Mathematics STRAIGHT LINE

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STRAIGHT LINE

SHIFTING OF ORIGIN :

Let `OX` and `OY` be the original axes and let the new axes, parallel to the original, be `O'X'` and `O'Y'`

Let the coordinates of the new origin `O',` referred to the original axes be `h` and `k,` so that, if `O'L` be
perpendicular to `OX`, we have
`OL = h` and `LO' = k`

Let `P` be any point in the plane of the paper, and let its coordinates, referred to the original axes, be `x` and
`y` and referred to the new axes let them `x'` andy'.

Draw `PN` perpendicular to `OX` to meet `O'X'` in `N'.`
then, `ON = x, NP = y, O'N' = x'` and `N'P = y'`
we therefore, have

`x = ON = OL + O'N' = h + x'`

and `y = NP = LO' + N'P = k + y'`

The origin is therefore, transferred to the point `(h, k)`
when we substitute for the coordinates `x` and `y` the
quantities.

`x' + h andy' + k`

The above article is true whether the axes be oblique or rectangular.

Let `OX` and `OY` be the original axes and let the new axes, parallel to the original, be `O'X'` and `O'Y'`

Let the coordinates of the new origin `O',` referred to the original axes be `h` and `k,` so that, if `O'L` be
perpendicular to `OX`, we have
`OL = h` and `LO' = k`

Let `P` be any point in the plane of the paper, and let its coordinates, referred to the original axes, be `x` and
`y` and referred to the new axes let them `x'` andy'.

Draw `PN` perpendicular to `OX` to meet `O'X'` in `N'.`
then, `ON = x, NP = y, O'N' = x'` and `N'P = y'`
we therefore, have

`x = ON = OL + O'N' = h + x'`

and `y = NP = LO' + N'P = k + y'`

The origin is therefore, transferred to the point `(h, k)`
when we substitute for the coordinates `x` and `y` the
quantities.

`x' + h andy' + k`

The above article is true whether the axes be oblique or rectangular.

ROTATION OF AXES :

ROTATION OF AXES WITHOUT CHANGING THE ORIGIN:

Let `OX` and `OY` be the original system of axes and `OX'` and `OY'` the new system, and let the angle,
`XOX',` through which the axes are turned be called `theta.`
Take any point `P` in the plane of the paper.

Draw `PN` and `PN'` perpendicular to `OX` and `OX',` and also `N'L` and `N'M` perpendicular to `OX` and `PN`.
lfthe coordinates of `P,` referred to the original axes, be `x` and `y,` and referred to the new axes, be `x'` and
`y',` we have

`ON = x, N P = y, ON' = x'` and `N'P = y'`

The angle,

`MPN' = 90" - angle LMN'P = angle LMN'O =angle L XOX' = theta`

We then have

`x = ON = OL - MN' = ON' cos theta - N'P sin theta`

`= x' costheta - y' sin theta`

and `y = NP = LN' + MP = ON' sine+ N'P cos theta`

`= x' sin theta + y' cos theta`

If therefore, in any equation we wish to tum the axes, being rectangular, through an angle `theta` we must
substitute

`x = x' cos theta - y' sin theta` and `y = x' sin theta + y' cos theta`

ROTATION OF AXES WITHOUT CHANGING THE ORIGIN:

Let `OX` and `OY` be the original system of axes and `OX'` and `OY'` the new system, and let the angle,
`XOX',` through which the axes are turned be called `theta.`
Take any point `P` in the plane of the paper.

Draw `PN` and `PN'` perpendicular to `OX` and `OX',` and also `N'L` and `N'M` perpendicular to `OX` and `PN`.
lfthe coordinates of `P,` referred to the original axes, be `x` and `y,` and referred to the new axes, be `x'` and
`y',` we have

`ON = x, N P = y, ON' = x'` and `N'P = y'`

The angle,

`MPN' = 90" - angle LMN'P = angle LMN'O =angle L XOX' = theta`

We then have

`x = ON = OL - MN' = ON' cos theta - N'P sin theta`

`= x' costheta - y' sin theta`

and `y = NP = LN' + MP = ON' sine+ N'P cos theta`

`= x' sin theta + y' cos theta`

If therefore, in any equation we wish to tum the axes, being rectangular, through an angle `theta` we must
substitute

`x = x' cos theta - y' sin theta` and `y = x' sin theta + y' cos theta`

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