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# Perpendicular Distance from a Point to a Line

Example using perpendicular distance formula

(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.

## 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.

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

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.

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.

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

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.

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.

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

### Example 2

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

y=6/5x+2

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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