The bisector of the acute angle formed between the lines $4x- 3y + 7 = 0$ and $3x - 4y + 14 = 0$ has the equation 

The equations of given straight lines, by making constant terms positive, are

$4x-3y+7=0\text{ and }3x-4y+14=0$

$\because 4\times 3+(-3)(-4)=24>0\text{ i.e. }{a}_1{a}_2+b_1{b}_2>0$

So, the bisector of the acute angle is given by

$\frac{4x-3y+7}{\sqrt{4^2+(-3{)}^2}}=-\frac{3x-4y+14}{\sqrt{3^2(-4{)}^2}}\Rightarrow x-y+3=0$