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A line $L_1$ passing through $A(3,4)$ and having slope 1 cuts another line $L_2$ passing through $C$ at $B$, such that $A B=A C$. If the equation of line $B C$ is $2 x-y+4=0$, then the equation of $A C$ is
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Verified Answer
The correct answer is:
$7 x-y-17=0$
Equation of $L_1$ :
$\begin{aligned}
y-4 & =1(x-3) \\
y-4 & =x-3 \\
x-y+1 & =0
\end{aligned}$
Let slope of $A C$ is $m$.
Since $A C=B C$
$\therefore \triangle A B C$ is isosceles
$\begin{aligned}
\therefore \quad \angle C & =\angle B \\
\left|\frac{m-2}{1+2 m}\right| & =\left|\frac{2-1}{1+2 \times 1}\right| \quad\left[\because \text { slope of } B C=\frac{-2}{-1}=2\right]
\end{aligned}$
$\begin{aligned}
& \Rightarrow \quad \frac{m-2}{1+2 m}= \pm \frac{1}{3} \\
& \Rightarrow \quad \frac{m-2}{1+2 m}=\frac{1}{3} \text { or } \frac{m-2}{1+2 m}=\frac{-1}{3} \Rightarrow m=7 \text { or } m=1
\end{aligned}$
But for $m=1$
AC become $L_1$
$\therefore \quad m=7$
Equation of $A C$ is
$\begin{gathered}
y-4=7(x-3) \\
y-4=7 x-21 \\
7 x-y=17=0
\end{gathered}$
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