Given **D** **ABC**, let **lines*** l* and

(1) The slope of a line can be calculated using two points(a, b)and(c, d)on the line.

(2) The slopes of perpendicular lines are negative reciprocals of each other.

(3) The point-slope form of the equation of a line isy - y1 =mwhere(x - x1)

(x1 , y1)is a point on the line andis the slope of the line.m

(4) The slope-intercept form of the equation of a line isy =m+xbwhereis them

slope of the line andbis the y-intercept.

Let the vertices have the following coordinates: **A** (b,
c); **B** (0, 0); **C** (a, 0).

The slope of the line containing **BC**
is

Therefore, the slope of the line perpendicular
to line **BC **through** A **at **(b, c) **is undefined.
The equation of the perpendicular line is **x = b**. The slope
of the line containing **AC** is

Therefore, the slope of the line perpendicular
to line **AC** through **B** at **(0, 0) **is

The equation of the perpendicular line is

The slope of the line containing **AB**
is

Therefore, the slope of the line perpendicular
to line **AB** through **C** at **(a, 0) **is **-(b/c)**.
The equation of the perpendicular line is

This information is summarized in the following chart:

To find the coordinates of **H**, the intersection
point of lines ** l** and

The y-coordinate can be found by substituting
for x:

Therefore, the coordinates of **H** are

To determine if the line containing the altitude
from **C **to **AB**, substitute the coordinates for **H**
into the equation

.

This implies that **H** does in fact also
lie on the line containing the third altitude of the triangle.
Therefore, the three altitudes of the triangle are concurrent
at point **H.** This point is called the ** orthocenter**
and is labeled as

This proof holds for obtuse and right triangles
(See figures** **below) as well. Notice that if the triangle
is obtuse, the orthocenter lies outside the triangle. If the triangle
is right, the orthocenter is the vertex of the right angle since
the two sides that form the right angle lie on two altitudes.