Change of Origin: Locus of P is Straight Line ax+by=k

[revitgaur]

Homework Statement

On shifting the origin to a point P, the axes remaining parallel to the old axes,the equation ax+by+c=0 is transferred to ax+by+c+k=0.Show that the locus of P is the straight line ax+by=k.

2. Relevant equation
1. ax+by+c=0
2. ax'+by'+c+k=0

The Attempt at a Solution

In the first equation ax+by+c=0 i put x=x'+h and y=y'+k'and then i subtract first and second equation but only one equation is obtained now I don't know how to find the value of h and k'.

 

[OmneBonum]

Have you tried graphing it? I just used https://www.desmos.com/calculator. Not that I'm affiliated with the site but it's a good resource. To keep it simple I let both a and b = 1 and c=0. Then for the second line, plug in whatever value you wish for k. Remember that k becomes negative when it passes the equal sign. That might stimulate some ideas for you.

 

[BvU]

All you have to show is that P is on that line; you don't have to find P.
Note that your h has something to do with the x coordinate of P and the (unhappily chosen) k' has something to do with the y coordinate of P

 

[Ray Vickson]

Homework Statement

On shifting the origin to a point P, the axes remaining parallel to the old axes,the equation ax+by+c=0 is transferred to ax+by+c+k=0.Show that the locus of P is the straight line ax+by=k.

2. Relevant equation
1. ax+by+c=0
2. ax'+by'+c+k=0

The Attempt at a Solution

In the first equation ax+by+c=0 i put x=x'+h and y=y'+k'and then i subtract first and second equation but only one equation is obtained now I don't know how to find the value of h and k'.

Part of the problem (and a possible source of confusion) is the problem's use of the same names x and y to stand for two different things. It might have been better if the problem stated that the new coordinate axes x' and y' are parallel to the old axes x and y, and the equation ax + by + c = 0 is transferred to ax' + by' + c + k = 0.

Note that parallel axes means that x' = x - u and y' = y - v for some constants u and v. The origin of the (x',y') system is at x = u and y = v in the old (x,y) system.