Parallelogram ABCD There's a parallelogram $ABCD$. I'm given point $A(3,12)$ and point  $B(-1,5)$. Given the equations of the lines $BC$ and $AC$ are $y=8x+13$ and $y=3x+3$ respectively. 
How to find the coordinates of the point of intersection between the diagonals $BD$ and $AC$? And the coordinates of $D$? 
I've no idea. Can someone explain it ? Thanks!

Let the coordinated of $D$ be $(a, b)$. Since $AD\parallel BC$ hence the lines $AD$ & $BC: y=8x+13$ have equal gradients  $$\frac{b-12}{a-3}=8$$ $$b-12=8a-24$$ $$\implies b=8a-12\tag 1$$ Now, solving $BC: y=8x+13$ & $AC: y=3x+3$ , we get coordinates of the vertex $C(-2, -3)$
Since $CD\parallel AB$ hence the lines $CD$ & $AB$ have equal gradients  $$\frac{b-(-3)}{a-(-2)}=\frac{5-12}{-1-3}$$  $$4b+12=7a+14$$ setting the value of $b$ from (1), we get $$4(8a-12)+12=7a+14\implies 25a=50 \implies  a=2$$ $$b=8(2)-12=4$$ Hence the coordinates of the point $D$ are $(2, 4)$ Since, the diagonals $BD$ & $AC$ are bisecting each other hence their intersection point will the mid-point of $BD$ joining $B(-1, 5)$ & $D(2, 4)$ given as follows $$\left(\frac{-1+2}{2}, \frac{5+4}{2}\right)\equiv \left(\frac{1}{2}, \frac{9}{2}\right)$$
Hence, $$\bbox[5px, border:2px solid #C0A000]{\text{intersection point of diagonals BD & AC}\equiv \color{blue}{\left(\frac{1}{2}, \frac{9}{2}\right)}}$$
$$\bbox[5px, border:2px solid #C0A000]{\color{blue}{D\equiv(2, 4)}}$$
