Perpendicular lines have negative reciprocal slopes

This explanation is guaranteed to be Mathematically Correct.* **
 

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A non-interactive version of this picture which will print. (This version also has some helpful markings that don't come out in the interactive diagram.)

Proof:

Consider two perpendicular lines, L and M,  intersecting at point P. Assume that the first line,L, has positive slope m.
This means that you can construct a triangle ABP, with hypoteneuse along the first line and legs parallel to the x and y axes, so that the horizontal leg PB has length 1 and the vertical leg AB has lengt vertical leg AB has length m.
Now extend the horizontal leg through p, and mark off a length of m (at point C) along it. Construct the perpendicular to this horizontal line through C, and call the point D where it intersects the second line M.

Claim: Triangles PBA and DCP are congruent.

Angle BAP + angle BPA = 90 degrees (since the angles of a triangle sum to 180, and one of the angles is 90)
Angle CPD + angle BPA = 90 degrees (since these two angles and right angle APD sum to 180)
So angle BAP = angle CPD.
Angles DCP and ABP are right angles, by construction.
Sides AB and CP have length m, by construction.
So by the ASA triangle congruence theorem, triangle PBA is congruent to triangle DCP, and all the other corresponding parts are congruent.

In particular, CD = BP = 1.

In computing slopes, the direction in which you subtract the points is important. In this picture,

(y coord of A - y coord of P) / (x coord of A - x coord of P) = m / 1
giving a positive slope for L. On the other triangle,
(y coord of D - y coord of P) / (x coord of D - x coord of P) = 1 / (-m)
giving a negative slope for M, which is the negative reciprocal of the slope of L.

If m is negative, the picture is just flipped vertically. (Or start with triangle DCP and construct triangle PBA.)

If the lines are horizontal and vertical, one has slope 0 and the other has slope 0 and the other has slope infinity. In some contexts (including this one), you can think of infinity as being 1/0 or -1/0, so the theorem still works in this case.



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