high school line equation question I hit stumbling block below.
But i try everything that I can think of but i failed to find any satisfactory answer.
my workout is that I find mid point between point B and C.
and then using A and earlier mid point to find slope which I know is perpendicular.
But line equation y = mx + c didn't match with answer at the back.  
the problem is as below:
Let A(1,1), B(4,5), and C(6,13) be points in the xy-plane.
Find the equation of the line which bisects $\angle$BAC.
The solution given is 7x - 4y = 3
your help is much appreciated.
 A: As pointed out in the comments, your method doesn't quite work. But it almost does.
If you extend $AB$ to a point $D$ so that $AD$ and $AC$ have the same length, you can find the midpoint of $CD$ and then continue as you did earlier. That same procedure should now produce the correct result.
A: Do you know (or perhaps, remember) that the angle bisector is the locus of all points that are equidistant from both angle's rays? In this case, the angle rays are the lines $\,AB\,,\,AC\,$:$$AB:\,4x-3y-1=0\,\,,\,AC:\,12x-5y-7=0$$So we want all the points $\,(x,y)\,$ which are at the same distance from both lines above: $$\frac{|4x-3y-1|}{\sqrt{4^2+3^2}}=\frac{|12x-5y-7|}{\sqrt{12^2+5^2}}\Longrightarrow 13^2(4x-3y-1)^2=5^2(12x-5y-7)^2\Longrightarrow$$Try to continue from here.
Added In fact you don't need the headache of the squares in the last line right: just check the left side with different values for the absolute values (once $\,|A|\,=A\,,\,\,another\,\,with\,\,|A|=-A\,$), taking into account that both lines $\,AB\,,\,AC\,$ have positive slope and thus the bisector line has also to have positive slope...and between the first two lines' slopes) 
Further added: Drop the squares but you can not drop the absolute values, so we get $$13|4x-3y-1|=5|12x-5y-7|\Longrightarrow 52x-39y-13=\pm\left(60x-25y-35\right)$$$$(i) \,\,52x-39y-13=60x-25y-35\Longrightarrow 8x+14y-22=0$$and this is not what we want as both given lines have positive slope (and thus their angle's bisector also has positive slope, so we get here the bisector of the obtuse angle between the lines, which is usually NOT "the angle between the lines", defined almost always as the acute (or straight) one between), so we go with $$(ii)\,\,52x-39y-13=-60x+25y+35\Longrightarrow 112x-64y=48\Longrightarrow 7x-4y=3$$and we get what we wanted.
Of course, there's also a well-known formula with tangents and difference of angles and stuff, but on purpose I left trigonometry out of the answer.
