Q&A #1895

Distance from a point to a line.

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From: Marielouise (for Teacher2Teacher Service)
Date: Aug 26, 1999 at 17:20:09
Subject: Re: Distance from a point to a line.

Hi Carolyn,

I worked on the general form of this problem myself and got myself in all
kinds of algebra difficulties.  I found a problem in the fifth edition of
Precalculus by David Cohen (p98 #106).  I will give you an outline of the
solution.  The difficulty is not being able to draw a picture and refer to

Consider a line:  y = mx + b.   Construct a right triangle so that the
hypotenuse PR lies on this line, the right angle is at Q, the vertical
distance PQ = m and the horizontal distance QR = 1

Choose a point A = (x0, y0) (read this as x sub zero, y sub zero) From A drop
a perpendicular to the line y = mx + b.  Let D be the point of intersection
with y = mx + b.  Therefore, distance = AD. From A drop a perpendicular to
x-axis with intersection point C.  The distance AC = y0.   The point of
intersection of AC and y = mx + b is the point B.

Since PQ and AC are both vertical lines they are parallel.  Therefore, angle
ABD and angle RPQ are the same measure.  Angle BDA and angle PQR are both
right angles.   This makes triangle ABD similar to triangle RPQ.

AD / RQ = AB / RP   or

(distance from A to y = mx + b) / 1 = AB / sqr root (1 + m^2).

Since AB = AC - BC, then  AB = y0 - (mx0 + b)

Substituting this into the above proportion:

dist from a pt to a line = (y0 - (mx0 + b)) /  sqr root (1 + m^2). Now this
does not look like the equation that you wish so you have to substitute the
standard form for the function form of the line.

Assume that you have the equation: Ax + By + C = 0 where m = (-A / B) and b =
(C/B).  Substituting in

dist from a pt to a line = (y0 - (mx0 + b)) /  sqr root (1 + m^2).

dist from a pt to a line = (y0 - ((-A/B)x0 + (C/B))) /  sqr root (1 + (-A/B)^

With squaring, clearing of fractions etc., you arrive at the formula:

dist from a pt to a line = (Ax0 + By0 +C) /  sqr root (A^2 + B^2).

This will take some time for younger students to do but I do think that it is
"do-able."   I think that the beauty of this approach is using m and b and
substituting at the end.

Thank David Cohen.

 -Marielouise, for the Teacher2Teacher service

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