What is the orthocenter of a triangle with corners at #(2 ,7 )#, #(1 ,2 )#, and (3 ,5 )#?

Answer 1

Orthocenter is at #(41/7,31/7)#

Slope of the line AB: #m_1= (2-7)/(1-2)=5# Slope of CF=perpendicular slope of AB: # m_2= -1/5# Equation of the line CF is #y-5= -1/5(x-3) or 5y-25 =-x+3 or x+5y=28 (1)# Slope of the line BC: #m_3= (5-2)/(3-1)=3/2# Slope of AE=perpendicular slope of BC: # m_4= -1/(3/2)=-2/3# Equation of the line AE is #y-7= -2/3(x-2) or 3y-21 =-2x+4 or 2x+3y=25 (2)# The intersection of CF & AE is the orthocenter of the triangle, which can be obtained by solving equation(1) & (2) #x+5y=28 (1)# ; #2x+3y=25 (2)# #2x+10y=56 (1)# obtained by multiplying 2 on both sides #2x+3y=25 (2)# subtracting we get #7y=31:. y=31/7; x=28-5*31/7=41/7:.#Orthocenter is at #(41/7,31/7)#[Ans]

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Answer 2

Carson Arnold

The orthocenter of the triangle with corners at (2, 7), (1, 2), and (3, 5) is approximately (1.4, 4.4).

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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.