Perpendicular Distance from a Point to a Line

(BTW - we don't really need to say 'perpendicular' because the distance from a point to a line always means the shortest distance.)

This is a great problem because it uses all these things that we have learned so far:

The distance from a point (m, n) to the line Ax + By + C = 0 is given by:

`d=(|Am+Bn+C|)/(sqrt(A^2+B^2`

There are some examples using this formula following the proof.

Continues below

Proof of the Perpendicular Distance Formula

Let's start with the line Ax + By + C = 0 and label it DE. It has slope `-A/B`.

straight line graph

We have a point P with coordinates (m, n). We wish to find the perpendicular distance from the point P to the line (that is, distance `PQ`).

perpendicular to straight line

We now do a trick to make things easier for ourselves (the algebra is really horrible otherwise). We construct a line parallel to DE through (m, n). This line will also have slope `-A/B`, since it is parallel to DE. We will call this line FG.

perpendicular and parallel constructions

Now we construct another line parallel to PQ passing through the origin.

This line will have slope `B/A`, because it is perpendicular to DE.

Let's call it line RS. We extend it to the origin `(0, 0)`.

We will find the distance RS, which I hope you agree is equal to the distance PQ that we wanted at the start.

perpendicular through origin

Since FG passes through (m, n) and has slope `-A/B`, its equation is `y-n=-A/B(x-m)` or `y=(-Ax+Am+Bn)/B`.

Line RS has equation `y=B/Ax.`

Line FG intersects with line RS when

`B/Ax=(-Ax+Am+Bn)/B`

Solving this gives us

`x=(A(Am+Bn))/(A^2+B^2`

So after substituting this back into `y=B/Ax,` we find that point R is

`((A(Am+Bn))/(A^2+B^2),(B(Am+Bn))/(A^2+B^2))`

Phone users

mobile phone

NOTE: If you're on a phone, you can scroll any wide equations on this page to the right or left to see the whole expression.

Point S is the intersection of the lines `y=B/Ax` and Ax + By + C = 0, which can be written `y=-(Ax+C)/B`.

This occurs when (that is, we are solving them simultaneously)

`-(Ax+C)/B=B/Ax`

Solving for x gives

`x=(-AC)/(A^2+B^2)`

Finding y by substituting back into

`y=B/Ax`

gives

`y=B/A((-AC)/(A^2+B^2))=(-BC)/(A^2+B^2`

So S is the point

`((-AC)/(A^2+B^2),(-BC)/(A^2+B^2))`

The distance RS, using the distance formula,

`d=sqrt((x_2-x_1)^2+(y_2-y_1)^2`

is

`d=sqrt(((-AC)/(A^2+B^2)-(A(Am+Bn))/(A^2+B^2))^2+((-BC)/(A^2+B^2)-(B(Am+Bn))/(A^2+B^2))^2)`

`=sqrt(({-A(Am+Bn+C)}^2+{-B(Am+Bn+C)}^2)/(A^2+B^2)^2)`

`=sqrt( ((A^2+B^2)(Am+Bn+C)^2)/(A^2+B^2)^2)`

`=sqrt( ((Am+Bn+C)^2)/(A^2+B^2))`

`=(|Am+Bn+C|)/(sqrt(A^2+B^2))`

The absolute value sign is necessary since distance must be a positive value, and certain combinations of A, m , B, n and C can produce a negative number in the numerator.

So the distance from the point (m, n) to the line Ax + By + C = 0 is given by:

`d=(|Am+Bn+C|)/(sqrt(A^2+B^2`

Example 1

Find the perpendicular distance from the point (5, 6) to the line −2x + 3y + 4 = 0, using the formula we just found.

Answer

`d=(|Am+Bn+C|)/(sqrt(A^2+B^2`

`=(|(-2)(5)+(3)(6)+4|)/(sqrt(4+9)`

`=3.328`

Here is the graph of the situation. We can see that our answer of just over 3 units is reasonable.

Graph line and perpendicular

So the required distance is `3.3` units, correct to 1 decimal place.

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Example 2

Find the distance from the point `(-3, 7)` to the line

`y=6/5x+2`

Answer

We first need to express the given line in standard form.

`y=6/5x+2`

`5y = 6x + 10`

`6x - 5y + 10 = 0`

Using the formula for the distance from a point to a line, we have:

`d=(|Am+Bn+C|)/(sqrt(A^2+B^2`

`=(|(6)(-3)+(-5)(7)+10|)/sqrt(36+25)`

`=|-5.506|`

`=5.506`

So the required distance is `5.506` units, correct to 3 decimal places.

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