(B) The given equation is $3 x^2-5 x y+4 y^2=0$. Comparing this with the general form $a x^2+2 h x y+b y^2=0$,we get $a=3$,$2 h=-5$,and $b=4$. The equ

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(B) The given equation is $3 x^2-5 x y+4 y^2=0$. Comparing this with the general form $a x^2+2 h x y+b y^2=0$,we get $a=3$,$2 h=-5$,and $b=4$. The equation of the bisectors of the angle between the pair of lines is given by $\frac{x^2-y^2}{a-b} = \frac{x y}{h}$. Substituting the values,we get $\frac{x^2-y^2}{3-4} = \frac{x y}{-5/2}$. This simplifies to $\frac{x^2-y^2}{-1} = \frac{2 x y}{-5}$. Multiplying both sides by $-5$,we get $5(x^2-y^2) = 2 x y$.

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 equation of the bisectors of the angle between the lines represented by $3 x^2-5 x y+4 y^2=0$ is

AP EAMCET 2021 All AP EAMCET

A
$9 x^2+6 y^2-2 x=0$
B
$5(x^2-y^2)=2 x y$
C
$3 x^2+2 x y-y^2=0$
D
$5 x^2+x y+4 y^2=0$

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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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