let ABC be a triangle such that the coordinates of vertex A are (-3,1).equation of the median of B is 2x+y-3=0 and equation of the angular bisector of C is 7x-4y-1=0. Find the slope of the line BC

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$$\begin{gathered} \mathrm{Let} \mathrm{co}-\mathrm{ordinates} \mathrm{of} \mathrm{C} \mathrm{be} \left(\mathrm{r},\mathrm{s}\right). \\ \mathrm{Co}-\mathrm{ordinates} \mathrm{of} \mathrm{A} \mathrm{are} \mathrm{A}\left(-3,1\right) \\ \mathrm{Equation} \mathrm{of} \mathrm{median} \mathrm{of} \mathrm{B} \mathrm{is} 2\mathrm{x}+\mathrm{y}-3=0 \mathrm{and} \mathrm{equation} \mathrm{of} \mathrm{angular} \mathrm{bisector} \mathrm{of} \\ \mathrm{C} \mathrm{is} 7\mathrm{x}-4\mathrm{y}-1=0. \\ \mathrm{Point} \mathrm{C} \mathrm{will} \mathrm{lie} \mathrm{on} \mathrm{the} \mathrm{angular} \mathrm{bisector} \mathrm{of} \mathrm{C}. \\ \Rightarrow 7\mathrm{r}-4\mathrm{s}-1=0 \\ \Rightarrow 7\mathrm{r}-4\mathrm{s}=1 ;\mathrm{equation}\left(\mathrm{i}\right) \\ \mathrm{Mid} \mathrm{point} \mathrm{of} \mathrm{AC} \mathrm{will} \mathrm{be} \left(\frac{\mathrm{r}-3}{2},\frac{\mathrm{s}+1}{2}\right). \mathrm{Now} \mathrm{this} \mathrm{mid} \mathrm{point} \mathrm{will} \mathrm{lie} \mathrm{on} \mathrm{median} \mathrm{through} \mathrm{B}. \\ \Rightarrow 2\left(\frac{\mathrm{r}-3}{2}\right)+\left(\frac{\mathrm{s}+1}{2}\right)-3=0 \\ \Rightarrow 2\mathrm{r}-6+\mathrm{s}+1-6=0 \\ \Rightarrow 2\mathrm{r}+\mathrm{s}=11 \\ \mathrm{equation}\left(\mathrm{i}\right)+4\times \mathrm{equation}\left(\mathrm{ii}\right), \mathrm{we} \mathrm{get}, \\ 7\mathrm{r}-4\mathrm{s}+8\mathrm{r}+4\mathrm{s}=1+44 \\ \Rightarrow 15\mathrm{r}=45 \\ \Rightarrow \mathrm{r}=3 \\ \mathrm{Putting} \mathrm{this} \mathrm{value} \mathrm{in} \mathrm{equation}\left(\mathrm{ii}\right), \mathrm{we} \mathrm{get}, \\ 2\times 3+\mathrm{s}=11 \\ \Rightarrow \mathrm{s}=5 \\ \Rightarrow \left(\mathrm{r},\mathrm{s}\right)=\left(3,5\right) \\ \mathrm{Slope} \mathrm{of} \mathrm{AC} \left({\mathrm{m}}_1\right)=\frac{5-1}{3-\left(-3\right)}=\frac{4}{6}=\frac{2}{3} \\ \mathrm{Slope} \mathrm{of} \mathrm{bisector} 7\mathrm{x}-4\mathrm{y}-1=0 \left({\mathrm{m}}_2\right) \mathrm{is} -\frac{\mathrm{Cofficient} \mathrm{of} \mathrm{x}}{\mathrm{Coefficient} \mathrm{of} \mathrm{y}}=-\frac{7}{\left(-4\right)}=\frac{7}{4} \\ \mathrm{Angle} \mathrm{between} \mathrm{AC} \mathrm{and} \mathrm{bisector} \mathrm{through} \mathrm{C} \mathrm{will} \mathrm{be}: \\ \mathrm{tan\alpha }=\left|\frac{{\mathrm{m}}_1-{\mathrm{m}}_2}{1+{\mathrm{m}}_1.{\mathrm{m}}_2}\right|=\left|\frac{{\displaystyle \frac{2}{3}-\frac{7}{4}}}{1+\frac{2}{3}.\frac{7}{4}}\right| \\ =\left|\frac{{\displaystyle \frac{8-21}{12}}}{{\displaystyle \frac{12+14}{12}}}\right|=\left|\frac{-13}{26}\right| \\ \Rightarrow \mathrm{tan\alpha }=\frac{1}{2} \\ \mathrm{Let} \mathrm{slope} \mathrm{of} \mathrm{BC} \mathrm{be} \mathrm{m}. \\ \mathrm{Angle} \mathrm{between} \mathrm{BC} \mathrm{and} \mathrm{bisector} \mathrm{through} \mathrm{C}=\mathrm{Angle} \mathrm{between} \mathrm{AC} \mathrm{and} \mathrm{bisector} \mathrm{through} \mathrm{C} \\ \Rightarrow \left|\frac{\mathrm{m}-{\mathrm{m}}_2}{1+\mathrm{m}.{\mathrm{m}}_2}\right|=\frac{1}{2} \\ \Rightarrow \left|\frac{\mathrm{m}-{\displaystyle \frac{7}{4}}}{1+\mathrm{m}.{\displaystyle \frac{7}{4}}}\right|=\frac{1}{2} \\ \Rightarrow \left|\frac{4\mathrm{m}-7}{7\mathrm{m}+4}\right|=\frac{1}{2} \\ \Rightarrow \frac{4\mathrm{m}-7}{7\mathrm{m}+4}=\pm \frac{1}{2} \\ \Rightarrow \frac{4\mathrm{m}-7}{7\mathrm{m}+4}=-\frac{1}{2} \mathrm{or} \frac{4\mathrm{m}-7}{7\mathrm{m}+4}=\frac{1}{2} \\ \Rightarrow 8\mathrm{m}-14=-7\mathrm{m}-4 \mathrm{or} 8\mathrm{m}-14=7\mathrm{m}+4 \\ \Rightarrow 8\mathrm{m}+7\mathrm{m}=14-4 \mathrm{or} 8\mathrm{m}-7\mathrm{m}=14+4 \\ \Rightarrow 15\mathrm{m}=10 \mathrm{or} \mathrm{m}=18 \\ \Rightarrow \mathrm{m}=\frac{2}{3} \mathrm{or} \mathrm{m}=18 \\ \mathrm{Hence} \mathrm{slope} \mathrm{of} \mathrm{BC} \mathrm{will} \mathrm{be} \mathrm{either} \mathrm{be} \frac{2}{3} \mathrm{or} 18. \\ \mathrm{But} \mathrm{as} \mathrm{slope} \mathrm{of} \mathrm{AC} \mathrm{is} \mathrm{also} \frac{2}{3} \mathrm{so} \mathrm{when} \mathrm{BC}=\frac{2}{3}, \mathrm{A},\mathrm{B},\mathrm{C} \mathrm{will} \mathrm{become} \mathrm{collinear} \mathrm{and} \mathrm{no} \mathrm{triangle} \mathrm{will} \mathrm{be} \mathrm{formed}. \\ \mathbf{Hence}\mathbf{ }\mathbf{slope}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{BC}\mathbf{ }\mathbf{for}\mathbf{ }\mathbf{given}\mathbf{ }\mathbf{case}\mathbf{ }\mathbf{will}\mathbf{ }\mathbf{be}\mathbf{ }\mathbf{18}\mathbf{.} \end{gathered}$$

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