Similar expressions may be found for (ẞ ̧ — ẞ2)3, (î1 — Y2)2.
Hence, r2 will be of the form

dicular to AB, PN, QN' to AC. Draw Qm perpendicular to PM, Qn to PN, and join mn.

Then

.. mn2=(ẞ1-ẞ1)2 + (Y1-2)2+2 (B1-B2) (Y1-Y2) cos A,
(B1 − B2)2 + (Y1 − Y1⁄2)2 + 2 (B1 − ẞ2) (Y1 — Y1⁄2) cos A
sin2 A

whence r2=

a rational integral function of the second degree.

where l, m, n are certain functions of a, b, c, which we proceed to determine.

Since the values of l, m, n are independent of the positions of the points, the distance of which we wish to find, suppose these points to be B and C.

Then

a2 = − 1 24.

2A = ; C

2A

b

C

abc

Hence r2 =

abc2

n = - 4A2

{a (B ̧ — ß2) (Y1 — Y1⁄2) + b (Y, − y2) (α, — d2) 4A*

This is one form of the expression for r2. It may also be proved in a similar manner that

4. We next proceed to investigate the equation of a straight line; and first, we shall consider the cases of certain straight lines bearing important relations to the triangle of reference.

To find the equation of the straight line drawn through one of the angular points of the triangle of reference, so as to bisect the opposite side.

Let D be the middle point of the side BC, we have then to investigate the equation of the straight line AD.

In AD take any point P, and let a, ẞ, y be its co-ordi

nates.

From D, P draw DE, PG perpendicular to AC, DF, PH perpendicular to AB. Then by similar triangles

for each is equal to the area of the triangle ABC.

Hence

or

PG.AC-PH. AB,

bB=cy.

This is a relation between the co-ordinates of any point on the line AD, it therefore is the equation of that line.

COR. It hence may be proved that the three straight lines, drawn through the angular points of a triangle to bisect the opposite sides, intersect in a point. For these straight lines will be represented by the equations

and, therefore, all pass through the point for which

aa=bB =cy.

In the next three propositions the reader will easily be able to draw a figure for himself, by comparison with fig. 5.

5. To find the equation of the straight line drawn through one of the angular points of the triangle of reference, perpendicular to the opposite side.

Making a construction similar to that in the last proposition, it will be seen that we have here

This will be the equation of the straight line, drawn through A, at right angles to BC.

COR. It may hence be shewn that the three straight lines drawn through the angular points of a triangle, perpendicular to the opposite sides, intersect in the point determined by the equations

a cos A = ẞ cos B = y cos C.

6. To find the equations of the internal and external bisectors of an angle of the triangle of reference.

For the internal bisector of the angle A, we shall have, making the same construction as before,

PG=PH.

The straight line will be therefore represented by the equation B = y.

For the external bisector we proceed as follows. Let Q be any point on the line, a, B, y its co-ordinates. Draw QK perpendicular to AC, QL to AB. Then, as before, we have

QK = QL.

It will however be observed, that if Q and B lie on the same side of AC, Q and C will lie on opposite sides of AB, and vice versa. Hence, if

QKB, QL-y.

We have therefore

B+7=0

as the equation of the line AQ, which externally bisects the angle A.

From the form of these equations we see, (1), That the three internal bisectors of the angles of a triangle intersect in a point; (2), That the internal bisector of any one angle, and the external bisectors of the other two, also intersect in a point.

These points may be shewn to be respectively the centres of the inscribed and escribed circles.

We shall hereafter prove that the points, in which the external bisectors of each angle respectively intersect the sides opposite to them, lie in the same straight line; and also that the points in which the external bisector of any one angle, and the internal bisectors of the other two angles, intersect the sides respectively opposite to them, lie in the same straight line.

7. We now proceed to investigate the general equation of a straight line.

Every straight line may be represented by an equation of the first degree.

Let be any point on the straight line AC, R on AB, and P any point on the straight line QR, then we have to