The joint equation of the pair of lines passi | vedclass

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(A) The given equation is $x^{2}-y^{2}=0$,which can be written as $(x-y)(x+y)=0$. Thus,the slopes of the given lines are $m_{1}=1$ and $m_{2}=-1$. The

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$x^{2}-y^{2}-4x+6y-5=0$

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(A) The given equation is $x^{2}-y^{2}=0$,which can be written as $(x-y)(x+y)=0$. Thus,the slopes of the given lines are $m_{1}=1$ and $m_{2}=-1$. The equations of the lines passing through $(2,3)$ and parallel to these lines are: $(y-3)=1(x-2) \Rightarrow x-y+1=0$ $(y-3)=-1(x-2) \Rightarrow x+y-5=0$ The joint equation is the product of these two lines: $(x-y+1)(x+y-5)=0$ Expanding this: $x(x+y-5) - y(x+y-5) + 1(x+y-5) = 0$ $x^{2}+xy-5x-xy-y^{2}+5y+x+y-5=0$ $x^{2}-y^{2}-4x+6y-5=0$

The joint equation of the pair of lines passing through $(2,3)$ and parallel to the lines represented by $x^{2}-y^{2}=0$ is:

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