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


By calculating the intersection point, we get is as $(2,5)$.
But I couldn't proceed because I don't know how to find the equation on the acute side. Please help.
I would prefer a more geometrical approach
 A: to find the equation of the bisector. then these points satisfy this equation:
$$\left|\frac{a_1x+b_1y+c_1}{\sqrt{a_1^2 + b_1^2}}\right|=\left|\frac{a_2x+b_2y+c_2}{\sqrt{a_2^2 + b_2^2}}\right|.$$
But which sign should we take to get the correct bisector?
I think it depends on the sign of the inner product of the normal vectors of the lines. That is if the inner product of the normal vectors is positive then you take the positive sign and other the sides of the equation should have the opposite signs.
But why when the inner products of the normals are positive you take the positive sign of the equations?
Because, each side of the equation the distance of the point with the direction of the line that makes the positive orientation with the normal of that line. And in this direction the angle between the lines is acute.
A: This question can be work out in a rigorous way through answers to this general question here.
However a much quicker way for this specific data, using geometrical approach is as follows:
Note that if $m_1, m_2$ are slopes of two lines such that $m_1 m_2=1$, these two lines are reflections in (or equally inclined to) some line of slope $1$.
It's easy to see a line of slope $1$ is the acute bisector since the slopes of both of given lines is positive.
Thus equation of bisector required, passing through $(2,5)$ is $$\boxed{y=x+3}$$
A: ${\bf n}_1=(4,-3)$ is a vector normal to the first line, and ${\bf n}_2=(3,-4)$ is normal to the second.
Their dot product ${\bf n}_1 \cdot {\bf n}_2=24$ is positive, so the angle between them is acute.
Normalize each of the vectors and take the sum ${\bf b}= \frac{1}{5}{\bf n}_1+\frac{1}{5}{\bf n}_2=(7/5,-7/5)$.
Then ${\bf b}$ is a vector normal to the line bisecting the acute angle, while
${\bf c}= \frac{1}{5}{\bf n}_1-\frac{1}{5}{\bf n}_2=(1/5,1/5)$, is normal to the bisector of the obtuse angle.
Then knowing the intersection point you know how to write the equation of the bisector.
A: Equation of angle bisector lines are given by normalized equations of the two lines.
So, $\frac{L1}{|L1|} \pm \frac{L2}{|L2|} = 0$
So, $\frac {4x-3y+7} {\sqrt{4^2+3^2}} \pm \frac {3x-4y+14} {\sqrt{4^2 + 3^2}} = 0$
That gives us both angle bisectors:  $x - y + 3 = 0, \, x + y - 7 = 0$
Now the slopes of original lines are both positive: $\frac{4}{3}, \frac{3}{4}$.
So the angle bisector that we have to choose on the acute angle side should have a positive slope between $\frac{4}{3}$ and $\frac{3}{4}$.
So equation of desired angle bisector is $x - y + 3 = 0$
A: since  their bisector pass the point $(2,5)$ too, so after shifting their bisectors to meet whose intersection points at axis $Y$, we can deduce, the signature in bisector equation $\frac{\vert{k-4/3}\vert}{1+(4/3)k}=\frac{\vert{3/4-k}\vert}{1+(3/4)k}$, is between $3/4$ and $4/3$. thus one is positive and another is negative, then computation lead $k=1$. lastly, use the fact that their bisector pass through $(2,5)$ too, to deduce $2a-5b=-c$, thus $c=3$ between $7/3$ and $14/4$.
