Straight line problem: Point $A (2,1)$ is translated parallel to line $x -y =3$ by a distance of $4$ units of new position... Straight line problem : 

Point $A(2,1)$ is translated parallel to line $x -y=3$ by a distance of $4$ units of new position $A'$ which is in third quadrant then find the coordinates of $A'$.

My approach : 
Slope of another line is same as of $x -y =3$ therefore slope of new line is also $1$. And angle $\theta = 45^{\circ}$.
Also knows the distance between a line $(ax +by+c=0)$ and a point $(x_1,y_1)$  (perpendicular distance) can be given by 
$$\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}.$$
But unable to understand the problem due to word translated but know the definition of this: "In a translation all the points in the object are moved in a straight line in the same direction. The size, the shape and the orientation of the image are the same as that of the original object. Same orientation means that the object and image are facing the same direction".
Could anybody please help on this... I will be grateful to you thanks.... 

Hint: Try to obtain an equation for your second line $l_2$ (that is parallel to $l_1:x-y=3$ and contains $A(2,1)$). Then use that: $A' \in l_2$, $AA'=4$ and $A'$ is in third quadrant.
