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

Question

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

Aaliyah Anderson · Accepted Answer

#(101/23, 91/23)#

Orthocenter of a triangle is a point where the three altitudes of a triangle meet. To find the orthocentre, it would be enough, if intersection of any two of the altitudes is found out. To do this, let the vertices be identified as A(5,7), B(2,3), C(7,2).

Slope of line AB would be #(3-7)/(2-5) = 4/3#. Hence the slope of the altitude from C(7,2) onto AB would be #-3/4#. The equation of this altitude would be #y-2= -3/4( x-7)#

Now consider the slope of line BC, it would be #(2-3)/(7-2)= -1/5#. Hence the slope of the altitude from A(5,7) on to BC would be 5. The equation of this altitude would be #y-7= 5 (x-5)#

Now eliminating y from the two equations of altitudes , by subtracting one eq from the other it would be #5=-(3x)/4 -5x +21/4+25# , #->(23x)/4=101/4 -> x= 101/23#. Then #y=7+5(101/23-5)= 91/23#

The orthocentre is thus #(101/23, 91/23)#

Carson Arnold · Answer

undefined The orthocenter of the triangle with corners at (5, 7), (2, 3), and (7, 2) is located at the point where the altitudes of the triangle intersect. To find the orthocenter, you need to find the equations of the altitudes, then solve for their intersection point.