Let A equiv (3, 2) and B equiv (5, 1). ABP is | vedclass

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Question

$m_{AB} = -1/2$

$M_{PM} = 2$

$h = \frac{{a\sqrt 3 }}{2}$ now $a =\sqrt{5} = AB$ ; $h = \frac{{\sqrt {15} }}{2}$

so point $P \frac{{

Vedclass · Accepted Answer

$\left( {4 + \frac{1}{6}\sqrt 3 ,\,\,\frac{3}{2} + \frac{1}{3}\sqrt 3 } \right)$

Vedclass · Answer

$m_{AB} = -1/2$

$M_{PM} = 2$

$h = \frac{{a\sqrt 3 }}{2}$ now $a =\sqrt{5} = AB$ ; $h = \frac{{\sqrt {15} }}{2}$

so point $P \frac{{x - y}}{{\cos 	heta }} = \frac{{y - 3/2}}{{\sin 	heta }} = h$

$P : x = 4 + \frac{{\sqrt 3 }}{2}$ , $y = \frac{3}{2} + \sqrt{3} $

so orthocentre/centroid is $\frac{{\sum {x_1}}}{3},\,\,\,\frac{{\sum {y_1}}}{3}$

$= 4 + \frac{{\sqrt 3 }}{6} , \frac{3}{2} + \frac{{\sqrt 3 }}{3}$

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

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