<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] So, the coordinate of $A(1,4)$
We can set the coordinates of $B(a,5-a);C(b,7b-3)$
We have $|AB|^2=|AC|^2\implies(a-1)^2+(a-1)^2=(b-1)^2+\{7(b-1)\}^2$
$\implies a-1=\pm(5b-5)=\pm5(b-1)$
$$\triangle ABC=\dfrac12\begin{vmatrix}1 & 4 & 1 \\ a & 5-a & 1 \\ b & 7b-3 &1\end{vmatrix}=4|(a-1)(b-1)|$$
$\implies\triangle ABC=4\cdot5|(b-1)^2|$
$\implies20(b-1)^2=5\iff b-1=\pm\dfrac12$
Can you take it home from here?
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