Transformation of Coordinates

Translation of Coordinate Axes

Graph in 2 dimensions illustrating Translation of Coordinate Axes. Points $P,P'$ are same point. Point $P$ is defined relative to origin $O,$ black arrows. Point $P'$ is defined relative to origin $\Omega ,$ red arrows. Any point in the 2 dimensional Cartesian plane is usually defined as $(x,y)$ meaning that the point is $x$ units horizontally from origin $O=(0,0)$ and $y$ units vertically from origin. It is always possible, and sometimes desirable, to give the point a new name or definition that reflects its position relative to another point in the 2 dimensional plane, for example $\Omega ,$ the position of which is defined as $(h,k)$ relative to origin $O.$ In the diagram point $P$ and point $P'$ are the same point. It's just that the point has the name or definition $P$ when referenced to origin $O$ (black arrows), and $P'$ when referenced to origin $\Omega$ (red arrows). By inspection: $h+x'=x$ or $x'=x-h$ $k+y'=y$ or $y'=y-k.$ Actual values: $\Omega =(h,k)=(6,4)$ $P=(x,y)=(9,10.5)$ $P'=(x',y')$$=(x-h,y-k)$$=(9-6,10.5-4)$$=(3,6.5).$ Point $P$ is defined as $(9,10.5)$ relative to origin $O,$ and point $P',$ the same point, is defined as $(3,6.5)$ relative to origin $\Omega .$ Examples Linear function Graph in 2 dimensions illustrating Translation of Coordinate Axes applied to linear function. line1 can have 2 equations: * $2x+3y-18=0$ relative to origin $O.$ * $2x'+3y'+6=0$ relative to origin $\Omega .$ In the diagram line1 has equation $2x+3y-18=0.$ What is equation of line1 relative to origin $\Omega =(6,4)?$ Using $x=x'+h,\ y=y'+k:$ $2(x'+6)+3(y'+4)-18=0$ $2x'+12+3y'+12-18=0$ $2x'+3y'+6=0$ When $x,y$ become $x',y',$ this means that equation $2x'+3y'+6=0$ is relative to origin $\Omega .$ line1 can have equation $2x+3y-18=0$ or equation $2x'+3y'+6=0.$ Equation $2x'+3y'+6=0$ relative to origin $\Omega$ is same as equation $2x+3y+6=0$ relative to origin $O.$ Note that in both cases: $X$ intercept relative to origin is $-3.$ $Y$ intercept relative to origin is $-2.$

Rotation of Coordinate Axes

Graph in 2 dimensions illustrating Rotation of Coordinate Axes. Point $P$ has 2 names or definitions: $(x,y)=(15,30)$ when defined relative to black arrows. $(x',y')=(30,15)$ when defined relative to red arrows. If $X$ axis, line $OX$, is rotated through angle $\theta$ so that $X$ axis of new system becomes $OX'$ then: Counter-clockwise rotation occurs when $\theta$ is positive. $Y$ axis of new system becomes $OY'.$ angle between $OX$ and $OX'$ is $\theta .$ angle between $OY$ and $OY'$ is $\theta .$ See diagram. By inspection: Angle $TPQ=\theta$ $QR=ST=x'\sin \theta$ $QT=RS=y'\sin \theta$ $PQ=y'\cos \theta$ $OS=x'\cos \theta =OR+RS=x+y'\sin \theta \ \dots \ (1)$ $PR=y=PQ+QR=y'\cos \theta +x'\sin \theta \ \dots \ (2)$ From $(1),\ x=x'\cos \theta -y'\sin \theta \ \dots \ (3)$ From $(2),\ y=x'\sin \theta +y'\cos \theta \ \dots \ (4)$ From $(3)$ and $(4):$ $x'=x\cos \theta +y\sin \theta \ \dots \ (5)$ $y'=y\cos \theta -x\sin \theta \ \dots \ (6)$ Actual values: Converting from $(x,y)$ to $(x',y')$ # Python code. >>> c,s = cosθ,sinθ = 4/5,3/5 ; c,s (0.8, 0.6) >>> x,y = 15,30 >>> x1 = x*c + y*s >>> y1 = y*c - x*s >>> x1,y1 (30.0, 15.0) Process reversed: # Python code. >>> c,s = cosθ,sinθ = 4/5,-3/5 ; c,s (0.8, -0.6) >>> x,y = 30,15 >>> x1 = x*c + y*s >>> y1 = y*c - x*s >>> x1,y1 (15.0, 30.0) Examples Linear function Graph in 2 dimensions illustrating Rotation of Coordinate Axes applied to linear function. line1 has 2 equations: * $6x+17y=300$ relative to $OX,OY$ (black system). * $3x'+2y'=60$ relative to $OX',OY'$ (red system). Let a line have equation: $ax+by+d=0.$ Let $c=\cos \theta ;\ s=\sin \theta .$ After rotation, equation of line relative to $OX',OY'$ (red arrows) is: $a(x'c-y's)+b(x's+y'c)+d$ $=ax'c-ay's+bx's+by'c+d$ $=ax'c+bx's+by'c-ay's+d$ $=x'ac+x'bs+y'bc-y'as+d$ $=x'(ac+bs)+y'(bc-as)+d$ $=Ax'+By'+d=0\ \dots \ (1)$ where: $A=a\cos \theta +b\sin \theta ;$$\ B=b\cos \theta -a\sin \theta .$ In the diagram line1 has equation $6x+17y-300=0,$ and $\cos \theta ={\frac {4}{5}}.$ What is equation of line1 relative to $OX',OY'$ (red system)? # python code. >>> a,b,d = 6,17,-300 >>> c,s = cosθ,sinθ = 4/5,3/5 ; c,s (0.8, 0.6) >>> A = a*c + b*s ; A 15.0 >>> B = b*c - a*s ; B 10.0 Equation of line1 relative to $OX',OY':\ 15x'+10y'-300=0$ or $3x'+2y'=60.$ $OP=OP';\ OQ=OQ'.$ Triangles $OPQ,OP'Q'$ are congruent. Line1: $3x'+2y'=60$ has same position in red system as line2: $3x+2y=60$ in black system. Quartic function Graph in 2 dimensions illustrating Rotation of Coordinate Axes applied to quartic function. Position of black curve in red system is same as that of red curve in black system. Let quartic function be defined as $y=f(x)={\frac {1.5625x^{4}-12.125x^{3}-14.75x^{2}+136.5x+114}{48}}.$ In diagram, $X$ and $Y$ axes are rotated through angle $\theta$ to produce new system of coordinates $OX',OY'$ (red system.) $\cos \theta ={\frac {4}{5}}.$ What is equation of $f(x)$ relative to red system? $g(x,y)=ax^{4}+bx^{3}+cx^{2}+dx+e-y$ After substituting appropriate values for $a,b,c,d,e,x,y$ code supplied to application grapher is: (0.0325520833333333) ((x(0.8) - y(0.6))^4) + (-0.252604166666667) ((x(0.8) - y(0.6))^3) + (-0.307291666666667) ((x(0.8) - y(0.6))^2) + (2.84375) ((x(0.8) - y(0.6)) ) + (2.375) - (x(0.6) + y(0.8)) = 0 Red curve in diagram has equation $g(x,y)=0.$ Ellipse Graph in 2 dimensions illustrating Rotation of Coordinate Axes applied to ellipse. Equation of ellipse in diagram is: $f(x,y)=55x^{2}-24xy+48y^{2}-2496=0.$ What is equation of $f(x,y)$ relative to minor and major axes (red system)? The most general equation of second degree in $x,y$ has form: $ax^{2}+bxy+cy^{2}+dx+ey+f=0.$ Rotation of coordinate axes when applied to the general equation produces the primed equation: $A(x')^{2}+Bx'y'+C(y')^{2}+Dx'+Ey'+F=0$ where: $A=an^{2}+bns+cs^{2}$ $B=bn^{2}-2ans+2cns-bs^{2}$$=bn^{2}+(2c-2a)ns-bs^{2}$ $C=cn^{2}-bns+as^{2}$ $D=dn+es$ $E=en-ds$ $F=f$ $n=\cos \theta$ $s=\sin \theta$ Choose values of $\sin \theta ,\cos \theta$ that make coefficient $B=0.$ From coefficient $B$ above: $bn^{2}+(2c-2a)ns-bs^{2}=0$ $bn^{2}-bs^{2}=-(2c-2a)ns$ Square both sides, substitute $(1-ss)$ for $nn,$ expand, gather like terms and result is: $PS^{2}+QS+R=0\ \dots \ (1)$ where: $S=\sin ^{2}\theta$ $P=4a^{2}+4b^{2}+4c^{2}-8ac$ $Q=-P$ $R=b^{2}$ See also: Solving ellipse at origin. From $(1)$ above, $S=0.36$ or $0.64.$ $\sin \theta ={\sqrt {S}}=\pm 0.6$ or $\pm 0.8.$ Graph in 2 dimensions showing 2 ellipses, f'(x,y), where f'(x,y) is f(x,y) relative to its minor and major axes. # 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 $\cos \theta ,\sin \theta =0.6,0.8,\ f'(x,y)=39x^{2}+64y^{2}-2496=0.$ When $\cos \theta ,\sin \theta =0.8,-0.6,\ f'(x,y)=64x^{2}+39y^{2}-2496=0.$ In this context, $f'(x,y)$ is not the derivative of $f(x,y).$ Expression $f'(x,y)$ means $f(x,y)$ relative to primed system $OX',OY'$ (red system.)