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If two equal sides of an isosceles triangle are given by the equations $7 x-y+3=0$ and $x+y-3=0$, then the equation of its third side passing through the point $(2,-5)$ is
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The correct answer is:
$x-3 y=17$
The third side of the isosceles triangle is parallel to the angle bisector of the given lines $7 x-y+3=0$ and $x+y-3=0$.
Now, the equation of angle bisectors are
$\frac{7 x-y+3}{\sqrt{50}}= \pm \frac{x+y-3}{\sqrt{2}}$
$\Rightarrow \quad 7 x-y+3= \pm 5(x+y-3)$
or $2 x-6 y+18=0$ and $12 x+4 y-12=0$
or $\quad x-3 y+9=0$ and $3 x+y-3=0$
Therefore, possible equation of third side passes through point $(2,-5)$ are
$x-3 y=17 \text { or } 3 x+y=1$
Now, the equation of angle bisectors are
$\frac{7 x-y+3}{\sqrt{50}}= \pm \frac{x+y-3}{\sqrt{2}}$
$\Rightarrow \quad 7 x-y+3= \pm 5(x+y-3)$
or $2 x-6 y+18=0$ and $12 x+4 y-12=0$
or $\quad x-3 y+9=0$ and $3 x+y-3=0$
Therefore, possible equation of third side passes through point $(2,-5)$ are
$x-3 y=17 \text { or } 3 x+y=1$
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