The equal sides of an isosceles triangle are | vedclass

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(A) Let $	riangle ABC$ be the isosceles triangle where $AB=AC$. The angle bisector of $\angle BAC$ is perpendicular to the base $BC$.The equations of

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-$3$

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(A) Let $	riangle ABC$ be the isosceles triangle where $AB=AC$. The angle bisector of $\angle BAC$ is perpendicular to the base $BC$.The equations of the angle bisectors of the lines $7x-y+3=0$ and $x+y-3=0$ are given by:$\frac{7x-y+3}{\sqrt{7^2+(-1)^2}} = \pm \frac{x+y-3}{\sqrt{1^2+1^2}}$$\frac{7x-y+3}{\sqrt{50}} = \pm \frac{x+y-3}{\sqrt{2}}$$\frac{7x-y+3}{5\sqrt{2}} = \pm \frac{x+y-3}{\sqrt{2}}$$7x-y+3 = \pm 5(x+y-3)$Case $1$: $7x-y+3 = 5x+5y-15$ $\Rightarrow 2x-6y+18=0$ $\Rightarrow x-3y+9=0$. The slope of this bisector is $m_1 = \frac{1}{3}$.Case $2$: $7x-y+3 = -5x-5y+15$ $\Rightarrow 12x+4y-12=0$ $\Rightarrow 3x+y-3=0$. The slope of this bisector is $m_2 = -3$.Since the third side $BC$ is perpendicular to the angle bisector,the slope $m$ of $BC$ satisfies $m \cdot m_{bisector} = -1$.For $m_1 = \frac{1}{3}$,$m = -3$.For $m_2 = -3$,$m = \frac{1}{3}$.Since $m$ is given as an integer,we have $m = -3$.

The equal sides of an isosceles triangle are given by the equations $7x-y+3=0$ and $x+y-3=0$. If the slope $m$ of the third side is an integer,then $m=$

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