# Transformation of Rectangular Axes: Translation, Rotation

### Translation: Change of origin without any change in the direction of the axes

Let the co-ordinates of a point P with respect to a set of rectangular axes OX and OY be (x, y), O being the origin.
Now, let the origin be shifted to the point O’ in the xy plane, and the new coordinate axes O’X’ and O’Y’ having the same direction as the original axes OX and OY respectively.
Also let the coordinates of O’, the new origin, be (h, k) with respect to the old axes, and (x’, y’) be the changed coordinates of P with reference to the new axes of reference.

Let PL and O’N be drawn perpendiculars on OX, also let PL cut O’X’ at M.
Obviously, x = OL, y = LP, x’ = O’M, y’ = MP, h = ON and k = ON = LM.
From the above figure,
x = OL = OM + NL = h + O’M = h + x’
y = LP = LM + MP = k + y’

Hence, the transformation formulae are
x = h + x’, y = k + y’ ………….(i)
x’ = x – h, y’ = y – k …………..(ii)

### Rotation: Transformation from one pair of rectangular axes to another with the same origin

Let the co-ordinate axes OX and OY be rotated about the origin O through and angle θ in the anticlockwise sense to take up position along OX’ and OY’, which are to be taken as new axes of co-ordinates.

Let (x, y) be the co-ordinates of any point P referred to OX and OY as axes of co-ordinates and (x’, y’) be the co-ordinates of the same point P with reference to new set of axes OX’, OY’, where ∠XOX’ = ∠YOY’ = θ

Hence, the transformation formulae are
x = x’ cos θ – y’ sin θ, y = x’ sin θ + y’ cos θ …………(i)
x’ = x cos θ + y sin θ, y’ = y cos θ – x sin θ …………..(ii)

The formulae (i) and (ii) for transformation of co-ordinates can be readily be obtained from the scheme given below

### Translation followed by a Rotation

When the origin is shifted to the point (h, k) without any change in the direction of the axes, the co-ordinates (x, y) of any point P become (x + h, y + k). If now the axes are turned through an angle θ in the anticlockwise sense, keeping the origin fixed in its new position, the co-ordinates of P referred to the new set of axes are obtained by putting (x cos θ – y sin θ) for x and (x sin θ + y cos θ) for y.

Hence the transformation formula for a rigid motion is
x’ = h + x cos θ – y sin θ
y’ = k + x sin θ + y cos θ
The formula will remain the same if rotation is followed by the translation.

### General Orthogonal Transformation

Let OX, OY be the old set of axes with origin at O. Let the origin the shifted to the point O’, and OX’, OY’ be the new set of axes. The equation of the new axes O’X’ and O’Y’ referred to the old axes OX, OY are respectively lx + my + n1 = 0 and mx – ly + n2 = 0

If the co-ordinates of any point P referred to the old system of axes be (x, y) and when referred to the new system of axes by (x’, y’), then from the figure O’N’ = x’, NP = y’,

But O’N and NP are the lengths of perpendiculars from P on O’X’ and O’X’ respectively; so

$x’=\frac{mx-ly+{{n}_{2}}}{\sqrt{{{m}^{2}}+{{l}^{2}}}}……..\left( i \right)$

$y’=\frac{mx+ly+{{n}_{1}}}{\sqrt{{{m}^{2}}+{{l}^{2}}}}……..\left( ii \right)$

These are the required transformation formulae. Solving equations (i) and (ii) for x and y, we get

$x=\frac{mx’+ly’}{\sqrt{{{m}^{2}}+{{l}^{2}}}}-\frac{l{{n}_{1}}+m{{n}_{2}}}{{{m}^{2}}+{{l}^{2}}}$

$y=\frac{-lx’+my’}{\sqrt{{{m}^{2}}+{{l}^{2}}}}-\frac{m{{n}_{1}}-l{{n}_{2}}}{\sqrt{{{m}^{2}}+{{l}^{2}}}}$

### Invariants in Orthogonal Transformation

It has been observed that some relations connecting the coefficients of an expression remain unchanged under an orthogonal transformation. Such relations are called invariants under that orthogonal transformation.

#### Theorem

If, by the orthogonal transformation without change of origin, the expression\left( a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c \right) be changed to \left( A{{X}^{2}}+2HXY+B{{Y}^{2}}+2GX+2FY+C \right), then

$\left( i \right)A+B=a+b\,\,\left( ii \right)AB-{{H}^{2}}=ab-{{h}^{2}}\,\,and\,\,\left( iii \right){{G}^{2}}+{{F}^{2}}={{g}^{2}}+{{f}^{2}}$

 Example 01

Find the equation to the curve 9x2 + 4y2 + 18x – 16y = 11 referred to parallel axes through the point (–1, 2)

Solution:

Here the transformation is due to the shifting of the origin to the point (–1, 2). If (X, Y) be the new coordinates under translation.
x = X – 1, y = Y + 2
So, the transformed equation is

$9{{\left( X-1 \right)}^{2}}+4{{\left( Y+2 \right)}^{2}}+18\left( X-1 \right)-16\left( Y+2 \right)=11$

$\Rightarrow 9{{X}^{2}}-18X+9+4{{Y}^{2}}+16Y+16+18X-18-16Y-32=11$

$\therefore \,\,9{{X}^{2}}+4{{Y}^{2}}=36$

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