Find the equation of base of Isosceles Traingle Given the two Legs $AB$ and $AC$ of an Isosceles Traingle as $7x-y=3$ and $x-y+3=0$ Respectively. if area of $\Delta ABC$ is $5$ Square units, Find the Equation of the base $BC$
My Try:
The coordinates of $A$ is $(1,4)$. Let the Slope of $BC$ is $m$. Since angle between $AB$ and $BC$ is same as Angle between $AC$ and $BC$ we have
$$\left|\frac{m-7}{1+7m}\right|=\left|\frac{m-1}{1+m}\right|$$ solving which we get $m=2$ or $m=\frac{-1}{2}$
Can any one help me further
 A: You have the vertex $A$ and the equations for the congruent sides.
So you can compute an apex angle $\theta$ with the dot product of two vectors from $A$.
Then, using that angle (or its complement) you can calculate the length of the congruent side $d$ from
$$\frac{d^2 \sin \theta}{2} = 5.$$
This gives you the length to go away from $A$ along the lines, from which you can define the equation for the base.
Note that as given you have four possible equations.
A: The problem can be solved using parametric equations.  Since you know that both equations pass through the point, $(1,4)^T$, convenient parameterizations for the lines are:
$\left(\begin{array}{c}x_1\\y_1\end{array}\right) = \left(\begin{array}{c}1\\7\end{array}\right)s+\left(\begin{array}{c}1\\4\end{array}\right)$
and
$\left(\begin{array}{c}x_2\\y_2\end{array}\right) = \left(\begin{array}{c}1\\1\end{array}\right)t+\left(\begin{array}{c}1\\4\end{array}\right)$
The lengths of the lines from the point $(1,4)^T$ must be equal and are given by 
$\left|\left|\left(\begin{array}{c}x_1\\y_1\end{array}\right)-\left(\begin{array}{c}1\\4\end{array}\right)\right|\right| = \left|\left|\left(\begin{array}{c}x_2\\y_2\end{array}\right)-\left(\begin{array}{c}1\\4\end{array}\right)\right|\right|$
or 
$\left|\left|\left(\begin{array}{c}1\\7\end{array}\right)s\right|\right| = \left|\left|\left(\begin{array}{c}1\\1\end{array}\right)t\right|\right|$
which reduces to $5s = \pm t$.
The area gives another equation.  The area of a parallelogram is given as the cross product of two vectors that meet at one corner, so the area of the triangle is half that much.  
$\left|\left|\left[\left(\begin{array}{c}x_1\\y_1\end{array}\right)-\left(\begin{array}{c}1\\4\end{array}\right)\right]\times\left[\left(\begin{array}{c}x_2\\y_2\end{array}\right)-\left(\begin{array}{c}1\\4\end{array}\right)\right]\right|\right|=\left|\left|\left(\begin{array}{c}1\\7\end{array}\right)s\times\left(\begin{array}{c}1\\1\end{array}\right)t\right|\right|=|st-7ts|=|6st|$
Therefore, $\frac{1}{2}6st=\pm 5$ or $st=\pm\frac{5}{3}$.  Combining the equations yields $s = \pm\frac{1}{\sqrt{3}}$, $t=\pm\frac{5}{\sqrt{3}}$.
Finally, the equation of the line defining the third edge can be given as a parametric equation passing through the points $(x_1,y_1)^T$ and $(x_2,y_2)^T$.
$\left(\begin{array}{c}x\\y\end{array}\right) = \left[\left(\begin{array}{c}x_2\\y_2\end{array}\right)-\left(\begin{array}{c}x_1\\y_1\end{array}\right)\right]u+\left(\begin{array}{c}x_1\\y_1\end{array}\right)=\left[\left(\begin{array}{c}1\\7\end{array}\right)s-\left(\begin{array}{c}1\\1\end{array}\right)t\right]u+\left(\begin{array}{c}1\\7\end{array}\right)s+\left(\begin{array}{c}1\\4\end{array}\right)$
Substituting in the values of $s$ and $t$ found above yields the equation.  Note that there are four solutions.
