Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
2204 Upvotes
Verified Answer
The correct answer is:
$\frac{\sqrt{5}}{6}$
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
$=\frac{\text { Hypotenuse }}{2}=\frac{\sqrt{5}}{6} .$
$\begin{aligned}
& 2 x^2-3 x y+y^2=0 \\
\Rightarrow & (x-y)(2 x-y)=0 \\
\Rightarrow & x-y=0 \text { and } 2 x-y=0
\end{aligned}$
$\begin{aligned}
& O A=\sqrt{\frac{1}{4}+\frac{1}{4}}=\frac{1}{\sqrt{2}} \\
& O B=\sqrt{\frac{1}{9}+\frac{4}{9}}=\frac{\sqrt{5}}{3} \\
& A B=\sqrt{\left(\frac{1}{3}-\frac{1}{2}\right)^2+\left(\frac{2}{3}-\frac{1}{2}\right)^2}=\frac{\sqrt{2}}{6} \\
& \because \quad O A^2+A B^2=\frac{1}{2}+\frac{2}{36}=\frac{5}{9}=\left(\frac{\sqrt{5}}{3}\right)^2=O B^2
\end{aligned}$
$\therefore \quad \triangle O A B$ is right angle triangle, where $O B$ is hypotenuse.
$\therefore$ Distance between orthocentre and circumcentre
$=\frac{\text { Hypotenuse }}{2}=\frac{\sqrt{5}}{6} .$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.