Mathematics Distance of a Point From a Line

Please Wait... While Loading Full Video

Distance of a Point From a Line

Length of the Perpendicular :

The length of the perpendicular from a point `(x_1, y_1)` to a line `ax + by+ c = 0` is

`(|ax_1+by_1+c|)/(sqrt(a^2+b^2))`

Proof :

The line `ax + by + c = 0` meets `x`-axis at `A(-c/a,0)` and `y`-axis at `B(0,-c/b)`

Let `P(x_1, y_1)` be the point. Draw `PN bot AB`.

Now, area of `Delta PAB`

`1/2|x_1(0+c/b)-c/a(-c/b-y_1)+0(y_1-0)|`

`1/2|(cx_1)/b+(cy_1)/a+c^2/(ab)|=|(ax_1+by_1+c)c/(2ab)|`.......................(i)

Also, area of `Delta PAB`

`=1/2 AB xx PN`

`=1/2sqrt(c^2/a^2+c^2/b^2) xx PN`

`= c/(2ab)sqrt(a^2+b^2) xx PN`....................(ii)

From equation (i) and (ii), we get

`|(ax_1+by_1+c)c/(2ab)|=c/(2ab) sqrt(a^2+b^2)xx PN`

`=> PN=(|ax_1+by_1+c|)/(sqrt(a^2+b^2))`

e.g. Find all points (4, y) that are 10 units from the point (–2, –1).

I'll plug the two points and the distance into the Distance Formula:

`10 = sqrt((-2 - 4)^2 + (-1 - y)^2) = sqrt{(-6)^2 + (1 + 2y + y^2)}`

`= sqrt{36 + (1 + 2y + y^2)}`

`= sqrt{y^2 + 2y + 37}`

Now I'll square both sides, so I can get to the variable:

`(10)^2 = sqrt{y^2 - 2y + 37}`

`100 = y^2 + 2y + 37100`

`0 = y^2 + 2y - 630=y`

`0 = (y + 9)(y - 7)`

This means y = –9 or y = 7, so:

the two points are (4, –9) and (4, 7).

The length of the perpendicular from a point `(x_1, y_1)` to a line `ax + by+ c = 0` is

`(|ax_1+by_1+c|)/(sqrt(a^2+b^2))`

Proof :

The line `ax + by + c = 0` meets `x`-axis at `A(-c/a,0)` and `y`-axis at `B(0,-c/b)`

Let `P(x_1, y_1)` be the point. Draw `PN bot AB`.

Now, area of `Delta PAB`

`1/2|x_1(0+c/b)-c/a(-c/b-y_1)+0(y_1-0)|`

`1/2|(cx_1)/b+(cy_1)/a+c^2/(ab)|=|(ax_1+by_1+c)c/(2ab)|`.......................(i)

Also, area of `Delta PAB`

`=1/2 AB xx PN`

`=1/2sqrt(c^2/a^2+c^2/b^2) xx PN`

`= c/(2ab)sqrt(a^2+b^2) xx PN`....................(ii)

From equation (i) and (ii), we get

`|(ax_1+by_1+c)c/(2ab)|=c/(2ab) sqrt(a^2+b^2)xx PN`

`=> PN=(|ax_1+by_1+c|)/(sqrt(a^2+b^2))`

e.g. Find all points (4, y) that are 10 units from the point (–2, –1).

I'll plug the two points and the distance into the Distance Formula:

`10 = sqrt((-2 - 4)^2 + (-1 - y)^2) = sqrt{(-6)^2 + (1 + 2y + y^2)}`

`= sqrt{36 + (1 + 2y + y^2)}`

`= sqrt{y^2 + 2y + 37}`

Now I'll square both sides, so I can get to the variable:

`(10)^2 = sqrt{y^2 - 2y + 37}`

`100 = y^2 + 2y + 37100`

`0 = y^2 + 2y - 630=y`

`0 = (y + 9)(y - 7)`

This means y = –9 or y = 7, so:

the two points are (4, –9) and (4, 7).

THE FOOT OF PERPENDICULAR DRAWN FROM THE POINT `(x_1 y_1)` TO THE LINE `ax + by + c = 0`

`text ( Method I :)`

Let `P = (x_1, y_1)` and let `M` be the foot of perpendicular drawn
from `P` on `ax + by + c = 0` .

In order to find the co-ordinates of `M` , find the equation of the line `PM` which is

perpendicular to `RS` and passes through

`P(x_1, y_1 )` , i.e., `bx- ay = bx_1 - ay_1`

or `b(x - x_1 ) - a(y - y_1) = 0` and solving it with

`ax + by + c = 0` , then we get co-ordinates of `M` .

`text ( Method II :)`

Let the co-ordinates of M are

`(x_2, y_ 2 )` then `M (x_2, y_ 2 )` lies on `ax+ by+ c = 0`

`=> ax_ 2 + by_ 2 + c = 0` ................................. `(1)`

and `PM` `bot` `RS`

then `( `slope of `PM ) (` slope of `RS) = - 1`

`((y_2-y_1)/(x_2-x_1)) xx (-a/b) = -1`

`bx_2 -ay_2 = bx_1 -ay_1.`....................`(2)`

Solving Eq. (1) and (2), we get `(x_2,y_2)`

`text ( Method I :)`

Let `P = (x_1, y_1)` and let `M` be the foot of perpendicular drawn
from `P` on `ax + by + c = 0` .

In order to find the co-ordinates of `M` , find the equation of the line `PM` which is

perpendicular to `RS` and passes through

`P(x_1, y_1 )` , i.e., `bx- ay = bx_1 - ay_1`

or `b(x - x_1 ) - a(y - y_1) = 0` and solving it with

`ax + by + c = 0` , then we get co-ordinates of `M` .

`text ( Method II :)`

Let the co-ordinates of M are

`(x_2, y_ 2 )` then `M (x_2, y_ 2 )` lies on `ax+ by+ c = 0`

`=> ax_ 2 + by_ 2 + c = 0` ................................. `(1)`

and `PM` `bot` `RS`

then `( `slope of `PM ) (` slope of `RS) = - 1`

`((y_2-y_1)/(x_2-x_1)) xx (-a/b) = -1`

`bx_2 -ay_2 = bx_1 -ay_1.`....................`(2)`

Solving Eq. (1) and (2), we get `(x_2,y_2)`

SiteLock