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....


2 Answers 2


Let $A (x, y)$ be your initial point and $A'(x', y')$ be your final point.
You know the line along which your point has been moved (since you know the slope and a point on it): $$ y = x-1$$

You also know that it has been moved a distance of 4 units. For problems of this type, assume the distance it has been moved as $r$.

Using simple trigonometry, the distance along the x-axis which your point has moved is $r\cos\theta$ (where $\theta$ is the angle the line makes with the x-axis).
Similarly, the distance moved along the y-axis is $r\sin\theta$.

enter image description here

So you now have two equations:

$$x' + r\cos\theta = x$$ $$y' + r\sin\theta = y$$

Substitute your values and you have your answer!

This is a convenient method to find the final point when a certain point is moved parallel to a line.


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.

  • $\begingroup$ Equation of second line $l_2$ which is parallel to given line x -y=3 and passing thru point (2,1) is given by $$ (y-1) = 1(x-2)$$ $$\Rightarrow y-x = -1$$ $\endgroup$
    – Sachin
    Commented Nov 3, 2013 at 11:13
  • $\begingroup$ Let $A'(x_1,y_1)$. Then $y_1-x_1=-1$ (since $A' \in l_2$) and $(x_1-2)^2+(y_1-1)^2=16$ (since $AA'=4$) and so on. $\endgroup$
    – user35603
    Commented Nov 3, 2013 at 11:16
  • $\begingroup$ Use $x_1=y_1+1$. It simplifies the distance formula. $\endgroup$
    – Tejas
    Commented Nov 3, 2013 at 11:19

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