The quadrilateral formed by the pairs of line | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $x=-1$

Vedclass · Accepted Answer

Square

Vedclass · Answer

(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

Similar Questions

If the slopes of the lines represented by $Kx^2 - 4xy + 5y^2 = 0$ differ by $2$,then $K = $

The value of $\lambda$ such that $\lambda x^2-10 x y+12 y^2+5 x-16 y-3=0$ represents a pair of straight lines is:

The slope of one of the pair of lines $2x^2 + hxy + 6y^2 = 0$ is thrice the slope of the other line. Then,$h = $

The joint equation of lines passing through the origin and having slopes $(1+\sqrt{2})$ and $\frac{-1}{1+\sqrt{2}}$ is ...

The equation of the pair of lines perpendicular to the lines represented by $x^2-5xy+4y^2=0$ and passing through the point $(2,1)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module