Question

The equation of pair of lines passing through $(1,-1)$ and parallel to the lines $2x^2+5xy+3y^2=0$ is

• A. $2(x-1{)}^2+5(x-1\left)\right(y+1)+3(y+1{)}^2=0$
• B. $(x-1{)}^2+5(x-1\left)\right(y+1)+2(y+1{)}^2=0$
• C. $2x^2+5xy+3y^2+x+y=0$
• D. $3x^2-5xy+2y^2-11x+9y+10=0$

Answer 1

Answer: (A), (C)

The correct options are
A $2(x-1{)}^2+5(x-1\left)\right(y+1)+3(y+1{)}^2=0$
C $2x^2+5xy+3y^2+x+y=0$

$2x^2+5xy+3y^2=0$

$\frac{x}{y}=t,2t^2+5t+3=0$

$t=\frac{-5\pm \sqrt{25-24}}{4}$

$t=\frac{-5\pm 1}{4}$

$t=\frac{-3}{2},-1$

$y=\frac{-2}{3}x$ and $y=-x$

Let the equations of the lines be $3y+2x+k=0$ and $y+x+m=0$

Since, both the lines pass through $(1,-1)$, substituting in the equations we get $k=1$ and $m=0$

So, the equation of the pair of lines is $(3y+2x+1)(y+x)=0$

$\Rightarrow 2x^2+5xy+3y^2+x+y=0$

The equation of the pair of lines can also be obtained by transformation of axes concept.

The coordinates of any point $(X,Y)$ on the pair of lines passing through $(1,-1)$ can be obtained from the corresponding points $(x,y)$ on the pair of lines through $(0,0)$ using

$X=x+1\Rightarrow x=X-1$

$Y=y-1\Rightarrow y=Y+1$

$\Rightarrow 2(X-1{)}^2+5(X-1\left)\right(Y+1)+3(Y+1)=0$

Hence, options A and C are correct.