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(A) The third side of an isosceles triangle is perpendicular to the angle bisector of the two equal sides. However,in this specific geometry,the third

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$x-3y=17$

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(A) The third side of an isosceles triangle is perpendicular to the angle bisector of the two equal sides. However,in this specific geometry,the third side is parallel to the angle bisector of the given lines $7x-y+3=0$ and $x+y-3=0$ if the vertex is the intersection point. First,find the intersection point of the two lines: $7x-y+3=0$ and $x+y-3=0$. Adding them gives $8x=0$,so $x=0$,which implies $y=3$. The vertex is $(0,3)$. The angle bisectors are given by $\frac{7x-y+3}{\sqrt{7^2+(-1)^2}} = \pm \frac{x+y-3}{\sqrt{1^2+1^2}}$. $\frac{7x-y+3}{\sqrt{50}} = \pm \frac{x+y-3}{\sqrt{2}} \Rightarrow 7x-y+3 = \pm 5(x+y-3)$. Case $1$: $7x-y+3 = 5x+5y-15$ $\Rightarrow 2x-6y+18=0$ $\Rightarrow x-3y+9=0$. Case $2$: $7x-y+3 = -5x-5y+15$ $\Rightarrow 12x+4y-12=0$ $\Rightarrow 3x+y-3=0$. The third side is perpendicular to these bisectors. For $x-3y+9=0$,the perpendicular line is $3x+y+k=0$. Passing through $(2,-5)$: $3(2)+(-5)+k=0$ $\Rightarrow 6-5+k=0$ $\Rightarrow k=-1$. So,$3x+y-1=0$. For $3x+y-3=0$,the perpendicular line is $x-3y+k=0$. Passing through $(2,-5)$: $2-3(-5)+k=0$ $\Rightarrow 2+15+k=0$ $\Rightarrow k=-17$. So,$x-3y=17$. Comparing with options,$x-3y=17$ is present.

If two equal sides of an isosceles triangle are given by the equations $7x-y+3=0$ and $x+y-3=0$,then the equation of its third side passing through the point $(2,-5)$ is

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