# Equation of the line given a translation vector (that is coincident) and a point

1. Aug 21, 2005

### singleton

Hey,

Usually I have a general idea how to tackle a problem when posting on here.

This time I am pretty darn lost :(

When asked to find the equation of the line that passes through a point (1,1) and maps onto itself under the translation vector a = (3,2).

The equation of the image is
A(x - 3) + B(y - 2) + C = 0
Ax - 3A + By - 2B + C = 0

The point (1, 1) satisfies the equation since the translation "maps onto itself" (I'm taking this as it means a parallel and coincident line? not heard that terminology before..."maps onto itself") So the point is on both lines...

Basically I know that (1, 1) satisfies the equation of the image, and that when I substitute I end up with
Ax - 3A + By - 2B + C = 0
A(1) - 3A + B(1) - 2B + C = 0
-2A -B + C = 0

I almost wanted to say the normal vector for the translation was (-2, -1) but that is incorrect, as those numbers (-2 and -1) are the coefficients I have for A and B AFTER I substitute not BEFORE...

Just a small hint or suggestion how to tackle this Q would be great :)

I know (rather, I think) that I should be exploring the equation of the image (the translation) but I do not know what more I can do to find the normal vector (so I can write the Cartesian equation of the unknown line)

Last edited: Aug 21, 2005
2. Aug 21, 2005

### AKG

If you translate the line by (3,2) then you get the same line back. For example, if you have the x-axis, and you translate it by (3,2), you get a different line, but if you translate it by (3,0) you get the same line. Translating a line will give you back the same line if and only if you translate in the direction that it's already "facing." So you can tell from this the direction vector of the line, and you already know one point that lies on the line, so you have all the necessary information to describe it (and give its Cartesian equation if you want).

3. Aug 21, 2005

### singleton

It is my understanding that if the line "maps onto itself under a translation along vector a = (3,2)" we know that what this says is that the new line is the same line as before, correct?

Then, if we find the image of point (1,1) under the translation vector (3,2), which is point (4,3) we know that this point is on the line. We also know that the original point, (1, 1) is on the line since it is still the same line as before...

So the translation vector is also the same as the direction vector, (3,2) ?
And the vector equation is: r = (1,1) + t(3,2) ? (then find a normal vector and go from there)

Last edited: Aug 21, 2005
4. Aug 21, 2005

### AKG

Yes to all questions.

5. Aug 21, 2005

### singleton

thanks very much :)

6. Aug 23, 2005

### HallsofIvy

Staff Emeritus
Essentially, saying that the vector <3,2> translates the line into itself just means that the vector <3,2> points "in the same direction" as the line. Even more helpful: if (x,y) is a point on the line then so is (x+3, y+2).

Here you are told that the point (1,1) is on the line and the vector <3,2> translates the line into itself. (1,1) translated by <3,2> is (1+3,1+2)= (4, 3) so we know that the two points (1,1) and (3,2) are on the line.

Write the line y= mx+ b. (1,1) is on the line means 1= m(1)+ b or m+ b= 1. (3,2) is on the line meas 2= m(3)+ b or 3m+ b= 2. Subtracting m+ b= 1 from that gives
2m= 1 or m= 1/2. m+ b= 1 becomes 1/2+ b= 1 so b= 1- 1/2= 1/2.

The equation of the line is y= (1/2)x+ 1/2 or 2y= x+ 1.

You could, of course, use the form Ax+ By= C. Putting x= 1, y= 1 gives A+ B= C.
Putting x= 3, y= 2 gives 3A+ 2B= C. That's two equations for 3 unknowns but that is because multiplying Ax+ By= C by any number gives another equation for the same line. We can choose one of the numbers to be anything (except 0) that we like and still have an equation for the same line.

Subtracting A+ B= C from 3A+ 2B= C eliminates C leaving 2A+ B= 0 or B= -2A. Choosing A= 1 gives B= -2 and C= A+ B= 1- 2= -1.

The equation is x- 2y= -1 which is the same as 2y= x+ 1.