In a $\Delta ABC$, orthocentre is $H(2354,981)$. If $A(2,1)$, $B(-10,6)$, then the distance between the orthocentres of $\Delta HBC$,$\Delta HAC$ is?
I don't even know how to proceed with the question. I tried solving it with the formula but it's getting complex since angles are also unknown. Can anyone please provide me with a solution?
Since $BH\perp AC$ and $BA\perp CH$, $B$ is the orthocentre of $\triangle HAC$.
Since $AH\perp BC$ and $AB\perp CH$, $A$ is the orthocentre of $\triangle HBC$.
The required distance is $AB=\sqrt{12^2+5^2}=13$.