Big Ideas Math Geometry, 2014
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Big Ideas Math Geometry, 2014 View details
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Exercise 11 Page 328

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

Location: Inside
Orthocenter: (0,4)

Practice makes perfect

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. The orthocenter describes the point of concurrency for the lines containing the altitudes of a triangle.
  2. An altitude of a triangle is 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 the system of these equations. Let's use the altitudes of TV and UV.

Equation of the Altitude of TV

Since TV is horizontal, its altitude will be vertical. From the diagram, we can see that PU is a vertical line through x=0. Therefore, the equation of the line for the line segment of the altitude is x=0.

Equation of the Altitude of UV

To find the equation for the second altitude, we need the slope of UV. We can use the Slope Formula and the coordinates of U and V to do this.

m = y_2 - y_1/x_2 - x_1
m = 5 - 1/2 - 0
â–¼
Simplify right-hand side
m = 4/2
m = 2

We found that the slope of UV is 2. The product of the slopes of two perpendicular lines is -1. This allows us to find the slope of the altitude, let's call it m_a. 2* m_a = -1 ⇒ m_a = -1/2 The slope of the altitude is - 12. y=-1/2x+b To complete its equation, we need the y-intercept. We can use the fact that the altitude passes through the vertex T(-2,5).

y=-1/2x+b
5=-1/2( -2)+b
â–¼
Solve for b
5=2/2+b
5=1+b
4=b
b=4

Therefore, the equation of the altitude of UV is y=- 12x+4.

Solving for the Coordinates

Finally, we can solve the system of the found equations to find the coordinates of their intersection.

x=0 & (I) y=-1/2x+4 & (II)
â–¼
Solve by substitution
x=0 y=-1/2( 0)+4
x=0 y=0+4
x=0 y=4

Therefore, the coordinates of the orthocenter are (0,4).