Question
Let $A \equiv (3,2)$ and $B \equiv (5,1)$. $ABP$ is an equilateral triangle is constructed one the side of $AB$ remote from the origin then the orthocentre of triangle $ABP$ is:
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Solution
Sol. [Key Concept: Equilateral triangle on $AB$ (far side): orthocentre $=$ centroid $=$ average of the three vertices.]
$${PD = ABsin60^{\circ} = \sqrt{5}\frac{\sqrt{3}}{2} }{HD = \frac{1}{3}\frac{\sqrt{5}\sqrt{3}}{2} = \frac{\sqrt{5}}{2\sqrt{3}} }$$Slope of $HD = 2,D \equiv \left( 4,\frac{3}{2} \right)$
Using parametric, form
$${H \equiv \left( 4 + \frac{\sqrt{5}}{2\sqrt{3}} \times \frac{1}{\sqrt{5}},\frac{3}{2} + \frac{\sqrt{5}}{2\sqrt{3}} \times \frac{2}{\sqrt{5}} \right) }{\Rightarrow H \equiv \left( 4 + \frac{\sqrt{3}}{6},\frac{3}{2} + \frac{\sqrt{3}}{3} \right)}$$
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