Exercise 14 Page 340

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 JL and JK.

Equation of the Altitude of JL

Since JL is horizontal , its altitude will be vertical. From the diagram, we can see that OK is a vertical line through x=5. Therefore, the equation of the line for the line segment of the altitude is x=5.

Equation of the Altitude of JK

To find the equation for the second altitude, we need the slope of JK. We can use the Slope Formula and the coordinates of J and K to do this.

We found that the slope of JK is 4. 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. 4* m_a = -1 ⇒ m_a = - 1/4 The slope of the altitude is - 14. From the diagram, we also know that the altitude passes through the point L(9,- 2). We will use point-slope form of a line to write its equation. y-y_1=m(x-x_1) Let's substitute - 14 for m and ( 9, - 2 ) for (x_1,y_1) in the formula.

This final equation is the equation of the line for the line segment of the altitude.

Solving for the Coordinates

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

Therefore, the coordinates of the orthocenter are (5,- 1).