In earlier classes, we have studied general equation of first degree in two variables, `Ax + By + C = 0,` where `A, B` and `C` are real constants such that `A` and `B` are not zero simultaneously.

Graph of the equation `color(red)(Ax + By + C = 0" is always a straight line.")`

Therefore, any equation of the form `color(blue)(Ax + By + C = 0),` where `A ` and `B` are not zero simultaneously is called `color(blue)("general linear equation or general equation of a line.")`

`\color{green} ✍️` if `color(red)(B ne 0)` , them `Ax +By+ C = 0` can be written as

`color(blue)(y = - A/B x - C/B)` or `\ \ \ \ \ y = mx+c` ......................(1)

where `m = - A/B` and `c = - C/B`

We know that Equation (1) is the slope-intercept form of the equation of a line

whose slope is `-A/B` , and y-intercept is `-C/B`

`\color{green} ✍️` if `color(red)(B = 0)`, then `color(blue)(x = -C/A)` which is a vertical line whose slope is undefined and x-intercept is `- C/A`

`\color{green} ✍️` If `color(red)(C ≠ 0,)` then `Ax + By + C = 0` can be written as

`color(blue)(x/(-C/A) +y/(-C/B) = 1)` or `x/a+y/b = 1`

where `a = -C/A` and `b = -C/B`

We know that equation (1) is intercept form of the equation of a line whose

x-intercept is `-C/A` and y-intercept is `- C/B`

`\color{green} ✍️` If `color(red)(C = 0,)` then `Ax + By + C = 0` can be written as `color(blue)(Ax + By = 0,)` which is a line passing through the origin and, therefore, has zero intercepts on the axes.

`\color{green} ✍️` Let `color(blue)(x cos ω + y sin ω = p)` be the normal form of the line represented by the equation `color(blue)(Ax + By + C = 0)` or `Ax + By = – C.`

Thus, both the equations are same and therefore `A/(cos omega) = B/(sin omega) = - C/p`

which gives `cosomega = - (Ap)/C` and `sin omega = - (Bp)/C`

Now `sin^2 omega +cos^2 omega = (-(Ap)/C)^2+(-(Bp)/C)^2 = 1`

or `p^2 = C^2/(A^2+B^2) ` or `p = pm C/sqrt(A^2+B^2)`

Therefore `cosomega = pm A/sqrt(A^2+B^2) ` and `sinomega = pm B/sqrt(A^2+B^2)`

`\color{green} ✍️` Thus, the normal form of the equation `Ax + By + C = 0` is `x cos ω + y sin ω = p,`

where `color(red)(cosomega = pm A/sqrt(A^2+B^2) , \ \ sinomega = pm B/sqrt(A^2+B^2)) ` and `color(red)(p = pm C/sqrt(A^2+B^2))`

Proper choice of signs is made so that `color(blue)(p" should be positive.")`

`\color{green} ✍️` The distance of a point from a line is the length of the perpendicular drawn from the point to the line.

Let `color(blue)(L : Ax + By + C = 0` be a line, whose distance from the point `P (x_1, y_1)` is d. Draw a perpendicular PM from the point `P` to the line `L` (Fig10.19).

If the line meets the x-and y-axes at the points Q and R, respectively. Then, coordinates of the points are `Q (-C/A , 0)` and `R (0 , - C/R)`.

Thus, the area of the triangle PQR is given by area `color(blue)((Delta PQR) = 1/2 P M . QR)`

which gives `color(red)(P M = (2 text{area} (Delta PQR))/(QR))` .............(1)

Also, area `(Delta PQR ) = 1/2 | x_1 (0+C/B) +(-C/A) (-C/B - y_1) +0 (y_1-0) |`

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1/2 | x_1 C/B +y_1 C/A+C^2/(AB) |`

or `color(blue)(2 area \ \Delta PQR) = | C/(AB) | . | A_(x_1) +B_(y_1) +C |`

`color(blue)(QR) = sqrt((0+C/A)^2+(C/B -0)^2) = | C/(AB)| sqrt(A^2+B^2)`

Substituting the values of area (ΔPQR) and QR in (1), we get

`color(red)(P M = (| A_(x_1) +B_(y_1) + C|)/sqrt(A^2+B^2))`

or `color(blue)(d = (| A_(x_1) + B_(y_1) +C|)/sqrt(A^2+B^2))`

`\color{green} ✍️` Thus, the perpendicular distance (d) of a line `Ax + By+ C = 0` from a point `(x_1, y_1)` is given by

`color(red)(d = (| A_(x_1) +B_(y_1) + C|)/sqrt(A^2+B^2))`

`\color{green} ✍️` We know that slopes of two parallel lines are equal. Therefore, two parallel lines can be taken in the form

`color(blue)(y = mx+c_1)` ..............(1)

`color(blue)(y = mx+c_2)` ..............(2)

Line `(1)` will intersect `x`-axis at the point `color(blue)(A (-c_1/m , 0))` as shown in Fig `10.20`

Distance between two lines is equal to the length of the perpendicular from point `A ` to line `(2).`

Therefore, distance between the lines `(1)` and `(2)` is

`(| (-m) (-c_1/m) +(-c_2)|)/sqrt(1+m^2) ` or `color(blue)(d = (| c_1-c_2|)/sqrt(1+m^2))`

`\color{green} ✍️` Thus, the distance `d` between two parallel lines `color(red)(y = mx + c_1)` and `color(red)(y = mx + c_2)` is given by

`color(red)(d = ( | c_1-c_2|)/sqrt(1+m^2))`

`\color{green} ✍️` If lines are given in general form, i.e., `color(blue)(Ax + By + C_1 = 0)` and `color(blue)(Ax + By + C_2 = 0,)`

then above formula will take the form `color(blue)(d = (| C_1-C_2|)/sqrt(A^2+B^2))`