The equation of the pair of bisectors of the | vedclass

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(C) Given the pair of straight lines: $3x^2+7xy+2y^2+5x+5y+2=0$. Factorizing the equation,we get $(3x+y+2)(x+2y+1)=0$. The lines are $L_1: 3x+y+2=0$ a

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$7(5x+3)^2-2(5x+3)(5y+1)-7(5y+1)^2=0$

Vedclass · Answer

(C) Given the pair of straight lines: $3x^2+7xy+2y^2+5x+5y+2=0$. Factorizing the equation,we get $(3x+y+2)(x+2y+1)=0$. The lines are $L_1: 3x+y+2=0$ and $L_2: x+2y+1=0$. Solving for the intersection point,we get $x = -3/5$ and $y = -1/5$. The equation of the angle bisectors is given by $\frac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}} = \pm \frac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}$. Substituting the lines and shifting the origin to $(-3/5, -1/5)$,the equation becomes $\frac{3(x+3/5)+(y+1/5)}{\sqrt{3^2+1^2}} = \pm \frac{(x+3/5)+2(y+1/5)}{\sqrt{1^2+2^2}}$. This simplifies to $\frac{5x+3}{\sqrt{10}} = \pm \frac{5y+1}{\sqrt{5}}$. Squaring both sides: $\frac{(5x+3)^2}{10} = \frac{(5y+1)^2}{5} \Rightarrow (5x+3)^2 = 2(5y+1)^2$. However,using the standard formula for angle bisectors of $ax^2+2hxy+by^2=0$,we have $\frac{x^2-y^2}{a-b} = \frac{xy}{h}$. Applying this to the shifted lines,we get $7(5x+3)^2-2(5x+3)(5y+1)-7(5y+1)^2=0$.

The equation of the pair of bisectors of the angles between the pair of straight lines $3x^2+7xy+2y^2+5x+5y+2=0$ is

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