An example tells a good tale.

Translation of a line in an x-y coordinate system:

Take a line y = 2x -3, and translate it by 4 up and 5 to the right.

Simple approach : The point P = (2, 1) is on the line (so are some others!). Let us translate the point to get Q = (2+5, 1+4), which is Q = (7, 5), and find the line through Q parallel to the original line. The only thing that changes is the c value, so the new equation is y = 2x + c, and it must pass through Q. So we require 5 = 14 + c, giving the value of c as -9.

Not much algebra there, but a horrible question remains – “What happened to all the other points on the line ?”

We try a more algebraic approach – with any old line ax + by + c = 0, and any old translation, q up and p to the right.

First thing is to find a point on the line – “What ? We don’t know ANYTHING about the line.”

This is where algebra comes to the rescue. Let us suppose (state) that a point P = (d, e) IS on the line.

Then **ad + be + c = 0**

Now we can move the point P to Q = (d + p, e + q) (as with the numbers earlier), and make the new line pass through this point: This requires a new constant c (call it newc) and we then have a(d + p) + b(e + q) + ‘newc’ = 0

Expand the parentheses (UK brackets, and it’s shorter) to get ad + ap + be + bq + ‘newc’ = 0

Some inspired rearranging gives ‘newc’ = -ap – bq – ad – be, which is equal to -(ap + bq) – (ad + be)

“Why did you do that last step ?” – “Because I looked back a few lines and figured that (ad + be) = -c, which not only simplifies the expression, it also disposes of the unspecified point P.

**End result is**: Translated line equation is ax + by + ‘newc’ =0, that is, **ax + by + c – (ap + bq) = 0**

and the job is done for ALL lines, even the vertical ones, and ALL translations. Also we can be sure that we know what has happened to ALL the points on the line.

I am not going to check this with the numerical example, you are !

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