What is the orthocenter of a triangle with corners at #(4 ,3 )#, #(9 ,1 )#, and (8 ,3 )#?

Answer 1

#(9, 5.5)#

Let #O(a, b)# be the ortho-center of #\Delta ABC# with the vertices #A(4,3)#, #B(9, 1)# & #C(8, 3)#

Ortho-center of a triangle is the point of intersection of its altitudes drawn from the vertices to opposite sides.

By property of ortho-center, the line #AO# will be perpendicular to the opposite side #BC# i.e. the product of the slopes of lines #AO# & #BC# will be #-1# hence we have

#(\frac{b-3}{a-4})(\frac{3-1}{8-9})=-1#

#a-2b=-2\ ..........(1)#

Similarly, the line #BO# will be perpendicular to the opposite side #AC# i.e. the product of the slopes of lines #BO# & #AC# will be #-1# hence we have

#(\frac{b-1}{a-9})(\frac{3-3}{8-4})=-1#

#a-9=0#

#a=9#

setting #a=9# in (1), we get

#9-2b=-2#

#b=5.5#

hence, the ortho-center of given triangle is

#(9, 5.5)#

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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.