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

3 Answers
Feb 10, 2016

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 #

enter image source here

Feb 10, 2016

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 #

enter image source here

Feb 10, 2016

Answer:

Orthocenter(16/2, 11/3)

Explanation:

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 #

enter image source here