Question

how do you find the orthocenter of a triangle with the vertices A(4,0) B(-2,4)
C(0,6)?
Please explain to me how you get x after doing point slope form for the two slopes.

Answer 1

the orthocentre is the intersection of the altitudes.

slope AB = (4-0)/(-2-4) = -2/3
so slope of altutude from C is 3/2 , and y-intercept is 6
equation of altitude from C is y = (3/2)x + 6
slope AC = 6/-4 = -3/2
slope of altitude from B is 2/3
equation: y = (2/3)x + b
but B(-2,4) lies on it, so
4 = (2/3)(-2) + b
12 = -4 + 3b
b = 16/3 , so altitude from B to AC is y = (2/3)x + 16/3

the orthocentre is the intersection of these two lines
(2/3)x + 16/3 = (3/2)x + 6
multiply by 6
4x + 32 = 9x + 36
-4 = 5x
x = -4/5
y = (3/2)(-4/5) + 6 = 24/5

The orthocentre is (-4/5, 24/5)