HSC Science (General) 11thMaharashtra State Board
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Distance of a Point from a Line

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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)`.

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