Question

No one can solve I challenge

Answer 1

$\bold {\huge {Solutions :-}}$

$\bold {\underline {In~ a~ Triangle }}$

The orthocenter lies at the vertex containing the right angle.

It is clear from the vertices given that ABC is a right angled triangle,right angled at A.

$\bold {\underline {Hence,}}$

Its orthocenter should lie on the vertex A (1,2)

$\bold { ANSWER:-~ (1,2) }$

$<marquee><i>Thanks$

Answer 2

Let the origin at  O  as  shown in  figure,

Now,

OA² = (a-0)² + (b-0)² = a² + b²

OB² = (c-0)² + (d-0)² = c² + d²

AB² =  (c-a)² + (d-b)² = c² + d² +  a² + b² - 2ac -2bd

Using  Pythagoras  Theorem,

AB² = OB² + OA²

c² + d² +  a² + b² - 2ac -2bd = c² + d² +  a² + b²

-ac - bd = 0

ac + bd = 0