`\color{green} ✍️ ` If a horizontal line `L` is at a distance a from the `x`-axis then ordinate of every point lying on the line is either `a` or `– a` [Fig 10.11 (a)].

`\color{green} ✍️ ` Therefore, equation of the line `L` is either `y = a` or `y = – a.` Choice of sign will depend upon the position of the line according as the line is above or below the `y`-axis.

`\color{green} ✍️ ` Similarly, the equation of a vertical line at a distance `b` from the `x`-axis is either `x = b` or `x = – b` [Fig 10.11(b)].

Suppose that `P_0 (x_0, y_0)` is a fixed point on a non-vertical line `L`, whose slope is `m.`

Let `P (x, y)` be an arbitrary point on `L` (Fig 10.13).

Then, by the definition, the slope of `L` is given by

`color(red)(m=( y- y_0)/(x-x_0))` i.e., `color(blue)(( y- y_0) =m(x-x_0))` ..........................(1)

Since the point `P_0 (x_0 , y_0)` along with all points `(x, y)` on `L` satisfies `(1)` and no other point in the plane satisfies` (1).`

Equation `(1)` is indeed the equation for the given line `L.`

Thus, the point `(x, y)` lies on the line with slope `m` through the fixed point `(x_0, y_0),` if and only if, its coordinates satisfy the equation

`color(red)(y – y_0 = m (x – x_0))`

Let the line `L` passes through two given points `P_1 (x_1, y_1)` and `P_2 (x_2, y_2).`

Let `P (x, y)` be a general point on `L` (Fig 10.14).

The three points `color(blue)(P_1, P_2 \ \ "and" \ \P" are collinear")`, therefore, we have

`color(red)("slope of " \ \ P_1P =" slope of" \ \ P_1P_2)`

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

Thus, equation of the line passing through the points `(x_1, y_1)` and `(x_2, y_2)` is given by

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

Sometimes a line is known to us with its slope and an intercept on one of the axes. We will now find equations of such lines.

`color(red)("Case I")` Suppose a line `L` with slope `m` cuts the `y`-axis at a distance `c` from the origin (Fig10.15).

The distance `c` is called the `y` intercept of the line `L.` Obviously, coordinates of the point where the line meet the `y`-axis are `(0, c).`

Thus, `L` has slope `m` and passes through a fixed point `(0, c).`

Therefore, `color(blue)("by point-slope form")`,

The equation of `L` is `color(blue)(y − c = m( x − 0 ))` or `color(red)(y = mx + c)`

Thus, the point `(x, y)` on the line with slope `m` and `y`-intercept c lies on the line if and only if

`color(red)(y = mx + c)`

`color(red)"Note : "` that the value of `c` will be positive or negative according as the intercept is made
on the positive or negative side of the `y`-axis, respectively.

`color("Case II")` Suppose line L with slope m makes x-intercept `d.`

Then equation of `L` is `color(red)(y = m(x − d ))`

Suppose a line `L` makes `x`-intercept `a` and `y`-intercept `b` on the axes.

Obviously `L` meets `x`-axis at the point `(a, 0)` and `y`-axis at the point `(0, b)` (Fig .10.16).

By two-point form of the equation of the line, we have

`y-0=(b-0)/(0-a) *(x-a)` or `ay=-bx+ab`

`color(blue)(x/a+y/b=1)`

Thus, equation of the line making intercepts `a` and `b` on `x`-and `y`-axis, respectively, is `color(red)(x/a+y/b=1)`

Suppose a non-vertical line is known to us with following data:

`(i)` Length of the perpendicular (normal) from origin to the line.

`(ii)` Angle which normal makes with the positive direction of `x`-axis.

Let `L` be the line, whose perpendicular distance from origin `O` be `OA = p` and the angle between the positive `x`-axis and `OA` be `∠XOA = ω.`

The possible positions of line `L` in the Cartesian plane are shown in the Fig 10.17.

Now, our purpose is to find slope of `L` and a point on it. Draw perpendicular `AM` on the `x`-axis in each case.

In each case, we have `OM = p cos ω` and `MA = p sin ω,` so that the coordinates of the point `A` are `(p cos ω, p sin ω).`
Further, line `L` is perpendicular to `OA.` Therefore

The slope of the line `color(blue)(L =- 1/(Slope of OA)=-1/(tan ω)= -(cos ω)/(sin ω)).`

Thus, the line `L` has slope `(− cosω )/(sinω)` and point `A(pcosω, p sinω)` on it.

Therefore, `color(blue)("by point-slope form")`, the equation of the line `L` is

`color(blue)(x cos ω + y sin ω = p.)`

Hence, the equation of the line having normal distance `p` from the origin and angle `ω` which the normal makes with the positive direction of `x`-axis is given by `color(red)(x cos ω + y sin ω = p)`