If 2x^2 - 3xy + y^2 = 0 represents two sides | vedclass

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(A) The given pair of lines is $2x^2 - 3xy + y^2 = 0$. Factorizing,we get $(x - y)(2x - y) = 0$,which represents the lines $L_1: x - y = 0$ and $L_2:

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$\frac{\sqrt{5}}{6}$

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(A) The given pair of lines is $2x^2 - 3xy + y^2 = 0$. Factorizing,we get $(x - y)(2x - y) = 0$,which represents the lines $L_1: x - y = 0$ and $L_2: 2x - y = 0$. The third side is $L_3: x + y = 1$. Finding the vertices of the triangle: Intersection of $L_1$ and $L_2$ is the origin $O(0, 0)$. Intersection of $L_1$ $(x = y)$ and $L_3$ $(x + y = 1)$: $2x = 1 \Rightarrow x = 1/2, y = 1/2$. So,$A(1/2, 1/2)$. Intersection of $L_2$ $(y = 2x)$ and $L_3$ $(x + y = 1)$: $x + 2x = 1$ $\Rightarrow 3x = 1$ $\Rightarrow x = 1/3, y = 2/3$. So,$B(1/3, 2/3)$. Calculating side lengths: $OA^2 = (1/2)^2 + (1/2)^2 = 1/4 + 1/4 = 1/2$. $OB^2 = (1/3)^2 + (2/3)^2 = 1/9 + 4/9 = 5/9$. $AB^2 = (1/3 - 1/2)^2 + (2/3 - 1/2)^2 = (-1/6)^2 + (1/6)^2 = 1/36 + 1/36 = 2/36 = 1/18$. Since $OA^2 + AB^2 = 1/2 + 1/18 = 9/18 + 1/18 = 10/18 = 5/9 = OB^2$,the triangle is a right-angled triangle at $A$. In a right-angled triangle,the distance between the orthocenter (vertex $A$) and the circumcentre (midpoint of the hypotenuse $OB$) is half the length of the hypotenuse. Distance $= \frac{OB}{2} = \frac{\sqrt{5/9}}{2} = \frac{\sqrt{5}}{3 	imes 2} = \frac{\sqrt{5}}{6}$.

If $2x^2 - 3xy + y^2 = 0$ 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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