Let OA = a, OB = b, OAˆ = a + k, then from the conditions of the problem, OB'b-k. The equations of AB', A'B are respectively

Subtracting, we eliminate k, and find for the equation of the locus

x + y = a + b.

Ex. 5. If on the base of a triangle we take any portion AT, and on the other side of the base another portion BS, in a fixed ratio to AT, and draw ET and FS parallel to a fixed line CR; to find the locus of O, the point of intersection of EB and FA. Take AB and CR for axes; let AT=k, BR = s, AR=s', CR = p, let the fixed ratio be m, then BS will=mk; the co-ordinates of S will be (s-mk, 0), and of T-(s′ — k), 0}.

C

F

E

Now form the equations of the transverse lines, and the equation of EB is

To eliminate k, subtract one equation from the other, and the result, divided by k, will be

mp

(m-1) y +

(m2 + 2) x + (mps - P3) = 0,

ps

which is the equation of a right line.

Ex. 6. PP' and QQ' are any two parallels to the sides of a parallelogram; to find the locus of the intersection of the lines PQ and P'Q'.

Let us take two of the sides for our axes, and let the lengths of the sides be a and b, and let AQ'm, APn. Then the equa

Q

D

sible to eliminate them from two equations. However, if we add the above equations, it will be found that both vanish together, and we get for our locus

bx - ay = 0,

the equation of the diagonal of the parallelogram.

Ex. 7. Given a point and two fixed lines: draw any two lines through the fixed point, and join transversely the points where they meet the fixed lines; to find the locus of intersection of the transverse lines.

Take the fixed lines for axes, and let the equations of the lines through the fixed point be

The condition that these lines should pass through the fixed point x'y' gives us

Now from this and the equation just found we can eliminate

the equation of a right line through the origin.

Ex. 8. At any point of the base of a triangle is drawn a line of given length, parallel to a given one, and so as to be cut in a given ratio by the base: find the locus of the intersection of the lines joining its extremities to those of the base.

49. The fundamental idea of Analytic Geometry is that every geometrical condition to be fulfilled by a point leads to an equation which must be satisfied by its co-ordinates. It is important that the beginner should quickly make himself expert in applying this idea, so as to be able to express by an equation any given geometrical condition. We add, therefore, for his further exercise some examples of loci which lead to equations of degrees higher than the first. The interpretation of such equations will be the subject of future chapters, but the method of arriving at the equations, which is all with which we are here concerned, is precisely the same as when the locus is a right line. In fact until the problem has been solved, we do not know what will be the degree of the resulting equation. The examples that follow are purposely chosen so as to admit of treatment similar to that pursued in former examples, according to the order of which they are arranged. In each of the answers given it is supposed that the same axes are chosen, and that the letters have the same meaning as in the corresponding previous example.

Ex. 1. Find the locus of vertex of a triangle, given base and sum of squares of sides. Ans. x2+ y2 m2 — c2.

Ex. 2. Given base and m squares of one side + n squares of the other.

=

Ans. (m±n) (x2 + y2) + 2 (m + n) cx + (m + n) c2 = p2.

Ex. 3. Given base and ratio of sides.

Ex. 4. Given base and product of tangents of base angles.

In this and the Examples next following, the learner will use the values of the tangents of the base angles given Ex. 2, Art. 46. Ans. y2+ m2x2 = m2c2.

Ex. 5. Given base and vertical angle, or, in other words, base and sum of base angles.

Ex. 6. Given base and difference of base angles.

Ans. x2 + y2 - 2cy cot C = c2

Ans. x2- y2+ 2xy cot D = c2,

Ex. 7. Given base, and that one base angle is double the other.

Ans. 3x2- y2 + 2cx = c2.

Ans. m (x2 + y2 — c2) = 2c (c − x).

Ex. 8. Given base, and tanC=m tan B. Ex. 9. PA is drawn parallel to OC, as in Ex. 4, p. 39, meeting two fixed lines in points B, B'; and PA2 is taken = PB.PB', find the locus of P.

Ans. mx (m'x + n') = y (mx + m’x + n').

Ans. 2mx (m'x + n') = y (mx + m'x + n').

Ex. 10. PA is taken the harmonic mean between AB and AB'.

Ex. 11. Given vertical angle of a triangle, find the locus of the point where the base is cut in a given ratio, if the area also is given.

Ex. 12. If the base is given.

constant. b2

Ans. xy 2xy cos w

n2

=

Ex. 13. If the base pass through a fixed point.

Ex. 14. Find the locus of P [Ex. 8, p. 40] if MN is constant.

Ex. 16. If MN pass through a fixed point, find the locus of the intersection of parallels to the axes through M and N.

x' y Ans. + = 1.

x y

Ex. 17. Find the locus of P [Ex. 1, p. 41] if the line CD be not parallel to AB. Ex. 18. Given base CD of a triangle, find the locus of vertex, if the intercept AB on a given line is constant.

Ans. (x'y — y'x) (y — y′′) — (x′′y — y′′x) (y — y') = c (y — y′) (y — y').

50. Problems where it is required to prove that a moveable

right line passes through a fixed point.

We have seen (Art. 40) that the line

Ax+By+C+k (A'x + B'y + C') =0;

or, what is the same thing,

(A+kA')x+(B+kB') y +C+kC' = 0,

where k is indeterminate, always passes through a fixed point, namely, the intersection of the lines

Ax+By+C=0, and A'x + B'y + C' = 0.

Hence, if the equation of a right line contain an indeterminate quantity in the first degree, the right line will always pass through a fixed point.

Ex. 1. Given vertical angle of a triangle and the sum of the reciprocals of the sides; the base will always pass through a fixed point.

Take the sides for axes; the equation of the base is the condition

818

214

+

=

1, and we are given

a

m

1

where

α

is indeterminate. Hence the base must always pass through the intersection

of the two lines x - y = 0, and y = m.

Ex. 2. Given three fixed lines OA, OB, OC, meeting in a point, if the three vertices of a triangle move one on each of these lines, and two sides of the triangle pass through fixed points, to prove that the remaining side passes through a fixed point.

line OC on which the vertex moves will have

x'y'

A

Take for axes the fixed lines OA, OB, on which the base angles move, then the

Now the length of the intercept OA is found by making x = 0 in equation AC, or

But since a is indeterminate, and only in the first degree, this line always passes through

a fixed point. The particular point is found by arranging the equation in the form

Hence the line passes through the intersection of the two lines

Ex. 3. If in the last example the line on which the vertex C moves do not pass through O, to determine whether in any case the base will pass through a fixed point.

We retain the same axes and notation as before, with the only difference that the equation of the line on which C moves will be y = mx + n, and the co-ordinates of the vertex in any position will be a, and ma+n. Then the equation of AC is

(x′- a) y - (y' - ma — n) x + a (y′ — mx′) — nx′ = 0.

The equation of BC is

Now when this is cleared of fractions, it will in general contain a in the second degree, and therefore, the base will in general not pass through a fixed point; if, however, the points x'y', x"y", lie in a right line (y = kx) passing through O, we may substitute in the denominators y" = kx", and y' = kx', and the equation becomes

which only contains a in the first degree, and, therefore, denotes a right line passing through a fixed point.

Ex. 4. If a line be such that the sum of the perpendiculars let fall on it from a number of fixed points, each multiplied by a constant, may = 0, it will pass through a fixed point.

Let the equation of the line be

x cos ay sina — p = 0,

then the perpendicular on it from x'y' is

x' cos a+y' sina - P,

and the conditions of the problem give us

m' (x' cosa + y' sin a − p) +m" (x" cosa +y" sina - p)

+m"" (x""' cosa + y"" sina - p) + &c. = 0.

Or, using the abbreviations Σ (mx') for the sum* of the mx, that is,

m'x' + m"x" + m'"'x''' + &c.,

and in like manner Σ (my') for

m'y' + m"y" + m""'y'"' + &c.,

and Σ (m) for the sum of the m's or

m' + m" + m'" + &c.,

* By sum we mean the algebraic sum, for any of the quantities m', m", &c. may be negative.

H