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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:


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


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

Line FG intersects with line RS when


Solving this gives us


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


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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)


Solving for x gives


Finding y by substituting back into




So S is the point


The distance RS, using the distance formula,





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

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


Example 1

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





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



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


`5y = 6x + 10`

`6x - 5y + 10 = 0`

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





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

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