The joint equation of the pair of lines passi | vedclass

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(A) Let $\phi(x, y) = 2x^{2}-xy-15y^{2}-7x+32y-9=0$ ...$(1)$To find the point of intersection,we take partial derivatives:$\frac{\partial \phi}{\parti

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

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(A) Let $\phi(x, y) = 2x^{2}-xy-15y^{2}-7x+32y-9=0$ ...$(1)$To find the point of intersection,we take partial derivatives:$\frac{\partial \phi}{\partial x} = 4x-y-7=0$ ...$(2)$$\frac{\partial \phi}{\partial y} = -x-30y+32=0$ ...$(3)$Solving equations $(2)$ and $(3)$:From $(2)$,$y = 4x-7$.Substitute into $(3)$: $-x - 30(4x-7) + 32 = 0$ $\Rightarrow -x - 120x + 210 + 32 = 0$ $\Rightarrow -121x + 242 = 0$ $\Rightarrow x = 2$.Then $y = 4(2)-7 = 1$.So,the point of intersection is $(2, 1)$.The lines passing through $(2, 1)$ parallel to the coordinate axes are $x=2$ and $y=1$.The joint equation is $(x-2)(y-1) = 0$.Expanding this,we get $xy - x - 2y + 2 = 0$.

The joint equation of the pair of lines passing through the point of intersection of the lines represented by $2x^{2}-xy-15y^{2}-7x+32y-9=0$ and parallel to the coordinate axes is:

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