The joint equation of the bisectors of the an | vedclass

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(B) The lines $x=5$ (a vertical line) and $y=3$ (a horizontal line) intersect at the point $(5, 3)$.Since these lines are perpendicular,the angle bise

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

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(B) The lines $x=5$ (a vertical line) and $y=3$ (a horizontal line) intersect at the point $(5, 3)$.Since these lines are perpendicular,the angle bisectors are lines passing through $(5, 3)$ that make angles of $45^{\circ}$ and $135^{\circ}$ with the $x$-axis.The slopes of these bisectors are $m = 	an 45^{\circ} = 1$ and $m = 	an 135^{\circ} = -1$.The equations of the bisectors are:$y - 3 = 1(x - 5) \Rightarrow y - x + 2 = 0$$y - 3 = -1(x - 5) \Rightarrow y + x - 8 = 0$The joint equation is $(y - x + 2)(y + x - 8) = 0$.Expanding this: $y^2 + xy - 8y - xy - x^2 + 8x + 2y + 2x - 16 = 0$.Simplifying,we get $-x^2 + y^2 + 10x - 6y - 16 = 0$,or $x^2 - y^2 - 10x + 6y + 16 = 0$.

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

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