If $2 x^2-3 x y+y^2=0$ represents two sides of a triangle and $x+y-1=0$ is its third side, then the distance between the orthocenter and the circumcentre of that triangle is

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Aicharya · Accepted Answer

Given pair of equations $\begin{aligned} & 2 x^2-3 x y+y^2=0 \\ \Rightarrow & (x-y)(2 x-y)=0 \\ \Rightarrow & x-y=0 \text { and } 2 x-y=0 \end{aligned}$

$\begin{aligned} & O A=\sqrt{\frac{1}{4}+\frac{1}{4}}=\frac{1}{\sqrt{2}} \\ & O B=\sqrt{\frac{1}{9}+\frac{4}{9}}=\frac{\sqrt{5}}{3} \\ & A B=\sqrt{\left(\frac{1}{3}-\frac{1}{2}\right)^2+\left(\frac{2}{3}-\frac{1}{2}\right)^2}=\frac{\sqrt{2}}{6} \\ & \because \quad O A^2+A B^2=\frac{1}{2}+\frac{2}{36}=\frac{5}{9}=\left(\frac{\sqrt{5}}{3}\right)^2=O B^2 \end{aligned}$ $\therefore \quad \triangle O A B$ is right angle triangle, where $O B$ is hypotenuse. $\therefore$ Distance between orthocentre and circumcentre $=\frac{\text { Hypotenuse }}{2}=\frac{\sqrt{5}}{6} .$

If $2 x^2-3 x y+y^2=0$ represents two sides of a triangle and $x+y-1=0$ is its third side, then the distance…

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