
Return to EMAT 6680 page
