(B) The equation of the pair of angle bisectors of the homogeneous equation $ax^2+2hxy+by^2=0$ is given by $\frac{x^2-y^2}{a-b} = \frac{xy}{h}$.Given

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(B) The equation of the pair of angle bisectors of the homogeneous equation $ax^2+2hxy+by^2=0$ is given by $\frac{x^2-y^2}{a-b} = \frac{xy}{h}$.Given the equation $x^2+3xy+2y^2=0$,we have $a=1$,$2h=3$ (so $h=\frac{3}{2}$),and $b=2$.Substituting these values into the formula:$\frac{x^2-y^2}{1-2} = \frac{xy}{3/2}$$\frac{x^2-y^2}{-1} = \frac{2xy}{3}$$3(x^2-y^2) = -2xy$$3x^2-3y^2 = -2xy$$3x^2+2xy-3y^2=0$.

Class 11 Mathematics Pair of straight lines Bisectors of the angle between the lines, Point of intersection of the lines

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The joint equation of the pair of lines which bisects the angles between the lines $x^2+3xy+2y^2=0$ is

MHT CET 2022 All MHT CET

A
$3x^2-2xy-3y^2=0$
B
$3x^2+2xy-3y^2=0$
C
$2x^2-3xy-2y^2=0$
D
$2x^2+3xy-2y^2=0$

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Pair of straight lines

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Bisectors of the angle between the lines, Point of intersection of the lines

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The joint equation of the pair of lines which bisects the angles between the lines $x^2+3xy+2y^2=0$ is

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