#### description

- Introduction of Distance of a Point from a Line
- Distance between two parallel lines

#### notes

Let 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 in the Fig.

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/B)`. Thus the area of the triangle PQR is given by

area

`(triangle PQR)=1/2PM .QR`, which gives

PM=`(2 area (triangle PQR))/(QR)` ...(1)

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

=`1/2|x_1C/B+y_1C/A +C_2/(AB)|`

or `2area(trianglePQR) = |C/(AB)|.|Ax_1 + By_1+C| `

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

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

PM = `|Ax_1+By_1+C|/sqrt(A_2+B_2)`

or

d=`|Ax_1+By_1+C|/sqrt(A_2+B_2)`.

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

d=`|Ax_1+By_1+C|/sqrt(A_2+B_2)`**Distance between two parallel lines :**

The slopes of two parallel lines are equal. Therefore, two parallel lines can be taken in the form

y = mx+c_1 ...(1)and

y = mx +c_2 ...(2)

Line (1) will intersect x-axis at the point

A`(-c_1/m,0)` as shown in fig.

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 d = |c_1-c_2|/sqrt(1+m_2)`

Thus, the distance d between two parallel lines y= mx+ `c_1` and ` y= mx c_2` = + is given by

d = `|c_1-c_2|sqrt(1+m_2)`.

If lines are given in general form, i.e., `Ax + By + C_1` = 0 and `Ax + By + C_2 `= 0,

then above formula will take the form d =` |c_1-C_2|/sqrt(A_2+B_2)`.