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 AB = AC. If the equation of line BC is $2x-y+4=0,$ then the slope of AC is ……..  

Equation of $L_1$ is $y-4=1(x-3)\Rightarrow x-y+1=0.$ Equation of BC is 2x - y+4=0

Now B is the point of intersection of $x-y+1=0,2x-y+4=0\Rightarrow B=(-3,-2)$

$AB=\sqrt{{(3+3)}^2+{(4+2)}^2}=\sqrt{36+36}=6\sqrt{2}$

C lies on $2x-y+4=0\Rightarrow C=(h+2h+4)$

$AB=AC\Rightarrow 6\sqrt{2}=\sqrt{{(h-3)}^2+{\left(2h\right)}^2}\Rightarrow 72=h^2-6h+9+4h^2$

$\Rightarrow 5h^2-6h-63=0\Rightarrow (5h-21)(h+3)=0\Rightarrow h=\frac{21}{5}orh=-3$

If h = -3 then $C=(-3,-2)=B.$ If $h=\frac{21}{5}$ then $C=\left(\frac{{\displaystyle 21}}{{\displaystyle 5}},\frac{{\displaystyle 62}}{{\displaystyle 5}}\right)$

Slope of $\overline{AC}$ is $\frac{62/5-4}{21/5-3}=\frac{42}{6}=7$