Pearson Geometry Common Core, 2011
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Pearson Geometry Common Core, 2011 View details
4. Medians and Altitudes
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Exercise 15 Page 313

The orthocenter describes the point of concurrency for the lines containing the altitudes of a triangle.

(6,4)

Practice makes perfect

We want to determine where the orthocenter lies and then find its exact location. Let's begin by drawing the triangle using the given coordinates.

To find the location of the orthocenter, we need to recall two definitions.

  1. Orthocenter: Describes the point of concurrency for the lines containing the altitudes of a triangle.
  2. Altitude of a Triangle: The perpendicular segment from a vertex to the opposite side of a triangle or to the line containing the opposite side.
Let's draw the altitudes of the vertices of our triangle.

We can see that the altitudes intersect inside the triangle. Therefore, the orthocenter lies inside the triangle. To find its coordinates, we should determine the equations for two of the altitudes and solve them as a system of equations. Let's use the altitudes of AB and AC.

Equation of the Altitude of AB

Since AB is horizontal, its altitude will be vertical. From the diagram, we can see that PC is a vertical line through x=6. Therefore, the equation of the line for the line segment of the altitude is x=6.

Equation of the Altitude of AC

To find the equation for the second altitude, we need the slope of AC. We can use the Slope Formula, and the coordinates of A and C, to do this. Let's find the slope first.
m = y_2 - y_1/x_2 - x_1
m = 2 - 6/6 - 2
â–Ľ
Simplify right-hand side
m = - 4/4
m = - 1
We found that the slope of AC is - 1. We can use the fact that the product of the slopes of two perpendicular lines is -1 to find the slope of the altitude. Let's call it m_a. -1* m_a = -1 ⇒ m_a = 1 The slope of the altitude is 1. From the diagram, we also know that the altitude passes through the point B(8,6). We can substitute this point into the above equation and solve for b.
y=x + b
6= 8+b
-2=b
b=-2
Therefore, the equation is y= x -2.

Solving for the Coordinates

Finally, we can solve the system of the two equations to find the coordinates of their intersection.
x=6 & (I) y=x-2 & (II)
x=6 y= 6-2
x=6 y=4
Therefore, the coordinates of the orthocenter are (6,4).