What is the orthocenter of a triangle with corners at #(4 ,9 )#, #(3 ,4 )#, and (5 ,1 )#?

Question

What is the orthocenter of a triangle with corners at #(4  ,9  )#, #(3 ,4   )#, and (5   ,1   )#?

Sadie Adams · Accepted Answer

The orthocenter of the triangle is #=(-5,3)# Let the triangle #DeltaABC# be #A=(4,9)# #B=(3,4)# #C=(5,1)# The slope of the line #BC# is #=(1-4)/(5-3)=-3/2# The slope of the line perpendicular to #BC# is #=2/3# The equation of the line through #A# and perpendicular to #BC# is #y-9=2/3(x-4)# #3y-27=2x-8# #3y-2x=19#...................#(1)# The slope of the line #AB# is #=(4-9)/(3-4)=-5/-1=5# The slope of the line perpendicular to #AB# is #=-1/5# The equation of the line through #C# and perpendicular to #AB# is #y-1=-1/5(x-5)# #5y-5=-x+5# #5y+x=10#...................#(2)# Solving for #x# and #y# in equations #(1)# and #(2)# #3y-2(10-5y)=19# #3y-20+10y=19# #13y=20+19=39# #y=39/13=3# #x=10-5y=10-15=-5# The orthocenter of the triangle is #=(-5,3)#

Aurora Ayala · Answer

undefined The orthocenter of a triangle is the point where the three altitudes of the triangle intersect. To find the orthocenter, you need to find the intersection point of the altitudes.

1. Find the slopes of the lines passing through each pair of vertices to determine the slopes of the altitudes.
2. Use the point-slope form to write the equations of the altitudes.
3. Find the intersection point of any two altitude lines to determine the orthocenter.

Given the coordinates of the vertices:
A (4, 9)
B (3, 4)
C (5, 1)

1. Find the slopes of the lines AB, BC, and CA using the formula: $m = \frac{{y_2 - y_1}}{{x_2 - x_1}}$.

$m_{AB} = \frac{{4 - 9}}{{3 - 4}} = 5$

$m_{BC} = \frac{{1 - 4}}{{5 - 3}} = -\frac{3}{2}$

$m_{CA} = \frac{{1 - 9}}{{5 - 4}} = -8$

2. Use the point-slope form $y - y_1 = m(x - x_1)$ to write the equations of the altitudes passing through each vertex.

Equation of the altitude from A:
$y - 9 = 5(x - 4)$
$y - 9 = 5x - 20$
$y = 5x - 11$

Equation of the altitude from B:
$y - 4 = -\frac{3}{2}(x - 3)$
$y - 4 = -\frac{3}{2}x + \frac{9}{2}$
$y = -\frac{3}{2}x + \frac{17}{2}$

Equation of the altitude from C:
$y - 1 = -8(x - 5)$
$y - 1 = -8x + 40$
$y = -8x + 41$

3. Now, find the intersection points of any two altitude lines. For example, let's find the intersection of altitudes from A and B:

$5x - 11 = -\frac{3}{2}x + \frac{17}{2}$

Solve for $x$:

$5x + \frac{3}{2}x = \frac{17}{2} + 11$

$x = \frac{29}{7}$

Substitute $x$ back into one of the altitude equations to find $y$:

$y = 5 \times \frac{29}{7} - 11$

$y = \frac{29}{7}$

So, the intersection point of the altitudes from A and B is $\left(\frac{29}{7}, \frac{29}{7}\right)$.

Similarly, find the intersection points of other pairs of altitude lines. The point where all three altitudes intersect is the orthocenter.