<s>[INST] 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.
 [/INST] 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
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