find the orthocentre of these 3 points, A(-9,-28) B(-57,-10) C (39,50)?

Question

i know how to find it..justu that i am not getting the right answer!!!!!

i know how to find it..jus that im not getting an orthocentre that works when i check it!!!

Answer 1

1. Determine the equation of the line from C perpendicular to the line AB.
2. Determine the equation of the line from B perpendicular to the line AC.
3. Determine the intersection point of these two lines. it is the orthocenter.

Answer 2

Let A(-9,-28), B(-57,-10) and C(39,50) is a triangle.

let the orthocenter, P =(x,y)

let AP and BP are altitude from A and B

the slope of BC = 50-(-10)/(39-(-57) = 60/96 = 5/8

(BP and BCare pendicular and their product of slopes =-1)

the slope of AP = -1/5/8 = -8/5

since A(-9,-28) and P(x,y)

slope of AP = (y +28)/(x+9) = -8/5

-8x - 72 = 5y + 140

-8x - 5y = 212 ------------eqn(1)

now slope of AC = (50+28)/(39+9) = 78/48=13/8

slope of BP = -8/13

the slope of BP = (y + 10)/(x+57)

so (y + 10)/(x+57) = -8/13

13y + 130= -8x - 456

8x + 13y = -586 ------------eqn(2)

adding (1) and (2)

8y = -374

y = -374/8 = -187/4

substituting y in -8x - 5y = 212

-8x -5(-187/4) = 212

-8x+ 935/4 = 212

-32x + 935 = 848

-32x = -935 + 848- = -87

x = -87/32

The ortho center is (-87/32, -187/4)