Co-ordinate Geometry: Equation of the line Find the equation of the line that passes through the point of intersection of the lines
$3 + 2 − 1 = 0$ and $2 −  + 7 = 0$ and is perpendicular to the line $4 + 4 = 7$.  
Solve for x:
$3 + 2 − 1 = 0$
$2 −  + 7 = 0$ (multiply by $2$)  
$3 + \not{2y} − 1 = 0$
$4 − \not{2} + 14 = 0$ 
$7x + 13 = 0$
$x =\frac{-13}{7}$.
Sub $x =\frac{-13}{7}$  into $3 + 2 − 1 = 0$
To get $3\left(\frac{-13}{7}\right) + 2 − 1 = 0$
Which reduces to $y=\frac{23}{7}$.
Line is perpendicular to $4 + 4 = 7$ so $m$ must be $1$.
Equation of line formula: $y-y_1=m(x-x_1)$.
$y-\frac{23}{7}=1(x-\frac{-13}{7})$
$y-1x-\frac{36}{7}=0$ $($Multiply by $7$$)$
$7y-7x-36=0$.  
Been at this for about 2 hours now so I'd appreciate any help or guidance, thank you.
 A: Looks like you found your error with the help of a comment or two. Here’s a perhaps simpler approach that doesn’t require finding the intersection point explicitly.  
Every line that passes through the the intersection point of the two given lines has an equation that is a linear combination of their equations. Thus, the required equation will be of the form $$\lambda(3x+2y-1)+\mu(2x-y+7)=0\tag1$$ with at least one of the coefficients nonzero. We want this to be orthogonal to $4x+4y-7=0$. This condition can be expressed as the vanishing of the dot product of the normals to the lines, i.e., $$(4,4)\cdot\left[\lambda(3,2)+\mu(2,-1)\right] = (4,4)\cdot(3\lambda+2\mu,2\lambda-\mu) = 4(5\lambda+\mu) = 0.$$ Pick any nonzero solution of this simple equation, such as $\lambda=1$ and $\mu=-5$, substitute into (1) and simplify.  
Another perhaps simpler way to approach this is to note that every line perpendicular to $4x+4y=7$ has an equation of the form $x-y=c$. We want the system that consists of the equations of the first two lines and this latter equation to have a unique solution, which will occur when the augmented matrix $$\left[\begin{array}{cc|c}3&2&-1\\2&-1&7\\1&-1&c\end{array}\right]$$ has rank 2. This occurs iff its determinant vanishes, which leads to a linear equation in $c$ to solve. Equivalently, row-reduce this matrix and impose the condition that the last row of the reduced matrix consist only of zeros so that the system is consistent. This again results in a simple linear equation in $c$ to solve.
A: The point of intersection is indeed $\left(-\dfrac{13}7,\dfrac{23}7\right)$. A line perpendicular to $4x+4y=7$ has its linear coefficients proportional to $(1,-1)$ (because $4\cdot1-4\cdot1=0$), hence
$$\left(x+\dfrac{13}7\right)-\left(y-\dfrac{23}7\right)=0$$
or
$$7x-7y+36=0.$$
