The quadrilateral formed by the pairs of line | vedclass

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(D) The first pair of lines is given by $xy+x+y+1=0$. Factoring this,we get $x(y+1)+1(y+1)=0$,which implies $(x+1)(y+1)=0$. Thus,the lines are $x=-1$

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Square

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(D) The first pair of lines is given by $xy+x+y+1=0$. Factoring this,we get $x(y+1)+1(y+1)=0$,which implies $(x+1)(y+1)=0$. Thus,the lines are $x=-1$ and $y=-1$.The second pair of lines is given by $xy+3x+3y+9=0$. Factoring this,we get $x(y+3)+3(y+3)=0$,which implies $(x+3)(y+3)=0$. Thus,the lines are $x=-3$ and $y=-3$.The four lines forming the quadrilateral are $x=-1, x=-3, y=-1,$ and $y=-3$.The distance between the vertical lines $x=-1$ and $x=-3$ is $|-1 - (-3)| = 2$.The distance between the horizontal lines $y=-1$ and $y=-3$ is $|-1 - (-3)| = 2$.Since the opposite sides are parallel and the distance between the vertical lines is equal to the distance between the horizontal lines,the quadrilateral is a square. Therefore,the correct option is $D$.

The quadrilateral formed by the pairs of lines $xy+x+y+1=0$ and $xy+3x+3y+9=0$ is

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