Find the point of intersection of the lines r | vedclass

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(D) Let $\varphi(x, y) = 2x^2 - 7xy - 4y^2 - x + 22y - 10 = 0$. To find the point of intersection,we take the partial derivatives with respect to $x$

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$(2, 1)$

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(D) Let $\varphi(x, y) = 2x^2 - 7xy - 4y^2 - x + 22y - 10 = 0$. To find the point of intersection,we take the partial derivatives with respect to $x$ and $y$ and set them to zero. $\frac{\partial \varphi}{\partial x} = 4x - 7y - 1 = 0$  (Equation $1$) $\frac{\partial \varphi}{\partial y} = -7x - 8y + 22 = 0$  (Equation $2$) From Equation $1$,$4x = 7y + 1 \implies x = \frac{7y + 1}{4}$. Substituting this into Equation $2$: $-7(\frac{7y + 1}{4}) - 8y + 22 = 0$ $-49y - 7 - 32y + 88 = 0$ $-81y + 81 = 0 \implies y = 1$. Substituting $y = 1$ into $4x - 7(1) - 1 = 0$: $4x - 8 = 0 \implies x = 2$. Thus,the point of intersection is $(2, 1)$.

Find the point of intersection of the lines represented by $2x^2 - 7xy - 4y^2 - x + 22y - 10 = 0$.

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