The combined equation of the bisectors of the | vedclass

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(A) Given equation of the curve: $x^2+xy+y^2+x+3y+1=0$ ...$(i)$Given equation of the line: $x+y+2=0 \Rightarrow \frac{x+y}{-2}=1$ ...(ii)Homogenizing

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$x^2+4xy-y^2=0$

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(A) Given equation of the curve: $x^2+xy+y^2+x+3y+1=0$ ...$(i)$Given equation of the line: $x+y+2=0 \Rightarrow \frac{x+y}{-2}=1$ ...(ii)Homogenizing equation $(i)$ using (ii):$x^2+xy+y^2+x(1)+3y(1)+1(1)^2=0$Substituting $1 = \frac{x+y}{-2}$:$x^2+xy+y^2+x(\frac{x+y}{-2})+3y(\frac{x+y}{-2})+(\frac{x+y}{-2})^2=0$Multiplying by $4$ to clear the denominator:$4x^2+4xy+4y^2-2x(x+y)-6y(x+y)+(x+y)^2=0$$4x^2+4xy+4y^2-2x^2-2xy-6xy-6y^2+x^2+2xy+y^2=0$$3x^2-2xy-y^2=0$Comparing with $ax^2+2hxy+by^2=0$,we get $a=3, 2h=-2, b=-1$.The equation of the angle bisectors is given by $\frac{x^2-y^2}{a-b} = \frac{xy}{h}$.$\frac{x^2-y^2}{3-(-1)} = \frac{xy}{-1}$$\frac{x^2-y^2}{4} = -xy$$x^2-y^2 = -4xy$$x^2+4xy-y^2=0$.

The combined equation of the bisectors of the angles between the lines joining the origin to the points of intersection of the curve $x^2+y^2+xy+x+3y+1=0$ and the line $x+y+2=0$ is

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