which the line joining the points is cut by the given locus; and we determine this unknown quantity from the condition, that the co-ordinates just written shall satisfy the equation of the locus. Thus, in the present example we have

and consequently the co-ordinates of the required point are

with a similar expression for y. This value for the ratio m: n might also have been deduced geometrically from the consideration that the ratio in which the line joining x'y', "y" is cut, is equal to the ratio of the perpendiculars from these points upon the given line; but (Art. 34) these perpendiculars are

Ax' + By +C Ax" + By" + C

√(A2 + B2)

and

√(A2 + B2)

The negative sign in the preceding value arises from the fact that in the case of internal section to which the positive sign of m: n corresponds (Art. 7), the perpendiculars fall on opposite sides of the given line, and must, therefore, be understood as having different signs (Art. 34).

If a right line cut the sides of a triangle BC, CA, AB, in the points LMN, then

BL.CM.AN

Let the co-ordinates of the vertices be x'y', x"y", ""y", then

*43. To find the ratio in which the line joining two points xy1, xy, is cut by the line joining two other points xY, XY The equation of this latter line is (Art. 29)

It is plain (by Art. 36) that

3

4

3

this is the ratio of the two triangles whose vertices are x,y,, y, xy, and x,y,, XY 31 x4Y47 as also is geometrically evident.

29

If the lines connecting any assumed point with the vertices of a triangle meet the opposite sides BC, CA, AB respectively, in D, E, F, then

BD.CE.AF
DC.EA.FB

= +1.

Let the assumed point be xy, and the vertices x11, XY 21

xy, then X3Y 37

44. To find the polar equation of a right line (see Art. 12). Suppose we take, as our fixed axis, OP the perpendicular on

the given line, then let OR be

any radius vector drawn from the pole to the given line

OR=p, ROP=0;

but, plainly,

R

If the fixed axis be OA making an angle a with the perpendicular, then ROA=0, and the equation is

p cos (0− a) = p.

This equation may also be obtained by transforming the equation with regard to rectangular co-ordinates,

Rectangular co-ordinates are transformed to polar by writing for x, p cose, and for y, p sine (see Art. 12); hence the equation becomes

p (cose cosa + sine sina)=p;

or, as we got before,

An equation of the form

p cos (0―a)=p.

p(4 cose + B sin 0)=C

can be (as in Art. 23) reduced to the form p cos(-a)=p, by dividing by √(A+B); we shall then have

Ex. 2. Find the polar co-ordinates of the intersection of the following lines, and

Ex. 3. Find the polar equation of the line passing through the points whose polar co-ordinates are p', '; p",

0".

Ans. p'p" sin (0'- 0") + p′′p sin (0" → 0) + pp' sin (0 — 0′) = 0.

CHAPTER III.

EXAMPLES ON THE RIGHT LINE.

45. HAVING in the last chapter laid down principles by which we are able to express algebraically the position of any point or right line, we proceed to give some further examples of the application of this method to the solution of geometrical problems. The learner should diligently exercise himself in working out such questions until he has acquired quickness and readiness in the use of this method. In working such examples our equations may generally be much simplified by a judicious choice of axes of co-ordinates: since, by choosing for axes two of the most remarkable lines on the figure, several of our expressions will often be much shortened. On the other hand, it will sometimes happen that by choosing axes unconnected with the figure, the equations will gain in symmetry more than an equivalent for what they lose in simplicity. The reader may compare the two solutions of the same question, given Ex. 1 and 2, Art. 41, where, though the first solution is the longest, it has the advantage that the equation of one bisector being formed, those of the others can be written down without further calculation.

Since expressions containing angles become more complicated by the use of oblique co-ordinates, it will be generally advisable to use rectangular axes in any question in which the consideration of angles is involved.

46. Loci.-Analytical geometry adapts itself with peculiar readiness to the investigation of loci. We have only to find what relation the conditions of the question assign between the co-ordinates of the point whose locus we seek, and then the statement of this relation in algebraical language gives us at once the equation of the required locus.

Ex. 1. Given base and difference of squares of sides of a triangle, to find the locus of vertex.

Let us take for axes the base and a perpendicular through its middle point. Let the half base = c, and let the co-ordinates of the vertex

be x, y. Then

AC2 = y2 + (c + x)2,* BC2 = y2 + (c − x)2,

C

is therefore a line perpendicular to the base at a dis- A

that the difference of squares of segments of base = difference of squares of sides.

Ex. 2. Find locus of vertex, given base and cot A + m cot B.

It is evident, from the figure, that

Ex. 3. Given base and sum of sides of a triangle, if the perpendicular be produced beyond the vertex until its whole length is equal to one of the sides, to find the locus of the extremity of the perpendicular.

Take the same axes, and let us inquire what relation exists between the co-ordinates of the point whose locus we are seeking. The x of this point plainly is MR, and the y is, by hypothesis, AC; and if m be the given sum of sides,

Now (Euclid II. 13),

or

=

Ex. 4. Given two fixed lines, OA and OB, if any line AB be drawn to intersect them parallel to a third fixed line OC, to find the locus of the point P where AB is cut in a given ratio; viz. PA = nAB,

Let us take the lines OA, OC for axes, and let the equation of OB be y = mx. Then since the point B lies on the latter line, its ordinate is m times its abscissa; or AB = mOA. Therefore PA = mnOA; but PA and OA are the co-ordinates of the point P, whose locus is therefore a right line through the origin, having for its equation,

C

B

P

* This is a particular case of Art. 4, and c + x is the algebraic difference of the abscissæ of the points A and C (see remarks at top of p. 4). Beginners often reason that since the line AR consists of the parts AM= == C1 and MR = =x, its length is -c+x, and not c+x, and therefore that AC2 = y2 + (x − c)2. It is to be observed that the sign given to a line depends not on the side of the origin on which it lies, but on the direction in which it is measured. We go from A to R by proceeding in the positive direction AM = c, and still further in the same direction MR = x, therefore the length AR=c+x: but we may proceed from R to B by first going in the negative direction RM=-x, and then in the opposite direction MB = c2 hence the length RB is c x.