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 P56·14, 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)P56·14 = 2cP5 + 2bP6 = (− a + b + c)P1 + (a + b + c)P4.
To find, now, the point P56·23 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 P57·13 where the line P5P7 is cut by the line P1P3 we have to consider that being on P1 and P3, the coefficient of P57·13 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)P57·13 = 2aP7 − 2cP5 = (a − b − c)P1 + (a + b − c)P3
and in like manner
2(a + c)P57·24 = 2aP7 + 2cP5 = (a − b + c)P2 + (a + b + c)P4.
To find P67·12 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)P67·12 = 2aP7 − 2bP6 = (a − b − c)P1 + (a − b + c)P2
and so
2(a + b)P67·34 = 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 P56·14, P57·24, P67·12,
we see that they can be so added as to give zero
2(b + c)P56·14 − 2(a + c)P57·24 + 2(a − b)P67·12 = 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)P67·34 + 2( − a + c)P57·13 + 2(− b − c)P56·14 = 0
2(a + b)P67·34 + 2( − a − c)P57·24 + 2(− b + c)P56·23 = 0
2(a − b)P67·12 + 2( − a + c)P57·13 + 2(b − c)P56·23 = 0
2(a − b)P67·12 + 2( − a − c)P57·24 + 2(b + c)P56·14 = 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. |