Teacher2Teacher |
Q&A #1895 |
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 it. 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 the 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)^ 2). 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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