The locus of the points equidistant from the | vedclass

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(D) The locus of points equidistant from two lines is the pair of angle bisectors of the lines represented by the equation $ax^2 + 2hxy + by^2 = 0$.Co

Vedclass · Accepted Answer

$x^2 - y^2 + 2xy \csc^2 	heta = 0$

Vedclass · Answer

(D) The locus of points equidistant from two lines is the pair of angle bisectors of the lines represented by the equation $ax^2 + 2hxy + by^2 = 0$.Comparing the given equation $x^2 \cos^2 	heta - xy \sin^2 	heta - y^2 \sin^2 	heta = 0$ with $ax^2 + 2hxy + by^2 = 0$,we have $a = \cos^2 	heta$,$2h = -\sin^2 	heta$,and $b = -\sin^2 	heta$.The equation of the angle bisectors is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$.Substituting the values,we get $\frac{x^2 - y^2}{\cos^2 	heta - (-\sin^2 	heta)} = \frac{xy}{-\frac{1}{2} \sin^2 	heta}$.$\frac{x^2 - y^2}{\cos^2 	heta + \sin^2 	heta} = \frac{2xy}{-\sin^2 	heta}$.Since $\cos^2 	heta + \sin^2 	heta = 1$,we have $x^2 - y^2 = -\frac{2xy}{\sin^2 	heta}$.$x^2 - y^2 = -2xy \csc^2 	heta$.Therefore,$x^2 - y^2 + 2xy \csc^2 	heta = 0$.

The locus of the points equidistant from the lines represented by $x^2 \cos^2 \theta - xy \sin^2 \theta - y^2 \sin^2 \theta = 0$ is

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