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A triangle $ABC$ with vertices $A(-1,0), B(-2,3/4),C(-3,-7/6)$ has orthocentre H. Find the orthocentre of triangle $BCH$

I know how to solve this question but my procedure is too long.

My Approach:
Find the slope of 2 sides and hence the slope of their perpendicular. Find the equation of the lines from the slope and the fact that orthocentre passes through the opposite vertex. Then find the intersection of the 2 sides. Thus we obtain H. Repeat procedure for the second triangle to obtain the orthocentre.

Since I'm preparing for competitive examinations I need a faster method to solve this question. It would be great if someone could help.

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    $\begingroup$ isn't $A$ the orthocenter of $\triangle BCH$? $\endgroup$ – Quang Hoang Apr 4 '17 at 4:27
  • $\begingroup$ @QuangHoang Oh right, how dumb of me! You can post this as the answer, Thank you. $\endgroup$ – oshhh Apr 4 '17 at 4:35
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ABC is any triangle

H is the orthocenter of the ABC triangle

J is the orthocenter of the BCH triangle

BC=common side

altitudes:

r is a line passing through B, perpendicular to AC.$( H \; ∈ \; r)$

s is a line passing through C, perpendicular to AB.$( H \; ∈ \; s)$

${ H }=r ∩ s$

BCH triangle:

t is a line passing through B, perpendicular to CH.$( J \; ∈ \; t)(t≡\overleftrightarrow{AB})$

u is a line passing through C, perpendicular to BH.$( J \; ∈ \; u)(u≡\overleftrightarrow{AC})$

$t ∩ u=J=\overleftrightarrow{AB}∩\overleftrightarrow{AC}=A$ ortocentreHJ

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