In Exercises 45 and 46, complete an analytic proof for each theorem.
The altitudes of a triangle are concurrent.
The altitudes of triangle are concurrent.
The triangle with vertices , and .is shown in the above figure.
Let , , be the altitudes of the triangle .
If the point of intersection of the altitudes is at the same point, then they are said to be concurrent.
If two lines are perpendicular, then the product of their slopes is or one slope is negative reciprocal of the other slope.
(i.e.) If , then or
Since is the horizontal line, slope is zero.
Obviously the altitude is vertical and it has the equation,
The slope of the line with vertices and is as follows.
Using the slope formula and choosing , , , and .
Thus the slope of the altitude which is perpendicular to the line is
The equation of the line that contains the point with slope is as follows:
Initially substitute in the equation ,
Substitute in the equation ,
The intersection of these altitudes is found by solving the two altitude equation.
The coordinates of orthocenter is
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