Triangle BAC was rotated 90° clockwise and dilated at a scale factor of 2 from the origin to create triangle XYZ. Based on these
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So, parallel lines have the same slope, therefore a parallel line to 2x+12, will have the same slope as that one, so which is it? 
so, we're really looking for the equation of a line whose slope is 2, and runs through -1,1.
![\bf \begin{array}{lllll} &x_1&y_1\\ % (a,b) &({{ -1}}\quad ,&{{ 1}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies 2 \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-1=2[x-(-1)] \\\\\\ y-1=2(x+1)\implies y-1=2x+2\implies y=2x+3](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Blllll%7D%0A%26x_1%26y_1%5C%5C%0A%25%20%20%20%28a%2Cb%29%0A%26%28%7B%7B%20-1%7D%7D%5Cquad%20%2C%26%7B%7B%201%7D%7D%29%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%0A%25%20slope%20%20%3D%20m%0Aslope%20%3D%20%7B%7B%20m%7D%7D%3D%20%5Ccfrac%7Brise%7D%7Brun%7D%20%5Cimplies%202%0A%5C%5C%5C%5C%5C%5C%0A%25%20point-slope%20intercept%0A%5Cstackrel%7B%5Ctextit%7Bpoint-slope%20form%7D%7D%7By-%7B%7B%20y_1%7D%7D%3D%7B%7B%20m%7D%7D%28x-%7B%7B%20x_1%7D%7D%29%7D%5Cimplies%20y-1%3D2%5Bx-%28-1%29%5D%0A%5C%5C%5C%5C%5C%5C%0Ay-1%3D2%28x%2B1%29%5Cimplies%20y-1%3D2x%2B2%5Cimplies%20y%3D2x%2B3)
so, we're really looking for the equation of a line whose slope is 2, and runs through -1,1.