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

Answer 1

Orthocenter(16/2, 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 #

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Answer 2

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 #

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Answer 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 #

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Answer 4

The orthocenter of a triangle is the point where the three altitudes of the triangle intersect. To find the orthocenter, you need to determine the equations of the altitudes and then find their point of intersection.

First, find the slopes of the sides of the triangle using the given coordinates. Then, find the equations of the perpendicular lines passing through each vertex (these are the altitudes). Finally, solve the system of equations formed by these perpendicular lines to find their point of intersection, which is the orthocenter.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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