i dont quite get the question but...
i guess this is how it is.
Take the mirror image of∆ABC Through the a line through the point y=3.
The new ∆ABC would have point C=(4,2)
B=(3,-6) A=(1,-3)
Now shifting the ∆ABC one unit (<em>i.e. 2 acc. to the graph as scale is 1 unit =2</em>) towards right ( or <em>adding 2 to the x coordinates of ∆ABC)</em>
We get the Coordinates of triangle ABC as A=(3,-3) B=(5,-6) C=(6,2).
This coordinate is the same coordinates of ∆A"B"C".
Hope it helps...
Regards;
Leukonov/Olegion.
First isolate the "y" in the equation.
2y - x = -12 Add x on both sides
2y - x + x = -12 + x
2y = -12 + x Divide 2 on both sides to get "y" by itself


Your slope is
.
For the equation of the line to be parallel to the given equation, the slopes have to be the same. So the parallel line's slope is also 
y = mx + b

To find "b", you plug in the point (18,2) into the equation


2 = 9 + b Subtract 9 on both sides
2 - 9 = 9 - 9 + b
-7 = b
Your equation is:

The correct answer would be D -13
-10
-|a-b|=-|7-(-3)|=-|10|=-10
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