And CE, BE meet in the point (1,0,

a"

The three points will lie on a straight line if

But, since the three points D, E, F are on the conic (i),

we have

By the elimination of λ, μ, v we see that the condition (ii) is satisfied, which proves the proposition. [See also Art. 319, Ex. 3.]

Since six points can be taken in order in sixty different ways, there are sixty hexagons corresponding to six points on a conic; and, since Pascal's Theorem is true for every

one of these hexagons, there are sixty Pascal lines corresponding to six points on a conic.

288. If a hexagon circumscribe a conic, the points of contact of its sides will be the angular points of a hexagon inscribed in the conic. Each angular point of the circumscribed hexagon will be the pole of the corresponding side of the inscribed hexagon; therefore a diagonal of the circumscribing hexagon, that is a line joining a pair of its opposite angular points, will be the polar of the point of intersection of a pair of opposite sides of the inscribed hexagon. But the three points of intersection of pairs of opposite sides of the inscribed hexagon lie on a straight line by Pascal's Theorem; hence their three polars, that is the three diagonals of the circumscribing hexagon, will meet in a point. This proves Brianchon's Theorem :-if `a hexagon be described about a conic, the three diagonals will meet in a point.

289. If we are given five tangents to a conic we can find their points of contact by Brianchon's Theorem. For, let A, B, C, D, E be the angular points of a pentagon formed by the five given tangents; then, if K be the point of contact of AB, Ă, K, B, C, D, E are the angular points of a circumscribing hexagon, two sides of which are coincident. By Brianchon's Theorem, DK passes through the point of intersection of AC and BE; hence K is found. The other points of contact can be found in a similar manner.

Similarly, by means of Pascal's Theorem, we can find the tangents to a conic at five given points. For, let A, B, C, D, E be the five given points, and let F be the point on the conic indefinitely near to A; then, by Pascal's Theorem, the three points of intersection of AB and DE; of BC and EF; and of CD and FA lie on a straight line. Hence, if the line joining the point of intersection of AB and DE to the point of intersection of BC and EA meet CD in H, AH will be the tangent at A. The other tangents can be found in a similar manner.

AREAL CO-ORDINATES.

290. The position of any point P is determined if the ratios of the triangles PBC, PCA, PAB to the triangle of reference ABC be given. These ratios are denoted by x, y, z respectively, and are called the areal co-ordinates of the point P.

The areal co-ordinates of any point are connected by the relation x + y + z = 1.

bB
24'

the equation in areal co-ordinates which corresponds to any given homogeneous equation in trilinear co-ordinates, by

substituting in the given equation 2

α

y

b

2

for a, B, Y

respectively; for example the equation of the line at infinity is x+y+z = 0. We will however find the areal equation of the circumscribing circle independently.

291. To find the equation in areal co-ordinates of the circle which circumscribes the triangle of reference.

If P be any point on the circle circumscribing the triangle ABC, then by Ptolemy's Theorem (Euclid vi. D.) we have

.........

PA.BC + PB. CA ± PC. AB= 0 ........... .(i). But since the angles BPC and BAC are equal, we have PB. PC =x, and similarly for y and z; hence, paying regard to the signs of x, y, z, we have from (i) PA.PB.PC PA.PB.PC PA.PB.PC

АВ. АС

a.

+b.

+c.

=

= 0,

bcx

cay

abz

292. If the conic represented by the general equation of the second degree in trilinear co-ordinates, viz.

ua2 + vß2 + wy2 + 2u' By + 2v'ya +2w'aß = 0,

be the same as that represented in areal co-ordinates by the equation

Xx2 + μy2+ vz2 + 2X'yz + 2μ'zx + 2v'xy = 0;

Hence we can obtain the relation between the coefficients in the areal equation which corresponds to any given relation between the coefficients in the trilinear equation.

For example, the condition that uz2 + vß2 + wy2 = 0 may be a rectangular hyperbola is u+v+w=0; hence the condition that λx2 + μy2 + vz2=0 may be a rectangular hyperbola is λa2 + μb2 + vc2 = 0.

TANGENTIAL CO-ORDINATES.

293. If l, m, n be the three constants in the trilinear or areal equation of any straight line, the position of the line will be determined when I, m and n are given; and by changing the values of l, m, and n the equation may be made to represent any straight line whatever.

The quantities l, m, and n which thus define the position of a straight line are called the co-ordinates of the line. If the equation of a straight line in areal co-ordinates lx+my+nz= 0,

be the lengths of the perpendiculars on the line from the angular points of the triangle of reference will be proportional to l, m, n. This follows at once from Art. 257; we will however give an independent proof.

Let the lengths of the perpendiculars from the angular points A, B, C of the triangle of reference be p, q, r

respectively. Let the line cut BC in the point K, and let the co-ordinates of K be 0, y', z'.

But, since K is on the line, my' +nz′ = 0; therefore

q r :: m : n.

294. The lengths of the perpendiculars on a straight line from the angular points of the triangle of reference may be called the co-ordinates of the line. If any two of these perpendiculars be drawn in different directions they must be considered to have different signs.

From the preceding Article we see that the equation of a line whose co-ordinates are p, q, r is px + qy +rz= 0.

When the lengths of two of the perpendiculars on a straight line are given, there are two and only two positions of the line; so that, when two of the co-ordinates of the line are given, the third has one of two particular values. Hence there must be some identical relation connecting the three co-ordinates of a line, and that relation must be of the second degree.

295. To find the identical relation which exists between the co-ordinates of any line.

Let be the angle the line makes with BA, then we have q-p=csin 0, and q — r = a sin (0 + B). The elimination of gives the required relation, viz.

or

a2 (q− p)2 - 2ac cos B (q − p) (q − r) + c2 (q − r)2 = 4A3,

a2 (p − q) (p − r) + b2 (q− r) (q − p) + c2 (r − p) (r− q) = 4A3.

296. If the line px+qy+rz=0 pass through a fixed point (f, g, h), then

pf+qg+rh = 0........

..(i).

So that the co-ordinates of all the lines which pass through the point whose areal co-ordinates are f, g, h satisfy the relation (i).

Hence the equation of a point is of the first degree.