Question

The distance between the orthocentre and circumcentre of a triangle with vertices \[(1,0),\left( \dfrac{-1}{2},\dfrac{\sqrt{3}}{2} \right),\left( \dfrac{-1}{2},\dfrac{-\sqrt{3}}{2} \right)\] is
A) \[\dfrac{1}{2}\]
B) \[\dfrac{\sqrt{3}}{2}\]
C) \[\dfrac{1}{3}\]
D) \[0\]

Answer 1

Hint: If the lengths of all three sides of a triangle are equal, then the triangle is equilateral. For such triangles, the centroid, orthocentre, and the circumcentre coincide with each other.

Complete step-by-step answer:
Consider the given vertices to be \[P(1,0),Q\left( \dfrac{-1}{2},\dfrac{\sqrt{3}}{2} \right)\] and \[R\left( \dfrac{-1}{2},\dfrac{-\sqrt{3}}{2} \right)\] .

First, we will find the length of the sides of the triangle.
We know, the distance between two points \[\left( {{a}_{1}},{{b}_{1}} \right)\] and \[\left( {{a}_{2}},{{b}_{2}} \right)\] is given as
\[\sqrt{{{\left( {{a}_{1}}-{{a}_{2}} \right)}^{2}}+{{\left( {{b}_{1}}-{{b}_{2}} \right)}^{2}}}\]
So, the length of the side \[PQ\] is given as \[PQ=\sqrt{{{\left( 1-\left( \dfrac{-1}{2} \right) \right)}^{2}}+{{\left( 0-\dfrac{\sqrt{3}}{2} \right)}^{2}}}\] .
\[\begin{align}
  & =\sqrt{{{\left( \dfrac{3}{2} \right)}^{2}}+\dfrac{3}{4}} \\
 & =\sqrt{\dfrac{9}{4}+\dfrac{3}{4}}=\sqrt{\dfrac{12}{4}} \\
 & =\sqrt{3} \\
\end{align}\]
And, the length of the side \[QR\] is given as \[QR=\sqrt{{{\left( \dfrac{-1}{2}+\dfrac{1}{2} \right)}^{2}}+{{\left( \dfrac{\sqrt{3}}{2}-\left( -\dfrac{\sqrt{3}}{2} \right) \right)}^{2}}}\] .
\[\begin{align}
  & =\sqrt{0+3} \\
 & =\sqrt{3} \\
\end{align}\]
And, the length of the side \[PR\] is given as \[PR=\sqrt{{{\left( 1-\left( \dfrac{-1}{2} \right) \right)}^{2}}+{{\left( 0-\left( -\dfrac{\sqrt{3}}{2} \right) \right)}^{2}}}\] .
\[\begin{align}
  & =\sqrt{{{\left( \dfrac{3}{2} \right)}^{2}}+\dfrac{3}{4}} \\
 & =\sqrt{\dfrac{9}{4}+\dfrac{3}{4}}=\sqrt{\dfrac{12}{4}}=\sqrt{3} \\
\end{align}\]
Clearly, we can see \[PQ=QR=PR\] , i.e. the length of the sides of the triangle are equal. Hence, the triangle is equilateral.
Now, we know, in an equilateral triangle, the centroid, the circumcentre, and the orthocentre coincide.
Hence, the distance between the circumcentre and the orthocentre is equal to \[0\] units.
Note: We know coordinates \[\left( x,y \right)\] can be expressed in polar form as \[x=r\cos \theta \And y=r\sin \theta \] where \[r=\sqrt{{{x}^{2}}+{{y}^{2}}}\] and \[\theta \] is the angle made by the line joining \[\left( x,y \right)\] and the origin \[O\left( 0,0 \right)\] with the \[x\] -axis.
So, \[P\text{ }\left( 1,0 \right)\] can be written as \[P\left( \cos 0,\sin 0 \right)\] .
\[Q\left( \dfrac{-1}{2},\dfrac{\sqrt{3}}{2} \right)\] can be written as \[Q\left( \cos \dfrac{2\pi }{3},\sin \dfrac{2\pi }{3} \right)\] .
\[R\left( \dfrac{-1}{2},\dfrac{-\sqrt{3}}{2} \right)\] can be written as \[R\left( \cos \dfrac{4\pi }{3},\sin \dfrac{4\pi }{3} \right)\] .
So, clearly \[OP\] is inclined with the \[x\] -axis at \[{{0}^{\circ }}\] .
\[OQ\] is inclined with \[x\] -axis at \[{{120}^{\circ }}\] and \[OR\] is inclined with \[x\] -axis at \[{{240}^{\circ }}\] (or \[-{{120}^{\circ }}\]).

So, \[PR\] is inclined with \[PQ\] at \[{{60}^{\circ }}\], and with \[QR\]at \[{{60}^{\circ }}\].
So, \[\Delta PQR\] is equilateral.
Now, we know, in an equilateral triangle, the centroid, the circumcentre, and the orthocentre coincide.
Hence, the circumcentre and the orthocentre will coincide and the distance between the circumcentre and the orthocentre is equal to \[0\] units.