The equation of a line passing through two points `(x_1, y_1)` and `(x_2, y_2)` is

`y-y_1=((y_2-y_1)/(x_2-x_1))(x-x_1)`

`text(Proof:)`

Let `m` be the slope of the line passing through `(x_1 , y_1)` and `(x_2, y_2)` then

`m=(y_2-y_1)/(x_2-x_1)`

So, the equation of the line is

`y- y_1 = m(x - x_1)` (Using point - slope form)

Substituting the value of `m`, we obtain

`y-y_1=(y_2-y_1)/(x_2-x_1)(x-x_1)`

This is the required equation of the line in two-point form.

The equation of a line with slope `m` that makes an intercept `c` on `y`-axis is

`y = mx + c`

`text(Proof:)`

Since `y` intercept `= c`

Hence it passes through `(0, c)`.

Equation of line with slope `m` and passing through `(0, c)` is given by

`y - c = m(x - 0)`

`y = mx + c`

The equation of a line which cut-off intercepts `a` and `b`, respectively from the `x` andy-axes is

`x/a+y/b`

`text(Proof:)`

Line cut-off intercepts `a` and `b` from the `x` and `y`-axes respectively,

Equation of line passes through the points

`(a, 0)` and `(0, b)` is `y-0 = -b/a(x-a)`

`=> bx+ay=ab`

`=> x/a+y/b=1`

This is the equation of the line in the intercept form.

The equation of the straight line upon which the length of the perpendicular from the origin is `p` and this perpendicular makes an angle alpha with positive direction of `x` axis is

`x cos alpha + y sin alpha = p`

`text(Proof:)`

Let the line `AB` be such that the length of the perpendicular `OQ` from the origin `O` to the line be `p`.

and `angle XOQ = alpha`.

From the diagram, using the intercept form, we get

Equation of line `AB` is

`x/(p sec alpha)+y/(p cosec alpha)=1`

or `x cos alpha+y sin alpha=p`

Two mutually perpendicular lines meeting at `'O'` (origin) are called axes. Horizontal line `X' OX` is known as `x`-axis and vertical line `Y'OY` is called `y`-axis. These two perpendicular lines divide the plane into four quadrants, viz., as follows their names are given in anti-clockwise sense.

`XOY =` First quadrant, `X'OY =` Second quadrant
`X'OY' =` Third quadrant, `Y'OX =` fourth quadrant

Co-ordinates of a point are given by ordered pair `(x, y)` whose first entry `(x)` denotes the `x`-coordinate or abscissa of the point and second entry `(y)` denotes they-coordinate or ordinate of the point.

For ` x`-coordinate (Abscissa) of the point, `|x |` is the perpendicular distance of the point from `y`-axis.

For `y`-coordinate (ordinate) of the point, `|y|` is the perpendicular distance of the point from `x`-axis.

The equation of straight line passing through a given point `A ( x_1,y_1)` and making an angle `theta` from positive direction of `x`-axis in anticlockwise sence is -

`(x-x_1)/(cos theta)=(y-y_1)/(sin theta)= pm r`

where, `r` is the distance of any point on the line from the given point `A(x_1, y_1)`.

Explanation:

Let `P(x, y)` be taken on the line above the given point `(x_1 , y_1)`

then from the `Delta PAN`.

`x - x_1=r cos theta`

`y - y_1 = r sin theta`

`(x-x_1)/(cos theta)=(y-y_1)/(sin theta)=r`..................(1)

Again, if point is taken on the line below the given point `A(x_1, y_1)` then from the `Delta APN'`

`x_1 - x = rcos theta`

`y_1 - y = r sin theta`

`:. (x-x_1)/(cos theta)=(y-y_1)/(sin theta)=-r`................(2)

Combining `(1) ` & `(2)`

` (x-x_1)/(cos theta)=(y-y_1)/(sin theta)=pm r`

Here, `x = x_1 pm r cos theta, y = y_1 pm r sin theta` are the co-ordinates of the points situated on the line at a distance r from the given point `A(x_1, y_1)` .