What is the orthocenter of a triangle with corners at #(7 ,8 )#, #(3 ,2 )#, and (5 ,6 )#?

Question

What is the orthocenter of a triangle with corners at #(7 ,8  )#, #(3 ,2   )#, and (5   ,6   )#?

Wyatt Anderson · Accepted Answer

The orthocenter is #=(-13,18)# Let the triangle #DeltaABC# be #A=(7,8)# #B=(3,2)# #C=(5,6)# The slope of the line #BC# is #=(6-2)/(5-3)=4/2=2# The slope of the line perpendicular to #BC# is #=-1/2# The equation of the line through #A# and perpendicular to #BC# is #y-8=-1/2(x-7)#...................#(1)# #2y=-x+7+16=-x+23# The slope of the line #AB# is #=(2-8)/(3-7)=-6/-4=3/2# The slope of the line perpendicular to #AB# is #=-2/3# The equation of the line through #C# and perpendicular to #AB# is #y-6=-2/3(x-5)# #3y-18=-2x+10# #3y=-2x+28#...................#(2)# Solving for #x# and #y# in equations #(1)# and #(2)# #-3x+69=-4x+56# #4x-3x=56-69# #x=-13# #y=(13+23)/2=18# The orthocenter of the triangle is #=(-13,18)#

James Atkins · Answer

undefined To find the orthocenter of a triangle, you need to find the point of intersection of the altitudes of the triangle. The altitude of a triangle is a line segment from a vertex perpendicular to the opposite side.

Here's how you can find the orthocenter:

1. Find the slopes of the lines passing through each pair of vertices.
2. Determine the slopes of the altitudes by taking negative reciprocals of the slopes of the corresponding sides.
3. Use the point-slope form to find the equations of the altitudes.
4. Find the intersection point of the altitudes, which is the orthocenter.

Let's go through these steps:

1. Slope of the line passing through (7, 8) and (3, 2):
$$ m_1 = \frac{2 - 8}{3 - 7} = -\frac{3}{2} $$

2. Slope of the line passing through (3, 2) and (5, 6):
$$ m_2 = \frac{6 - 2}{5 - 3} = 2 $$

3. Slope of the line passing through (5, 6) and (7, 8):
$$ m_3 = \frac{8 - 6}{7 - 5} = 1 $$

4. The slopes of the altitudes are the negative reciprocals of the slopes of the corresponding sides:
   - Altitude from (7, 8) to the line passing through (3, 2) and (5, 6): slope $m_1' = \frac{1}{2}$
   - Altitude from (3, 2) to the line passing through (7, 8) and (5, 6): slope $m_2' = -\frac{1}{2}$
   - Altitude from (5, 6) to the line passing through (3, 2) and (7, 8): slope $m_3' = -1$

5. Using the point-slope form, find the equations of the altitudes:
   - Altitude passing through (7, 8) with slope $m_1'$: $y - 8 = \frac{1}{2}(x - 7)$
   - Altitude passing through (3, 2) with slope $m_2'$: $y - 2 = -\frac{1}{2}(x - 3)$
   - Altitude passing through (5, 6) with slope $m_3'$: $y - 6 = -1(x - 5)$

6. Solve the system of equations to find the intersection point, which is the orthocenter.

By solving the system of equations formed by the three altitude lines, you'll find the coordinates of the orthocenter.