<s>[INST] What will be the equation of side $BC$. 
The equation of two equal sides $AB$ and $AC$ of an isosceles triangle $ABC$ are $x+y=5$ and $7x-y=3$ respectively . What will be the equation of the side $BC$ if the area of the triangle $\triangle ABC$ is $5$ square units.

$a.)x+3y-1=0\\
b.)x-3y+1=0\\
c.)2x-y-5=0\\
\color{green}{d.)x+2y-5=0}\\$
$\quad\\~\\~\\$
 
Let the slope of the required line be $m$.
I used the slope formula between two lines 
$\angle B=\angle C\\~\\\left|{\dfrac{(-1-m)}{(1-m)}}\right|=\left|{\dfrac{(7-m)}{(1+7m)}}\right|\\
\implies m=-3,\dfrac{1}{3}$
But the book is giving right answer as option $d.)$.
By finding $m$ i only found slope not the whole equation.
Also i would like to know if their is clean simple short way , and also using $Area=5$ .
I have studied maths upto $12th$ grade.
Update : by using Geogebra I found that $x-3y+1=0$ fits perfectly

 [/INST] the two two lines that bisects the $\angle BAC$ are given by 
$$\frac{7x-y-3}{\sqrt{50}} = \pm\frac{x+y-5}{\sqrt2}.$$ they are 
$$x-3y+11 = 0, \quad 3x + y - 7=0 \text{ and } A = (1, 4).$$
we will pick a point $$D = (1+2k, 4-6k), k \text{ to be fixed later} $$ on the angle bisector  $3x+y-7 = 0.$  the line through $D$  parallel to the other bisector is $$x - 3y = 1+2k -3(4-6k) = 20k-11.\tag 1$$  the point $C$ is the intersection of $(1)$ and $$7x - y = 3 \tag 2 $$ therefore $$C = (1-k, 4-7k). $$  we can now compute $$AD^2 = 4k^2 + 36k^2 = 40k^2\\
CD^2 = 9k^2 + k^2 =10k^2$$  the constraint that the area of the triangle $ABC = 5$ translates to $$AD^2 \times CD^2 = 25\implies 400k^2 = 25, k = \pm \frac14 $$  the line $BC$  are $$x-3y = -6, x-3y = -16. $$
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