What is the orthocenter of a triangle with corners at #(4 ,1 )#, #(7 ,4 )#, and (3 ,6 )#?

The trick to this little problem is to find the slope between two points from there find the slope of perpendicular line which simply given by:
1) #m_(perp) = -1/m_("original") # then
2) find the equation of line that passes through the angle opposite the original line for you case give: A(4,1), B(7, 4) and C(3,6)
step1:
Find the slope of #bar(AB) => m_(bar(AB)) #
#m_(bar(AB)) = (4-1)/(7-4) = 3 :. m_(perp) = m_(bar(CD)) = -1/1 = -1 #
To get the equation of line write:
#y = m_bar(CD)x + b_bar(CD); #use point C(3, 6) to determine #barB#
#6 = -3 + b_bar(CD); b_bar(CD) = 9 :. #
#y_bar(CD) = color(red)(-x + 9) # #color(red)" Eq. (1)"#

step2
Find the slope of #bar(CB) => m_(bar(CB)) #
#m_(bar(AB)) = (6-4)/(3-7) = -1/2 :. m_(perp) = m_(bar(AE)) = 2#
To get the equation of line write:
#y = m_bar(AE)x + b_bar(AE); #use point A(4, 1) to determine #barB#
#1 = 8 + b_bar(AE); b_bar(CD) = -7 :. #
#y_bar(AE) = color(blue)(2x - 7) # #color(blue)" Eq. (2)"#
Now equate #color(red)" Eq. (1)"# = #color(blue)" Eq. (2)"#
Solve for => #x = 16/3#
Insert #x=2/3# into #color(red)" Eq. (1)"#
#y = -2/3 + 9 = 11/3 #

The trick to this little problem is to find the slope between two points from there find the slope of perpendicular line which simply given by:
1) #m_(perp) = -1/m_("original") # then
2) find the equation of line that passes through the angle opposite the original line for you case give: A(4,1), B(7, 4) and C(3,6)
step1:
Find the slope of #bar(AB) => m_(bar(AB)) #
#m_(bar(AB)) = (4-1)/(7-4) = 3 :. m_(perp) = m_(bar(CD)) = -1/1 = -1 #
To get the equation of line write:
#y = m_bar(CD)x + b_bar(CD); #use point C(3, 6) to determine #barB#
#6 = -3 + b_bar(CD); b_bar(CD) = 9 :. #
#y_bar(CD) = color(red)(-x + 9) # #color(red)" Eq. (1)"#
step2
Find the slope of #bar(CB) => m_(bar(CB)) #
#m_(bar(AB)) = (6-4)/(3-7) = -1/2 :. m_(perp) = m_(bar(AE)) = 2#
To get the equation of line write:
#y = m_bar(AE)x + b_bar(AE); #use point A(4, 1) to determine #barB#
#1 = 8 + b_bar(AE); b_bar(CD) = -7 :. #
#y_bar(AE) = color(blue)(2x - 7) # #color(blue)" Eq. (2)"#
Now equate #color(red)" Eq. (1)"# = #color(blue)" Eq. (2)"#
Solve for => #x = 16/3#
Insert #x=2/3# into #color(red)" Eq. (1)"#
#y = -2/3 + 9 = 11/3 #

The trick to this little problem is to find the slope between two points from there find the slope of perpendicular line which simply given by:
1) #m_(perp) = -1/m_("original") # then
2) find the equation of line that passes through the angle opposite the original line for you case give: A(4,1), B(7, 4) and C(3,6)
step1:
Find the slope of #bar(AB) => m_(bar(AB)) #
#m_(bar(AB)) = (4-1)/(7-4) = 3 :. m_(perp) = m_(bar(CD)) = -1/1 = -1 #
To get the equation of line write:
#y = m_bar(CD)x + b_bar(CD); #use point C(3, 6) to determine #barB#
#6 = -3 + b_bar(CD); b_bar(CD) = 9 :. #
#y_bar(CD) = color(red)(-x + 9) # #color(red)" Eq. (1)"#
step2
Find the slope of #bar(CB) => m_(bar(CB)) #
#m_(bar(AB)) = (6-4)/(3-7) = -1/2 :. m_(perp) = m_(bar(AE)) = 2#
To get the equation of line write:
#y = m_bar(AE)x + b_bar(AE); #use point A(4, 1) to determine #barB#
#1 = 8 + b_bar(AE); b_bar(CD) = -7 :. #
#y_bar(AE) = color(blue)(2x - 7) # #color(blue)" Eq. (2)"#
Now equate #color(red)" Eq. (1)"# = #color(blue)" Eq. (2)"#
Solve for => #x = 16/3#
Insert #x=2/3# into #color(red)" Eq. (1)"#
#y = -2/3 + 9 = 11/3 #