What is the orthocenter of a triangle with corners at (1 ,3 ), (5 ,7 ), and (9 ,8 )#?

1 Answer
Mar 25, 2016

(-10/3,61/3)

Explanation:

Repeating the points:
A(1,3)
B(5,7)
C(9,8)

The orthocenter of a triangle is the point where the line of the heights relatively to each side (passing through the opposed vertex) meet. So we only need the equations of 2 lines.

The slope of a line is k=(Delta y)/(Delta x) and the slope of the line perpendicular to the first is p=-1/k (when k!=0).

AB-> k_1=(7-3)/(5-1)=4/4=1 => p_1=-1
BC-> k=(8-7)/(9-5)=1/4 => p_2=-4

Equation of line (passing through C) in which lays the height perpendicular to AB
(y-y_C)=p(x-x_C) => (y-8)=-1*(x-9) => y=-x+9+8 => y=-x+17 [1]

Equation of line (passing through A) in which lays the height perpendicular to BC
(y-y_A)=p(x-x_A) => (y-3)=-4*(x-1) => y=-4x+4+3 => y=-4x+7[2]

Combining equations [1] and [2]
{y=-x+17
{y=-4x+7 => -x+17=-4x+7 => 3x=-10 => x=-10/3

->y=10/3+17=(10+51)/3 => y=61/3

So the orthocenter P_"orthocenter" is (-10/3,61/3)