IDENTICAL RELATION.

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A similar equation may be proved to hold for any point without the triangle, BOC being considered negative, if A and O be on opposite sides of BC.

The following are the equations of some important points connected with the triangle of reference :

Centre of gravity,

p+q+r=0.

Centre of circumscribing circle, p sin 2A+qsin 2B+rsin 2 C=0.

4. We proceed to investigate the identical relation which holds between the co-ordinates of any straight line.

Let any straight line cut the sides AB, AC of the triangle of reference in D, E. From A, B, C let fall AP, BQ, CR, perpendiculars on the line, then BQ=q, CR=r, AP=-p.

Let the triangular equation of RPQ referred to ABC as triangle of reference, be

lx+my+nz = 0.

Then, as shewn in the last article,

(7. 2A)2 = {(l—m) (l—n) a2+(m—n) (m− 1)b2 + (n − 1) (n−m) c2} p2.

Similar equations holding for q2 and r2, we get

= (1 — m) (1 − n) a2 + (m − n) (m − 1) b2 + (n − 1) (n − m) c2'

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

the required relation between the co-ordinates of any straight line whatever.

This may also be expressed in the form

as may be found by evaluating the determinant.

COR. Since the line at infinity may be considered as equidistant from A, B, and C, it will be represented by the equations p=q=r.

5. To find the distance from the point lp+mq+nr = 0, to the line (P1, q1, 1).

By what has been shewn above, it appears that the triangular co-ordinates of the point are

EQUATION OF THE SECOND DEGREE.

135

=

Hence, if ♪ be the distance between them,

p1l+q,m+r,n

(l+m+n){(P1−q1)(P1—r1) a2+(q1—r1) (%1—p1)b2+(r1—P1)(r,−91) c2 } $

COR. Hence, if p be the radius of a circle, lp+mq+nr=0 the equation of its centre, the circle, being the envelope of a line whose distance from the centre is constant, will be represented by the equation

(p − q) (p − r) a2 + (q − r) (q − p) b2 + (r − p) (r − q) c2

2A.

6. An equation of a degree, higher than the first, may be regarded as representing the curve which is touched by all the straight lines, the co-ordinates of which satisfy the equation of the curve. Adopting this mode of interpretation, the values of the ratios p q r which simultaneously satisfy two given equations will be the co-ordinates of the common tangents to the two curves represented by these equations, and the values obtained by combining any given equation with an equation of the first degree, will represent all the straight lines which pass through the point represented by the equation of the first degree, and which touch the curve. From this it follows, that an equation of the nth degree will represent a curve such that n tangents, real or imaginary, can be drawn to it from any point, that is, a curve of the nth class.

It will hence follow that every equation of the second degree represents a conic. We may proceed to consider some of its more interesting special forms.

7. To find the equation of a conic which touches the three sides of the triangle of reference.

are

The co-ordinates of the sides of the triangle of reference

q=0, r=0 for BC,

r=0, p=0 for CA,

p=0, q=0 for AB.

Hence, the equation of the required conic must be satisfied whenever two out of the three co-ordinates p, q, r are = 0. It must therefore be of the form

Lqr+Mrp+Npq=0.

The equations of the points of contact are

These may be established as follows: If in the given. equation we make Mr+ Nq=0, we obtain either q=0, or r0. It hence appears that the tangents drawn through the point Mr+ Nq=0, pass either through the point q = 0, or through the point r=0. But the three points

lie in the same straight line; hence the tangents drawn from Mr+Nq=0 coincide, that is, it is the point of contact of the tangent for which q=r=0. Similarly for the other two points of contact.

It will hence appear, by reference to the equations of the points of contact of the inscribed circle, given in Art. 2, that that circle is represented by the equation

CIRCUMSCRIBING CONIC.

The escribed circles will be represented as follows:

− sqr + (s − c) rp + (s − b) pq = 0,

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8. To find the equation of a conic circumscribed about the triangle of reference.

The equations of the angular points of the triangle of reference are p=0, q = 0, r = 0. Now, since each of these points lies on the curve, the two tangents drawn through any one of them must coincide, hence when any one of these quantities is put =0, the remaining equation must have two equal roots. The required equation will therefore be of the

form

L3p2+M2q2 + N22 — 2MNqr — 2NLrp – 2LMpq = 0.

The co-ordinates of the several tangents at the angular points will be given by the equations

p=0, Mq-Nr 0,

q=0, Nr - Lp = 0,

r=0, Lp - Mq = 0.

If the conic be a circle, the tangent at A will be determined by the equations

which last is equivalent to b'q-c3r = 0.

Similar equations holding for the other two tangents, the equation for the circumscribing circle will be

a*p2 + b‘q2 + c*r2 — 263c'qr - 2c3a3rp - 2a2b3pq = 0,

which may be reduced to

± ap±bq±cr*= 0.