Writings of Charles S. Peirce: A Chronological Edition, Volume 8: 1890–1892

16. Promptuarium of Analytical Geometry

57 16 Promptuarium of Analytical Geometry c. 1890 Houghton Library Let P1 and P2 be any two points. Now consider this expression lP1  (1  l)P2 where l is a number. P1 and P2 are not numbers, and therefore the binomial cannot be understood exactly as in ordinary algebra; but we are to seek some meaning for it which shall be somewhat analogous to that of algebra. If l  0, it becomes 0P1  1P2 and this we may take as equal to P2, making 0P1  0 and 1P2  P2. Then if l  1, the expression will become equal to P1. When l has any other value, we may assume that the expression denotes some other point, and as l varies continuously we may assume that this point moves continuously. As l passes through the whole series of real values , the point will describe a line; and the simplest assumption to make is that this line is straight. That we will assume; but at present we make no further assumption as to the position of the point on the line when l has values other than 0 and 1. We may write lP1  (1  l)P2  P3. Transposing P3 to the first side of the equation and multiplying by any number, we get an equation of the form a1P1  a2P2  a3P3  0 where a1  a2  a3  0. Such an equation will signify that P1, P2, P3 are in a straight line; for it is equivalent to P1 P2 • • Writings of C. S. Peirce 1890–1892 58 . Let P1, P2, P3 be any three points, not generally in a straight line. Then lP1 ⫹ (1 ⫺ l)P2 may be any point in the straight line through P1 and P2 and m(lP1 ⫹ (1 ⫺ l)P2) ⫹ (1 ⫺ m)P3, where m is a second number, will be any point in a line with that point and with P3. But that plainly describes any point in the plane through P1, P2, P3 so that P4 ⫽ a1P1 ⫹ a2P2 ⫹ a3P3 where a1 ⫹ a2 ⫹ a3 ⫽ 1 denotes any point in that plane. By transposing and multiplying by any number, we can give this the form b1P1 ⫹ b2P2 ⫹ b3P3 ⫹ b4P4 ⫽ 0 where b1 ⫹ b2 ⫹ b3 ⫹ b4 ⫽ 0. To avoid the necessity of the second equation, we may put for b1, b2, b3, b4, four algebraical expressions which identically add up to zero; and may write (a ⫺ b ⫺ c)P1 ⫹ (⫺ a ⫹ b ⫺ c)P2 ⫹ (⫺ a ⫺ b ⫹ c)P3 ⫹ (a ⫹ b ⫹ c)P4 ⫽ 0. This expression will signify that P1, P2, P3, P4 lie in one plane. I will now give an example to show the utility of this notation. Let P1, P2, P3, P4 be any four points in a plane. Assume for the equation connecting them (a ⫺ b ⫺ c)P1 ⫹ (⫺ a ⫹ b ⫺ c)P2 ⫹ (⫺ a ⫺ b ⫹ c)P3 ⫹ (a ⫹ b ⫹ c)P4 ⫽ 0. P3 a1 a1 a2 ⫹ ------------------ - P1 a2 a1 a2 ⫹ ------------------ - P2 ⫹ ⫽ P1 P2 P5 P7 P4 P3 P6 16. Promptuarium, c. 1890 59 Let a line 1 be drawn through P1 and P2 and another through P3 and P4. How shall we express the point P5 where these two lines meet? Very simply; for the above equation gives by transposition (a  b  c)P1  ( a  b  c)P2  (a  b  c)P3  ( a  b  c)P4, and therefore for the point P5, which is equal to some binomial in P1 and P2 and also to some binomial in P3 and P4, we must have the equation  2cP5  (a  b  c)P1  ( a  b  c)P2  (a  b  c)P3  ( a  b  c)P4. For this equation requires it to be on a line with P1 and P2 and also on a line with P3 and P4; and there is but one point so situated. In like manner , the mere inspection of the equation shows us that P6 where the lines P1P3 and P2P4 intersect is represented by the equation  2bP6  (a  b  c)P1  ( a  b  c)P3  (a  b  c)P2  ( a  b  c)P4. And in like manner, the point P7, where P1P4 and P2P3 intersect, is represented by  2aP7  ( a  b  c)P2  ( a  b  c)P3  ( a  b  c)P1  ( a  b  c)P4. Let us now join the points P5 and P6. How are we to find the point, which I will call P56 14, where the line P5P6 is cut by the line P1P4? For this purpose, we have to...