<s>[INST] 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
 [/INST] 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 $x - y + 3 = 0$
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