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16

Promptuarium of Analytical Geometry

c. 1890

Houghton Library

Let P1 and P2 be any two points.

Now consider this expression

λP₁ + (1 − λ)P₂

where λ 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 λ = 0, it becomes

0P₁ = 1P₂

and this we may take as equal to P₂, making 0P₁ = 0 and 1P₂ = P₂. Then if λ = 1, the expression will become equal to P₁. When λ has any other value, we may assume that the expression denotes some other point, and as λ varies continuously we may assume that this point moves continuously. As λ 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 λ has values other than 0 and 1. We may write

λP₁ + (1 − λ)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

Let P₁, P₂, P₃ be any three points, not generally in a straight line. Then λP₁ + (1 − λ)P₂ may be any point in the straight line through P₁ and P₂ and μ(λP₁ + (1 − λ)P₂) + (1 − μ)P₃, where μ 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.

Let a line¹ 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

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

And in like manner, the point P₇, where P₁P₄ and P₂P₃ intersect, is represented by

Let us now join the points P₅ and P₆. How are we to find the point, which I will call P_56·14, where the line P₅P₆ is cut by the line P₁P₄? For this purpose, we have to find an equation between the four points P₁, P₄, P₅, P₆. The first two members of the equation of P₅ give

2cP₅ + (a − b − c)P₁ + (− a + b − c)P₂ = 0.

The first and third member of the equation of P₆ give

2bP₆ + (a − b + c)P₂ + (− a − b − c)P₄ = 0.

Adding these we get

2cP₅ + 2bP₆ + (a − b − c)P₁ + (− a − b − c)P₄ = 0.

We now see, at once, that

2(b + c)P_56·14 = 2cP₅ + 2bP₆ = (− a + b + c)P₁ + (a + b + c)P₄.

To find, now, the point P_56·23 where the line P₅P₆ is cut by the line P₂P₃, we have only to reflect that its coefficient must be composed of 2b and 2c, and also of − a − b + c and − a + b − c. In order to eliminate a between the last, we must take their difference, which is 2b − 2c. Hence, we see that

A little attentive practice will enable the student to write down the equations for the intersections without difficulty. Thus, to find the equation of the point P_57·13 where the line P₅P₇ is cut by the line P₁P₃ we have to consider that being on P₁ and P₃, the coefficient of P_57·13 must be composed of (a − b − c) and (− a − b + c), and being on P₅P₇ it must be composed of 2c and 2a. Hence the equation is

2(a − c)P_57·13 = 2aP₇ − 2cP₅ = (a − b − c)P₁ + (a + b − c)P₃

and in like manner

2(a + c)P_57·24 = 2aP₇ + 2cP₅ = (a − b + c)P₂ + (a + b + c)P₄.

To find P_67·12 we simply consider that the coefficient must be composed of a and b and that in the coefficients of P₁ and P₂ these letters have opposite signs, so that

2(a − b)P_67·12 = 2aP₇ − 2bP₆ = (a − b − c)P₁ + (a − b + c)P₂

and so

2(a + b)P_67·34 = 2aP₇ + 2bP₆ = (a + b − c)P₃ + (a + b + c)P₄.

We now come upon a very curious theorem. Namely, if we take the equations of the three points P_56·14, P_57·24, P_67·12,

we see that they can be so added as to give zero

2(b + c)P_56·14 − 2(a + c)P_57·24 + 2(a − b)P_67·12 = 0

and therefore these three points lie in a straight line. And we can find several other sets, for if the coefficients add up to zero, the points will be certain to do so. Thus we have

2(a + b)P_67·34 + 2( − a + c)P_57·13 + 2(− b − c)P_56·14 = 0

2(a + b)P_67·34 + 2( − a − c)P_57·24 + 2(− b + c)P_56·23 = 0

2(a − b)P_67·12 + 2( − a + c)P_57·13 + 2(b − c)P_56·23 = 0

2(a − b)P_67·12 + 2( − a − c)P_57·24 + 2(b + c)P_56·14 = 0.

Compare the relations of these 4 lines with those of the 4 points originally taken.

These 4 lines intersect in 6 points P67·34, P67·12, P57·13, P57·24, P56·14, P56·23.	The original 4 points lie in pairs on the 6 lines P1P2, P3P4, P1P3, P2P4, P1P4, P2P3.
Those 6 points lie in pairs on three other lines besides the 4 with which we started. Namely P6P7, P5P7, P5P6.	Those 6 lines intersect in three other points besides the original 4, namely P5, P6, P7.
These 3 lines intersect in 3 points P5, P6, P7.	These 3 points lie in pairs on 3 lines P6P7, P5P7, P5P6.
These 3 points lie in pairs with the other 6 points on 6 new lines, namely P5P67·34, P5P67·12, P6P57·13, P6P57·24, P7P56·14, P7P56·23.	These 3 lines intersect the other 6 lines on 6 new points namely P67·12, P67·34, P57·13, P57·24, P56·14, P56·23.
These 6 lines intersect in threes in 4 points P1, P2, P3, P4.	These 6 points lie in threes on 4 lines.

Figure 2 shows all these points and lines.

If L₁ and L₂ are any two lines in a plane, they have a point of intersection, and we may write

L₃ = λL₁ + (1 − λ)L₂

for any third line passing through that point. Or if L₁, L₂, L₃ have not a common intersection,

μ(λL₁ + (1 − λ)L₂) + (1 − μ)L₃

may represent any line in the plane. Thus, the whole theory of lines is exactly like that of points.

1. In analytical geometry, a line always means a straight line.