Question

Exercises: COORDINATE GEOMETRY Find the coordinates of the orthocenter of the triangle with the given vertices. 1. J(1, 0), H(6, 0), I(3, 6) 2. S(1, 0), T(4, 7), U(8, -3)

Answer 1

We have two triangles and we have to find the orthocenter.

The orthocenter is the intersection of the three segments that start in each vertex and are perpendicular to its opposite side:

1) The triangle has vertices J(1, 0), H(6, 0), I(3, 6).

We need to find the equations of the perpendicular lines to each side. Then, we have to first know the slope of each side (in fact, with only two sides, we can find the otrhocenter).

We start with the slope of segment JH. We can calculate the slope as:

`\begin{gathered} m_(JH)=(y_H-y_J)/(x_H-x_J) \\ m_(JH)=(0-0)/(6-1)=(0)/(5)=0 \end{gathered}`

This segment, because of the y-coordinates of this two points, is an horizontal line (slope 0).

The perpendicular line will be then a vertical line, with slope undefined.

We know that it pass through the other vertex, I(3,6), so it will have the same x-coordinate. The perpendicular line is then x = 3.

We know calculate the slope of another sement, like JI:

`\begin{gathered} m_(JI)=(y_I-y_J)/(x_I-x_J) \\ m_(JI)=(6-0)/(3-1)=(6)/(2)=3 \end{gathered}`

As the slope of the segment is m = 3, the slope of the perpendicular line will be its negative reciprocal:

`m_(JI\perp)=-(1)/(m_(JI))=-(1)/(3)`

Then, the perpendicular line has slope -1/3 and it pass through the other vertex, H(6,0), so we can write the equation of this line as:

`\begin{gathered} y-y_0=m(x-x_0) \\ y-0=-(1)/(3)(x-6) \\ y=-(1)/(3)x+2 \end{gathered}`

We then have a system of equation with the two perpendicular lines. The point where they intersect will be the coordinates of the orthocenter:

`\begin{gathered} x=3 \\ y=-(1)/(3)x+2 \end{gathered}`

As we know that x = 3, we can calculate y as:

`y=-(1)/(3)(3)+2=-1+2=1`

The orthocenter is at (3,1).

b) We have a triangle with vertices S(1, 0), T(4, 7), U(8, -3)​.

We start by calculating the slope of the segment ST:

`\begin{gathered} m_(ST)=(y_T-y_S)/(x_T-y_S) \\ m_(ST)=(7-0)/(4-1)=(7)/(3) \end{gathered}`

The perpendicular line, that will pass through the other vertex U, will have a slope that is the negative reciprocal, so knowing U=(8,-3) and the slope of the line, we can write the equation of the perpendicular line as:

`\begin{gathered} y-y_U=-(1)/(m_(ST))(x-x_U) \\ y-(-3)=-(3)/(7)(x-8) \\ y+3=-(3)/(7)(x-8) \end{gathered}`

We know look at the slope of another segment, like TU:

`\begin{gathered} m_(TU)=(y_U-y_T)/(x_U-x_T) \\ m_(TU)=(-3-7)/(8-4)=-(10)/(4)=-(5)/(2) \end{gathered}`

The perpendicular line will pass through S(1,0), so we can write the equation of the perpendicular line as:

`\begin{gathered} y-y_S=-(1)/(m_(TU))(x-x_S) \\ y-0=(2)/(5)(x-1) \\ y=(2)/(5)(x-1) \end{gathered}`

We now have a system of equations: