The joint equation of the bisectors of the an | vedclass

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(B) The given lines are $x=5$ and $y=3$. These lines are perpendicular to each other. The bisectors of the angles between the lines $x=h$ and $y=k$ ar

Vedclass · Accepted Answer

$x^2-y^2-10x+6y+16=0$

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(B) The given lines are $x=5$ and $y=3$. These lines are perpendicular to each other. The bisectors of the angles between the lines $x=h$ and $y=k$ are given by the lines $x-h = \pm(y-k)$. Substituting $h=5$ and $k=3$,we get $x-5 = \pm(y-3)$. This gives two equations: $x-5 = y-3 \implies x-y-2=0$ and $x-5 = -(y-3) \implies x+y-8=0$. The joint equation is $(x-y-2)(x+y-8) = 0$. Expanding this: $x(x+y-8) - y(x+y-8) - 2(x+y-8) = 0$. $x^2 + xy - 8x - xy - y^2 + 8y - 2x - 2y + 16 = 0$. $x^2 - y^2 - 10x + 6y + 16 = 0$.

The joint equation of the bisectors of the angles between the lines $x=5$ and $y=3$ is

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