Perpendicular Distance from a Point to a Line
Later, on this page...
(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:
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- distance formula
- slope of parallel and perpendicular lines
- rectangular coordinates
- different forms of the straight line
- solving simultaneous equations
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.
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`.
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`).
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
Solving this gives us
So after substituting this back into `y=B/Ax,` we find that point R is
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,
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:
Find the perpendicular distance from the point (5, 6) to the line −2x + 3y + 4 = 0, using the formula we just found.
Find the distance from the point `(-3, 7)` to the line