What is the orthocenter of a triangle with corners at #(3 ,1 )#, #(1 ,3 )#, and (5 ,2 )#?

Question

What is the orthocenter of a triangle with corners at #(3   ,1   )#, #(1   ,3   )#, and (5    ,2   )#?

Aurora Ayala · Accepted Answer

The orthocenter is at the point #(4/3, -17/3)#

Let's begin by writing the equation of the line that goes through point #(3,1)# and perpendicular to the line going through points #(1,3)# and #(5,2)#.

The slope, m, of the line going through points #(1,3)# and #(5,2)# is:

#m = (3 - 2)/(1 - 5) = -1/4#

Any perpendicular line's slope, n, is

#n = -1/m = 4#

To get the desired line equation, use the point slope form of the equation of a line:

#y - y_1 = n(x - x_1)#

#y - 1 = 4(x - 3)#

#y - 1 = 4x - 12#

#y = 4x - 11#

Write the equation of a line that goes through point #(5,2)# perpendicular to the line through the points #(3,1)# and #(1,3)#:

The slope, m, of the line that goes through the points #(3,1)# and #(1,3)# is:

#m = (3 - 1)/(1 - 3) = 2/-2 = -1#

Any perpendicular line has a slope, n, which is:

#n = -1/m = 1#

Again, Use the point slope form for the point #(5,2)#:

#y - y_1 = n(x - x_1)#

#y - 2 = 1(x - 5)#

#y = x - 7#

The point where these two lines intersect is the orthocenter:

#y = 4x - 11# #y = x - 7#

#x - 7 = 4x - 11#

#4 = 3x#

#x = 4/3#

#y = 4/3 - 7#

#y = -17/3#

The orthocenter is at the point #(4/3, -17/3)#

Aaliyah Anderson · Answer

undefined To find the orthocenter of a triangle with vertices (3,1), (1,3), and (5,2), follow these steps:

1. Determine the slopes of the lines passing through each pair of vertices.
2. Find the perpendicular bisectors of the sides of the triangle.
3. The point of intersection of the perpendicular bisectors is the orthocenter.

Let's find the slopes of the sides of the triangle:

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

- Slope of the line passing through (3,1) and (5,2):
  $ m_2 = \frac{{2 - 1}}{{5 - 3}} = \frac{1}{2} $

- Slope of the line passing through (1,3) and (5,2):
  $ m_3 = \frac{{2 - 3}}{{5 - 1}} = -\frac{1}{4} $

Now, find the equations of the perpendicular bisectors:

- For the line passing through (3,1) and (1,3), the midpoint is $ \left(\frac{{3 + 1}}{2}, \frac{{1 + 3}}{2}\right) = (2,2) $. Its slope will be the negative reciprocal of $ m_1 $, so it is $ m_1' = 1 $. The equation of the perpendicular bisector is $ y - 2 = 1(x - 2) $ or $ y = x $.

- For the line passing through (3,1) and (5,2), the midpoint is $ \left(\frac{{3 + 5}}{2}, \frac{{1 + 2}}{2}\right) = (4,1.5) $. Its slope will be the negative reciprocal of $ m_2 $, so it is $ m_2' = -2 $. The equation of the perpendicular bisector is $ y - 1.5 = -2(x - 4) $ or $ y = -2x + 9 $.

- For the line passing through (1,3) and (5,2), the midpoint is $ \left(\frac{{1 + 5}}{2}, \frac{{3 + 2}}{2}\right) = (3,2.5) $. Its slope will be the negative reciprocal of $ m_3 $, so it is $ m_3' = 4 $. The equation of the perpendicular bisector is $ y - 2.5 = 4(x - 3) $ or $ y = 4x - 10 $.

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

$ y = x $  (1)

$ y = -2x + 9 $ (2)

$ y = 4x - 10 $ (3)

Solving (1) and (2), we get $ x = 2 $ and $ y = 2 $.

Solving (1) and (3), we get $ x = \frac{5}{3} $ and $ y = \frac{5}{3} $.

Therefore, the orthocenter of the triangle is $ \left(\frac{5}{3}, \frac{5}{3}\right) $.