What is the orthocenter of a triangle with corners at #(4 ,7 )#, #(9 ,5 )#, and #(5 ,6 )#?

Question

What is the orthocenter of a triangle with corners at #(4  ,7  )#, #(9 ,5   )#, and #(5   ,6   )#?

Aurora Ayala · Accepted Answer

#color(blue)((5/3,-7/3)#

The orthocenter is the point where the extended altitudes of a triangle meet. This will be inside the triangle if the triangle is acute, outside of the triangle if the triangle is obtuse. In the case of the right angled triangle it will be at the vertex of the right angle. ( The two sides are each altitudes ).

It is generally easier is you do a rough sketch of the points so you know where you are.

Let #A=(4,7), B=(9,5), C=(5,6)#

Since the altitudes pass through a vertex and are perpendicular to the side opposite, we need the find the equations of these lines. It will be obvious from the definition that we only need to find two of these lines. These will define a unique point. It is unimportant which ones you choose.

I will use:

Line #AB# passing through #C#

Line #AC# passing through #B#

For #AB#

First find the gradient of this line segment:

#m_1=(6-7)/(5-4)=-1#

A line perpendicular to this will have a gradient that is the negative reciprocal of this:

#m_2=-1/m_1=-1/(-1)=1#

This passes through #C#. Using point slope form of a line:

#y-5=1(x-9)#

#y=x-4 \ \ \ \ [1]#

For #AC#

#m_1=(5-7)/(9-4)=-2/5#

#m_2=-1/(-2/5)=5/2#

Passing through #B#

#y-6=5/2(x-5)#

#y=5/2x-13/2 \ \ \ \ \[2]#

The intersection of #[1]# and #[2]# will be the orthocenter:

Solving simultaneously:

#5/2x-13/2-x+4=0=>x=5/3#

Substituting in #[1]#:

#y=5/3-4=-7/3#

Orthocenter:

#(5/3,-7/3)#

Notice the the orthocenter is outside the triangle because it is obtuse. The altitude lines passing through #C# and #A# have to be produced at D and E to allow for this.

James Atkins · Answer

undefined To find the orthocenter of a triangle, we first need to find the altitude of each side of the triangle. The altitude is a perpendicular line from one vertex to the opposite side.

1. Calculate the slopes of the lines containing each side of the triangle.
2. Determine the perpendicular slopes to each side.
3. Find the equations of the lines perpendicular to each side, passing through the opposite vertex.
4. Find the intersection point of these perpendicular lines. This point is the orthocenter.

Let's denote the vertices of the triangle as A(4, 7), B(9, 5), and C(5, 6).

The slope of line AB: m_AB = (5 - 7) / (9 - 4) = -2/5.
The slope of line BC: m_BC = (6 - 5) / (5 - 9) = 1/(-4) = -1/4.
The slope of line AC: m_AC = (6 - 7) / (5 - 4) = -1.

The perpendicular slopes are the negative reciprocals of the original slopes:
Perpendicular to AB: m_perp_AB = -1/m_AB = 5/2.
Perpendicular to BC: m_perp_BC = -1/m_BC = 4.
Perpendicular to AC: m_perp_AC = -1/m_AC = 1.

Now, we use point-slope form to find the equations of the lines passing through each vertex and perpendicular to the opposite side:
Line perpendicular to AB passing through C: y - 6 = 1(x - 5) --> y = x + 1.
Line perpendicular to BC passing through A: y - 7 = 5/2(x - 4) --> y = (5/2)x - 3.
Line perpendicular to AC passing through B: y - 5 = 4(x - 9) --> y = 4x - 31.

Now, we solve the system of equations formed by these three lines to find the intersection point, which is the orthocenter of the triangle. This point is (5, 4). Therefore, the orthocenter of the triangle with vertices (4, 7), (9, 5), and (5, 6) is at coordinates (5, 4).

Avery Allen · Answer

undefined The orthocenter of the triangle with corners at (4, 7), (9, 5), and (5, 6) is at the point (6.2, 5.8).