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Let $A \equiv (3, 2)$ and $B \equiv (5, 1)$. $ABP$ is an equilateral triangle is constructed on the side of $AB$ remote from the origin then the orthocentre of triangle $ABP$ is
$\left( {4 - \frac{1}{2}\sqrt 3 ,\,\,\frac{3}{2} - \sqrt 3 } \right)$
$\left( {4 + \frac{1}{2}\sqrt 3 ,\,\,\frac{3}{2} + \sqrt 3 } \right)$
$\left( {4 - \frac{1}{6}\sqrt 3 ,\,\,\frac{3}{2} - \frac{1}{3}\sqrt 3 } \right)$
$\left( {4 + \frac{1}{6}\sqrt 3 ,\,\,\frac{3}{2} + \frac{1}{3}\sqrt 3 } \right)$
Solution
$m_{AB} = -1/2$
$M_{PM} = 2$
$h = \frac{{a\sqrt 3 }}{2}$ now $a =\sqrt{5} = AB$ ; $h = \frac{{\sqrt {15} }}{2}$
so point $P \frac{{x – y}}{{\cos \theta }} = \frac{{y – 3/2}}{{\sin \theta }} = h$
$P : x = 4 + \frac{{\sqrt 3 }}{2}$ , $y = \frac{3}{2} + \sqrt{3} $
so orthocentre/centroid is $\frac{{\sum {x_1}}}{3},\,\,\,\frac{{\sum {y_1}}}{3}$
$= 4 + \frac{{\sqrt 3 }}{6} , \frac{3}{2} + \frac{{\sqrt 3 }}{3}$
