The point of intersection of the pair of line | vedclass

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(D) The point of intersection of the pair of lines $f(x, y) = x^2+xy+2y^2-3x+2y+4=0$ can be found by solving the partial derivatives of $f(x, y)$ with

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$(2,-1)$

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(D) The point of intersection of the pair of lines $f(x, y) = x^2+xy+2y^2-3x+2y+4=0$ can be found by solving the partial derivatives of $f(x, y)$ with respect to $x$ and $y$ equal to zero.Taking partial derivative with respect to $x$:$\frac{\partial f}{\partial x} = 2x + y - 3 = 0 \quad \dots (i)$Taking partial derivative with respect to $y$:$\frac{\partial f}{\partial y} = x + 4y + 2 = 0 \quad \dots (ii)$Solving the system of linear equations $(i)$ and $(ii)$:From $(i)$,$y = 3 - 2x$.Substituting into $(ii)$: $x + 4(3 - 2x) + 2 = 0$$x + 12 - 8x + 2 = 0$$-7x + 14 = 0 \implies x = 2$Substituting $x = 2$ into $y = 3 - 2x$:$y = 3 - 2(2) = 3 - 4 = -1$Thus,the point of intersection is $(2, -1)$.Hence,option $(D)$ is correct.

The point of intersection of the pair of lines $x^2+xy+2y^2-3x+2y+4=0$ is

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