Equation of a straight line passing through passing through a point and equally inclined to two other lines 
Find the equation of the straight line passing through the point $(4,5)$ and equally inclined to the lines $3x= 4y+7$ and $5y=12x+6$.

I know that the equation of the bisector is given by:
$\dfrac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}$=$\pm$$\dfrac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}$
but I am facing real difficulty in finding which sign I should choose and why? 
the answers are: $9x-7y=1 $ and $7x+9y=73$
Here is a Desmos graph of the above equation.
 A: By using your idea we can get two slops: $m=\frac{9}{7}$ or $m=-\frac{7}{9}$ and from here two equations:
$$y-5=\frac{9}{7}(x-4)$$ 
$$y-5=-\frac{7}{9}(x-4)$$
A: the straight line which passes through $$P(4;5)$$ has the equation
$$y=m(x-4)+5$$ converting the others into the Hessian Normalform we get
$$0=\frac{4y-3x+7}{\pm5}$$ and the other
$$0=\frac{5y-12x-6}{\pm 13}$$ the $$Point (0; -4m+5)$$ is situated on our line and we must compute
$$\frac{|5(-4m+5)-6|}{13}=\frac{|4(-3m+5)+7|}{5}$$
fromk here you will get $m$
A: You have TWO lines which form equal angles with the given straight lines
$3x-4y-7=0;\;12x-5y+6=0$
$\dfrac{3x-4y-7}{\sqrt{3^2+4^2}}=\pm\dfrac{12x-5y+6}{\sqrt{12^2+5^2}}$
$\dfrac{3x-4y-7}{5}=\pm\dfrac{12x-5y+6}{13}$
$13(3x-4y-7)=\pm 5(12x-5y+6)$
$99 x-77 y-61=0;\;21 x+27 y+121=0$
$y=\dfrac{9 x}{7}-\dfrac{61}{77};\;y=-\dfrac{7 x}{9}-\dfrac{121}{27}$
If the wanted lines gave to pass through the point $(4,5)$ then their equation is
$y-5=m(x-4)$ 
where $m_1=\dfrac{9}{7};\;m_2=-\dfrac{7}{9}$
Hope this helps
