If two pairs of straight lines with combined | vedclass

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

Question

(C) The given equations are $xy+4x-3y-12=0$ and $xy-3x+4y-12=0$.For the first equation: $x(y+4)-3(y+4)=0 \Rightarrow (x-3)(y+4)=0$. This represents th

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The given equations are $xy+4x-3y-12=0$ and $xy-3x+4y-12=0$.For the first equation: $x(y+4)-3(y+4)=0 \Rightarrow (x-3)(y+4)=0$. This represents the lines $x=3$ and $y=-4$.For the second equation: $x(y-3)+4(y-3)=0 \Rightarrow (x+4)(y-3)=0$. This represents the lines $x=-4$ and $y=3$.The four lines forming the square are $x=3, x=-4, y=3, y=-4$.The vertices of the square are $A(-4,-4), B(3,-4), C(3,3),$ and $D(-4,3)$.The diagonal $AC$ passes through $A(-4,-4)$ and $C(3,3)$. Its equation is $y-(-4) = \frac{3-(-4)}{3-(-4)}(x-(-4))$ $\Rightarrow y+4 = 1(x+4)$ $\Rightarrow x-y=0$.The diagonal $BD$ passes through $B(3,-4)$ and $D(-4,3)$. Its equation is $y-(-4) = \frac{3-(-4)}{-4-3}(x-3)$ $\Rightarrow y+4 = \frac{7}{-7}(x-3)$ $\Rightarrow y+4 = -x+3$ $\Rightarrow x+y+1=0$.The combined equation of the diagonals is $(x-y)(x+y+1)=0$.Expanding this,we get $x^2+xy+x-xy-y^2-y=0$,which simplifies to $x^2-y^2+x-y=0$.

If two pairs of straight lines with combined equations $xy+4x-3y-12=0$ and $xy-3x+4y-12=0$ form a square,then the combined equation of its diagonals is

Similar Questions

The equation of a line which makes an angle of $45^{\circ}$ with each of the pair of lines $xy-x-y+1=0$ is

If the equation of the pair of straight lines passing through the point $(1,1)$ and perpendicular to the pair of lines $3x^2+11xy-4y^2=0$ is $ax^2+2hxy+by^2+2gx+2fy+12=0$,then $2(a-h+b-g+f-12)=$

The equations to a pair of opposite sides of a parallelogram are $x^2 - 5x + 6 = 0$ and $y^2 - 6y + 5 = 0$. The equations to its diagonals are

The orthocenter of the triangle formed by the lines $x-2y=10$ and $6x^2+xy-y^2=0$ is

If $2x^2 - 3xy + y^2 = 0$ represents two sides of a triangle and $x + y - 1 = 0$ is its third side,then the distance between the orthocenter and the circumcentre of that triangle is

Vedclass Test Series

Exam Paper Generator

Online Exam Module