In the diagram line1 has equation ${\displaystyle 2x+3y-18=0.}$

What is equation of line1 relative to origin ${\displaystyle \Omega =(6,4)?}$

Using ${\displaystyle x=x'+h,\ y=y'+k:}$

${\displaystyle 2(x'+6)+3(y'+4)-18=0}$

${\displaystyle 2x'+12+3y'+12-18=0}$

${\displaystyle 2x'+3y'+6=0}$

When ${\displaystyle x,y}$ become ${\displaystyle x',y',}$ this means that equation ${\displaystyle 2x'+3y'+6=0}$ is relative to origin ${\displaystyle \Omega .}$ line1 can have equation ${\displaystyle 2x+3y-18=0}$ or equation ${\displaystyle 2x'+3y'+6=0.}$

Equation ${\displaystyle 2x'+3y'+6=0}$ relative to origin ${\displaystyle \Omega }$ is same as equation ${\displaystyle 2x+3y+6=0}$ relative to origin ${\displaystyle O.}$

Note that in both cases:

• ${\displaystyle X}$ intercept relative to origin is ${\displaystyle -3.}$

• ${\displaystyle Y}$ intercept relative to origin is ${\displaystyle -2.}$

Let a line have equation: ${\displaystyle ax+by+d=0.}$

Let ${\displaystyle c=\cos \theta ;\ s=\sin \theta .}$

After rotation, equation of line relative to ${\displaystyle OX',OY'}$ (red arrows) is:

${\displaystyle a(x'c-y's)+b(x's+y'c)+d}$ ${\displaystyle =ax'c-ay's+bx's+by'c+d}$ ${\displaystyle =ax'c+bx's+by'c-ay's+d}$ ${\displaystyle =x'ac+x'bs+y'bc-y'as+d}$ ${\displaystyle =x'(ac+bs)+y'(bc-as)+d}$ ${\displaystyle =Ax'+By'+d=0\ \dots \ (1)}$

where: ${\displaystyle A=a\cos \theta +b\sin \theta ;}$${\displaystyle \ B=b\cos \theta -a\sin \theta .}$

In the diagram line1 has equation ${\displaystyle 6x+17y-300=0,}$ and ${\displaystyle \cos \theta ={\frac {4}{5}}.}$

What is equation of line1 relative to ${\displaystyle OX',OY'}$ (red system)?

# python code.
>>> a,b,d = 6,17,-300
>>> c,s = cosθ,sinθ = 4/5,3/5 ; c,s
(0.8, 0.6)
>>> A = a*c + b*s ; A
15.0
>>> B = b*c - a*s ; B
10.0

Equation of line1 relative to ${\displaystyle OX',OY':\ 15x'+10y'-300=0}$ or ${\displaystyle 3x'+2y'=60.}$

${\displaystyle OP=OP';\ OQ=OQ'.}$ Triangles ${\displaystyle OPQ,OP'Q'}$ are congruent.

Line1: ${\displaystyle 3x'+2y'=60}$ has same position in red system as line2: ${\displaystyle 3x+2y=60}$ in black system.

Let quartic function be defined as ${\displaystyle y=f(x)={\frac {1.5625x^{4}-12.125x^{3}-14.75x^{2}+136.5x+114}{48}}.}$

In diagram, ${\displaystyle X}$ and ${\displaystyle Y}$ axes are rotated through angle ${\displaystyle \theta }$ to produce new system of coordinates ${\displaystyle OX',OY'}$ (red system.)

${\displaystyle \cos \theta ={\frac {4}{5}}.}$ What is equation of ${\displaystyle f(x)}$ relative to red system?

  (0.0325520833333333)  ((x(0.8) - y(0.6))^4) 
+ (-0.252604166666667)  ((x(0.8) - y(0.6))^3) 
+ (-0.307291666666667)  ((x(0.8) - y(0.6))^2) 
+            (2.84375)  ((x(0.8) - y(0.6))  )
+ (2.375)
- (x(0.6) + y(0.8)) = 0

Red curve in diagram has equation ${\displaystyle g(x,y)=0.}$

Equation of ellipse in diagram is: ${\displaystyle f(x,y)=55x^{2}-24xy+48y^{2}-2496=0.}$

What is equation of ${\displaystyle f(x,y)}$ relative to minor and major axes (red system)?

The most general equation of second degree in ${\displaystyle x,y}$ has form: ${\displaystyle ax^{2}+bxy+cy^{2}+dx+ey+f=0.}$

Rotation of coordinate axes when applied to the general equation produces the primed equation: ${\displaystyle A(x')^{2}+Bx'y'+C(y')^{2}+Dx'+Ey'+F=0}$ where:

${\displaystyle A=an^{2}+bns+cs^{2}}$

${\displaystyle B=bn^{2}-2ans+2cns-bs^{2}}$${\displaystyle =bn^{2}+(2c-2a)ns-bs^{2}}$

${\displaystyle C=cn^{2}-bns+as^{2}}$

${\displaystyle D=dn+es}$

${\displaystyle E=en-ds}$

${\displaystyle F=f}$

${\displaystyle n=\cos \theta }$

${\displaystyle s=\sin \theta }$

Choose values of ${\displaystyle \sin \theta ,\cos \theta }$ that make coefficient ${\displaystyle B=0.}$

From coefficient ${\displaystyle B}$ above:

${\displaystyle bn^{2}+(2c-2a)ns-bs^{2}=0}$

${\displaystyle bn^{2}-bs^{2}=-(2c-2a)ns}$

Square both sides, substitute ${\displaystyle (1-ss)}$ for ${\displaystyle nn,}$ expand, gather like terms and result is:

${\displaystyle PS^{2}+QS+R=0\ \dots \ (1)}$ where:

${\displaystyle S=\sin ^{2}\theta }$

${\displaystyle P=4a^{2}+4b^{2}+4c^{2}-8ac}$

${\displaystyle Q=-P}$

${\displaystyle R=b^{2}}$

See also: Solving ellipse at origin.

From ${\displaystyle (1)}$ above, ${\displaystyle S=0.36}$ or ${\displaystyle 0.64.}$

${\displaystyle \sin \theta ={\sqrt {S}}=\pm 0.6}$ or ${\displaystyle \pm 0.8.}$

# python code. values_of_cosθ_sinθ = ( (0.6,-0.8), (0.6,0.8), (0.8,-0.6), (0.8,0.6), ) a,b,c,d,e,f = 55,-24,48,0,0,-2496 for ns in values_of_cosθ_sinθ : n,s = ns B = b*n*n + (2*c-2*a)*n*s - b*s*s if (abs(B) < 1e-14) : print ('ns =',n,s) A = a*n*n + b*n*s + c*s*s C = c*n*n - b*n*s + a*s*s D = d*n + e*s E = e*n - d*s F = f print (' ',A,0,C,D,E,F) ns = 0.6 0.8 39.0 0 64.0 0.0 0.0 -2496 ns = 0.8 -0.6 64.0 0 39.0 0.0 0.0 -2496 When ${\displaystyle \cos \theta ,\sin \theta =0.6,0.8,\ f'(x,y)=39x^{2}+64y^{2}-2496=0.}$ When ${\displaystyle \cos \theta ,\sin \theta =0.8,-0.6,\ f'(x,y)=64x^{2}+39y^{2}-2496=0.}$ In this context, ${\displaystyle f'(x,y)}$ is not the derivative of ${\displaystyle f(x,y).}$ Expression ${\displaystyle f'(x,y)}$ means ${\displaystyle f(x,y)}$ relative to primed system ${\displaystyle OX',OY'}$ (red system.)