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(D) The equations of the two equal sides are:$7x-y+3=0 \quad \dots(i)$$x+y-3=0 \quad \dots(ii)$In an isosceles triangle,the third side is perpendicula

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$-3$

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(D) The equations of the two equal sides are:$7x-y+3=0 \quad \dots(i)$$x+y-3=0 \quad \dots(ii)$In an isosceles triangle,the third side is perpendicular to the angle bisector of the angle between the two equal sides. Alternatively,the third side makes equal angles with the two equal sides.Let the slope of the third side be $m$. The angle between the line with slope $m$ and the line $7x-y+3=0$ (slope $m_1=7$) is equal to the angle between the line with slope $m$ and the line $x+y-3=0$ (slope $m_2=-1$).Using the formula $	an 	heta = |\frac{m_1-m_2}{1+m_1m_2}|$:$|\frac{m-7}{1+7m}| = |\frac{m-(-1)}{1+m(-1)}|$$|\frac{m-7}{1+7m}| = |\frac{m+1}{1-m}|$$(m-7)(1-m) = \pm(m+1)(1+7m)$Case $1$: $(m-7)(1-m) = (m+1)(1+7m)$$m - m^2 - 7 + 7m = m + 7m^2 + 1 + 7m$$8m - m^2 - 7 = 7m^2 + 8m + 1$$8m^2 = -8 \Rightarrow m^2 = -1$ (No real solution).Case $2$: $(m-7)(1-m) = -(m+1)(1+7m)$$m - m^2 - 7 + 7m = -(m + 7m^2 + 1 + 7m)$$8m - m^2 - 7 = -7m^2 - 8m - 1$$6m^2 + 16m - 6 = 0$$3m^2 + 8m - 3 = 0$$(3m-1)(m+3) = 0$$m = \frac{1}{3}$ or $m = -3$.Since $m$ is an integer,$m = -3$.

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

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