Respuesta :

Ziexli
......ok.....the orthocenter would be at (1,6)

Solution: A triangle ABC  having vertices at  A(1, 2) , B(1, 6) , and C(5, 6).

AB= 4 units

BC= 4 units

CA = [tex]\sqrt{(5-1)^{2} +(6-2)^{2} }=\sqrt{32}=4\sqrt{2}[/tex]

The triangle ABC is right angled triangle.

And Orthocentre of a triangle is the point where perpendiculars of triangle meet.As two perpendiculars AB and AC meet at Point B , So Point B(1,6) is orthocenter of triangle.


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