Your objection is valid.
Unless there's some special definition in force (which is why I asked for the textbook), there's no vertex, hence no angle, hence no angle bisector.
Thus, assuming the standard definition, the answer you quoted is simply wrong.
The formula
$$ d_1(x,y) = \frac{\lvert a_1x+b_1y+c_1 \rvert}{\sqrt{a_1^2+b_1^2}} $$
gives you the distance from a point $(x,y)$ to the line $L_1$ whose equation
is $a_1x+b_1y+c_1 = 0.$
See
Distance Between A Point And A Line
for a proof of this.
For the line $L_2$ given by $a_2x+b_2y+c_2 = 0,$ the distance of a point to the line is given by
$$ d_2(x,y) = \frac{\lvert a_2x+b_2y+c_2 \rvert}{\sqrt{a_2^2+b_2^2}}. $$
A point $(x,y)$ on an angle bisector between two lines is equidistant from the two lines, that is, it satisfies the condition
$d_1(x,y) = d_2(x,y).$ Writing out the formulas for $d_1$ and $d_2$ in full,
$$ \frac{\lvert a_1x+b_1y+c_1 \rvert}{\sqrt{a_1^2+b_1^2}}
= \frac{\lvert a_2x+b_2y+c_2 \rvert}{\sqrt{a_2^2+b_2^2}}. $$
Now observe that $\lvert a_1x+b_1y+c_1 \rvert$ will be either
$a_1x+b_1y+c_1$ or $-(a_1x+b_1y+c_1),$ whichever of those two
expressions is positive.
In fact, $a_1x+b_1y+c_1$ will be positive for all points on one side
of the line and negative for all points on the other side.
Now if the lines $L_1$ and $L_2$ intersect,
they divide the plane into four regions.
Label each these regions as $+L_1$ or $-L_1$ depending on whether
$a_1x+b_1y+c_1$ is (respectively) positive or negative in that region.
Label each region as $+L_2$ or $-L_2$ depending on whether
$a_2x+b_2y+c_2$ is (respectively) positive or negative in that region.
One of the angle bisectors of $L_1$ and $L_2$ will go through the regions
labeled $+L_1,+L_2$ or $-L_1,-L_2.$
That is, on that line the signs of $a_1x+b_1y+c_1$ and $a_2x+b_2y+c_2$
are either both positive or both negative.
Points on this line therefore satisfy the formula
$$ \frac{a_1x+b_1y+c_1 }{\sqrt{a_1^2+b_1^2}}
= \frac{a_2x+b_2y+c_2 }{\sqrt{a_2^2+b_2^2}}. $$
(For points in the region $-L_1,-L_2,$ this formula gives negative values
on both sides, but their absolute values are equal.)
The other angle bisector goes through
$+L_1,-L_2$ and $-L_1,+L_2$ and has the formula
$$ \frac{a_1x+b_1y+c_1 }{\sqrt{a_1^2+b_1^2}}
= - \frac{a_2x+b_2y+c_2 }{\sqrt{a_2^2+b_2^2}}. $$
Best Answer
Once you know how to find those bisectors I think the problem is how to locate the point $P=(-1,4)$ in one of those for regions.
Note that you can distinguish those regions looking to up or down of each line:
For example, regions $2$ is above of line $s$ and below of line $r$. Taking as reference the axis $OY$.
First, consider the equation $(s)$ $3x+4y+12=0$. If you take $x=-1$ you get $y=-9/4$ which is below of $y_p=4$. So $P$ is above of the line $3x+4y+12=0$.
Samething for $(r)$ $12x-5y+7=0$. If you take $x=-1$ you get $y=1$ which is below of $y_p=4$. So $P$ is above of the line $12x-5y+7=0$.
So the point $P$ would be in the region $1$.