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

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Shifting the origin to point P while keeping the axes parallel results in the transformation of the equation ax + by + c = 0 to ax + by + c + k = 0. The locus of point P is determined to be the straight line ax + by = k. The discussion highlights the confusion arising from using the same variable names for different coordinate systems, suggesting clearer notation would improve understanding. It is noted that the new coordinates x' and y' should be defined in relation to the old coordinates as x' = x - u and y' = y - v. Ultimately, the focus is on demonstrating that point P lies on the line defined by ax + by = k.
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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'.
 
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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.
 
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 :smile:
 
revitgaur said:

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.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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