The point of intersection of the lines repres | vedclass

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(A) Given the equation of the pair of lines: $f(x, y) = 3x^2 - 11xy + 10y^2 - 7x + 13y + 4 = 0$. To find the point of intersection,we solve the partia

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$(3, 1)$

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(A) Given the equation of the pair of lines: $f(x, y) = 3x^2 - 11xy + 10y^2 - 7x + 13y + 4 = 0$. To find the point of intersection,we solve the partial derivatives $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$. $\frac{\partial f}{\partial x} = 6x - 11y - 7 = 0$ $\frac{\partial f}{\partial y} = -11x + 20y + 13 = 0$ Solving these simultaneous equations using the cross-multiplication method: $\frac{x}{(-11)(13) - (20)(-7)} = \frac{-y}{(6)(13) - (-11)(-7)} = \frac{1}{(6)(20) - (-11)(-11)}$ $\frac{x}{-143 + 140} = \frac{-y}{78 - 77} = \frac{1}{120 - 121}$ $\frac{x}{-3} = \frac{-y}{1} = \frac{1}{-1}$ $x = 3$ and $y = 1$. Thus,the point of intersection is $(3, 1)$.

The point of intersection of the lines represented by the equation $3x^2 - 11xy + 10y^2 - 7x + 13y + 4 = 0$ is

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