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(D) Step $1$: Find the point of intersection of the lines represented by $x^2 - 6xy + 5y^2 + 10x - 14y + 9 = 0$. Let $f(x, y) = x^2 - 6xy + 5y^2 + 10x

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$xy - 2x - y + 2 = 0$

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(D) Step $1$: Find the point of intersection of the lines represented by $x^2 - 6xy + 5y^2 + 10x - 14y + 9 = 0$. Let $f(x, y) = x^2 - 6xy + 5y^2 + 10x - 14y + 9 = 0$.Step $2$: Partial differentiate with respect to $x$: $\frac{\partial f}{\partial x} = 2x - 6y + 10 = 0 \implies x - 3y + 5 = 0$ $(i)$.Step $3$: Partial differentiate with respect to $y$: $\frac{\partial f}{\partial y} = -6x + 10y - 14 = 0 \implies -3x + 5y - 7 = 0$ (ii).Step $4$: Solve equations $(i)$ and (ii). From $(i)$, $x = 3y - 5$. Substitute into (ii): $-3(3y - 5) + 5y - 7 = 0 \implies -9y + 15 + 5y - 7 = 0 \implies -4y + 8 = 0 \implies y = 2$.Step $5$: Substitute $y = 2$ into $(i)$: $x = 3(2) - 5 = 1$. The point of intersection is $(1, 2)$.Step $6$: The lines parallel to the coordinate axes passing through $(1, 2)$ are $x = 1$ and $y = 2$.Step $7$: The combined equation is $(x - 1)(y - 2) = 0 \implies xy - 2x - y + 2 = 0$.

The combined equation of lines parallel to the coordinate axes and passing through the point of intersection of lines represented by $x^2 - 6xy + 5y^2 + 10x - 14y + 9 = 0$ is

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