Point by point, the transformation $T(x, y) = (4x - y, 3x -2y)$ sends the line $x + 2y = 6$ onto an image line. What is the slope of the image? The question is as follows:

Point by point, the transformation $T(x, y) = (4x - y, 3x -2y)$ sends the line $x + 2y = 6$ onto an image line. What is the slope of the image? 

I don't understand what this question basically. How is it that the transformation is able to send a line into an "image line" (a term that I am not even familiarized with)? Any help will be greatly appreciated. 
 A: Think of $T$ here as sending all points $(x,y)$ satisfying $x+2y=6$ to $(x',y')=(4x-y,3x-2y)$. Since there is an equation between $x$ and $y$, there must also be one between $x'$ and $y'$, and this latter equation will be the image line.
$$x'=4x-y$$
$$y'=3x-2y$$
$$-2x'=2y-8x$$
$$y'-2x'=-5x\qquad x=(2x'-y')/5\tag1$$
$$y'=3/5(2x'-y')-2y$$
$$2y=3/5(2x'-y')-y'=6/5x'-8/5y'$$
$$y=(3x'-4y')/5\tag2$$
Then substituting $(1)$ and $(2)$ into $x+2y=6$:
$$(2x'-y')/5+2(3x'-4y')/5=6$$
$$8x'-9y'=30$$
$$y'=(8x'-30)/9$$
This is the image line, and its slope is $\frac89$.
A: $T:\mathbb{R^2} \rightarrow \mathbb{R^2}$.
$T(x,y) \mapsto (x',y')$,
$x'=4x-y$;  $y'=3x-2y.$
Solving for $x,y$ in terms of $x',y'$.
$2x'-y'=5x$;  and $ 3x' -4y' =5y$.
The line $x+2y = 6$ is transformed:
$5x +10y =30 \rightarrow$
$(2x'-y' ) + 2 (3x'-4y') =30.$
$8x' - 9y' = 30.$
Slope of the image line:  $m' = 8/9.$
A: assume $h=4x-y$ and $k=3x-2y$   now take the image of the point $(h,k)$ with respect to the line $4x+2y=6$ you will get a point now you can get the slope easily
