Exercise 10 Page 351

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 altitude of the vertices of our triangle.

   

   We can see that the altitudes intersect outside the triangle. Therefore, the orthocenter lies on the outside of 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 KM and KL.

   Equation of the Altitude of KM

   Since KM is horizontal, its altitude will be vertical. From the diagram, we can see that it is a vertical line passing through the vertex L( -6,3). Therefore, the equation of the line for the line segment of the altitude is x= -6.

   

   Equation of the Altitude of KL

   To find the equation for the second altitude, we need the slope of KL. We can use the Slope Formula and the coordinates of A and C to do this.
   m = y_2 - y_1/x_2 - x_1

   SubstitutePoints

   Substitute ( -8,5) & ( -6,3)

   m = 3 - 5/-6 - ( -8)

   ▼

   Simplify right-hand side

   SubNeg

   a-(- b)=a+b

   m=3-5/-6+8

   AddSubTerms

   Add and subtract terms

   m = -2/2

   CalcQuot

   Calculate quotient

   m = - 1
   We found that the slope of KL is - 1. 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. -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 vertex M(0,5). This is a y-intercept. Therefore, the equation becomes y=x+5.
   

   Solving for the Coordinates

   Finally, we can solve the system of the found equations to find the coordinates of their intersection.
   x=-6 & (I) y=x+5 & (II)

   ▼

   Solve by substitution

   Substitute

   (II): x= -6

   x=-6 y= -6+5

   AddTerms

   (II): Add terms

   x=-6 y=-1
   Therefore, the coordinates of the orthocenter are (- 6,- 1).