In $ riangle \mathrm{ABC}$ the coordinates of the vertex $\mathrm{A}$ are $(-3,1)$. If the equation of the median through $B$ is $2 x+y-3=0$ and the equation of the bisector of angle $\mathrm{C}$ is $7 \mathrm{x}-4 \mathrm{y}-1=0$, then the equation of the side $\mathrm{BC}$ is

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In $	riangle \mathrm{ABC}$ the coordinates of the vertex $\mathrm{A}$ are $(-3,1)$. If the equation of the median through $B$ is $2 x+y-3=0$ and the equation of the bisector of angle $\mathrm{C}$ is $7 \mathrm{x}-4 \mathrm{y}-1=0$, then the equation of the side $\mathrm{BC}$ is

Aicharya · Accepted Answer

$M=$ Mid point of $A C=\left(\frac{-3+h}{2}, \frac{1+k}{2}\right)$ $\because(h, k)$ lies on $7 x-4 y-1=0$. $\Rightarrow 7 h-4 k-1=0 \Rightarrow h=\frac{4 k+1}{7}$ $\therefore M=\left(\frac{-3+\frac{4 k+1}{7}}{2}, \frac{1+k}{2}\right)=\left(-10+2 k, \frac{1+k}{2}\right)$ $\because M$ lies on $B M$. $\Rightarrow 2(-10+2 k)+\left(\frac{1+k}{2}\right)-3=0 \Rightarrow k=5$ from equation $(2) \Rightarrow h=\frac{4 \times 5+1}{7}=3$ $\therefore(h, k)=(3,5)$ Slope of line $A C=m_1=\frac{5-1}{3+3}=\frac{4}{6}=\frac{2}{3}$ Slope of line $\mathrm{CN}=m_2=\frac{7}{4}$. Let slope of $B C$ is $m$. Since $\angle B C N=\angle N C A$. $\Rightarrow\left|\frac{m-\frac{7}{4}}{1+\frac{7}{4} m}\right|=\left|\frac{\frac{7}{4}-\frac{2}{3}}{1+\frac{7}{4} \times \frac{2}{3}}\right|=\frac{1}{2}$ $\Rightarrow \frac{m-\frac{7}{4}}{1+\frac{7}{4} m}=\frac{1}{2} \Rightarrow m=18$ Equation of $B C$ is $\begin{aligned} & (y-5)=m(x-3) \Rightarrow(y-5)=18(x-3) \\ & \Rightarrow 18 x-y=49 . \end{aligned}$

In $\triangle \mathrm{ABC}$ the coordinates of the vertex $\mathrm{A}$ are $(-3,1)$. If the equation of the…

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