Orthocenter
Let
A(2,0)
B(3,2)
C(6,0)
step 1
Find out the slope AB
m=(2-0)/(3-2)
m=2/1
m=2
Remember that
If two lines are perpendicular, then, their slopes are negative reciprocal
so
the slope of the altitude that passes through C is -1/2
Find out the equation of the altitude that passes through C
y=mx+b
we have
m=-1/2
point C(6,0)
substitute
0=-(1/2)(6)+b
solve for b
0=-3+b
b=3
equation is y=-(1/2)x+3 -------> equation 1
step 2
Find out the slope BC
m=(0-2)/(6-3)
m=-2/3
so
the slope of the altitude that passes through A is 3/2
Find out the equation of the altitude that passes through A
y=mx+b
we have
m=3/2
point A(2,0)
0=(3/2)(2)+b
solve for b
0=3+b
b=-3
y=(3/2)x-3 ------> equation 2
step 3
the intersection point equation 1 and equation 2 is the orthocenter
y=-(1/2)x+3 -----> equation 1
y=(3/2)x-3 ------> equation 2
solve the system
solve by graphing
using a graphing tool
the intersection point is (3,1.5)
that means
the orthocenter is (3,1.5)
see the attached figure to better understand the problem