If the equations had been given in the form Ax+By+C=0, A'x + B'y +C′ = 0, the equation of a bisector would be

It is evident from the double sign that there are two bisectors: one such that the perpendicular on what we agree to consider the positive side of one line is equal to the perpendicular on the negative side of the other: the other such that the equal perpendiculars are either both positive or both negative.

If we choose that sign which will make the two constant terms of the same sign, it follows from Art. 34 that we shall have the bisector of that angle in which the origin lies; and if we give the constant terms opposite signs, we shall have the equation of the bisector of the supplemental angle.

Ex. 1. Reduce the equations of the bisectors of the angles between two lines, to the form x cos a + y sin a = p.

p-p'

Ans. x cos {(a + ß) + 90°} + y sin { (a + ß) + 90°} =
= 2 sin (a - ẞ)

Ex. 2. Find the equations of the bisectors of the angles between

3x+4y-9=0, 12x + 5y − 3 = 0.

Ans. 7x-9y+34 = 0, 9x + 7y = 12.

36. To find the area of the triangle formed by three points. If we multiply the length of the line joining two of the points, by the perpendicular on that line from the third point, we shall have double the area. Now the length of the perpendicular from x,y, on the line joining x,y,, x,y,, the axes being rectangular, is (Arts. 29, 34)

3

and the denominator of this fraction is the length of the line joining x,y,, xy, hence

Y 21

3

Y1 (x − x ̧) +у, (x − x ̧) + у ̧ (x ̧ − x ̧) represents double the area formed by the three points.

If the axes be oblique, it will be found on repeating the investigation with the formula for oblique axes, that the only change that will occur is that the expression just given is to be multiplied by sin w. Strictly speaking, we ought to prefix to

these expressions the double sign implicitly involved in the square root used in finding them. If we are concerned with a single area we look only to its absolute magnitude without regard to sign. But if, for example, we are comparing two triangles whose vertices xy, xy, are on opposite sides of the line joining the base angles x,y,, xy, we must give their areas different signs; and the quadrilateral space included by the four points is the sum instead of the difference of the two triangles.

27

COR. 1. Double the area of the triangle formed by the lines joining the points x,y,, x,y, to the origin, is y,x,y,, as appears by making x = 0, y,=0, in the preceding formula.

2

COR. 2. The condition that three points should be on one right line, when interpreted geometrically, asserts that the area of the triangle formed by the three points becomes = 0 (Art. 30).

37. To express the area of a polygon in terms of the co-ordinates of its angular points.

19

Take any point xy within the polygon, and connect it with all the vertices 11, xy...y; then evidently the area of the polygon is the sum of the areas of all the triangles into which the figure is thus divided. But by the last Article double these areas are respectively

y

When we add these together, the parts which multiply x and vanish, as they evidently ought to do, since the value of the total area must be independent of the manner in which we divide it into triangles; and we have for double the area

Ex. 1. Find the area of the triangle (2, 1),

(3, - 2), (— 4, − 1).

Ex. 2. Find the area of the triangle (2, 3), (4, 5), (— 3, — 6).

Ex. 3. Find the area of the quadrilateral (1, 1), (2, 3), (3, 3), (4, 1).

Ans. 10.

Ans. 29.

Ans. 4.

38. To find the condition that three right lines shall meet in a point.

Let their equations be

Ax+By+C=0, A'x+B'y+C' =0, A′′x+B'y+C" = 0. If they intersect, the co-ordinates of the intersection of two of them must satisfy the third equation. But the co-ordinates of BC-B'C CA'-C'A AB'-A'B' AB' — A'B'

the intersection of the first two are

Substituting in the third, we get, for the required condition, A" (BC' − B' C\ + B" ( CA' − C'A) +C" (AB' — A'B) = 0, which may be also written in either of the forms

A (B' C" — B" C') + B (C'A" -C"A")+C (A'B" - A"B') = 0, A (B' C" — B′′ C') + A′ (B" C – BC") + A" (BC'

B'C) = 0. *39. To find the area of the triangle formed by the three lines Ax+By+C=0, A'x+ B'y +C' = 0, A′′x+B"y+C" = 0.

By solving for x and y from each pair of equations in turn we obtain the co-ordinates of the vertices, and substituting them in the formula of Art. 36, we obtain for the double area the expression

But if we reduce to a common denominator, and observe that the numerator of the fraction between the first brackets is

{A" (BC' - B'C) + A (B'C" - B" C) + A' (B" C-C"B)} multiplied by A"; and that the numerators of the fractions between the second and third brackets are the same quantity multiplied respectively by A and A', we get for the double area the expression

{A (B'C" – B" C') + A' (B" C – BC") + A" (BC' − B'C')}2 (AB' — BA') (A'B" – B'A") (A′′B – B" A)

If the three lines meet in a point, this expression for the area vanishes (Art. 38); if any two of them are parallel, it becomes infinite (Art. 25).

40. Given the equations of two right lines, to find the equation of a third through their point of intersection.

The method of solving this question, which will first occur to the reader, is to obtain the co-ordinates of the point of intersection by Art. 31, and then to substitute these values for x'y' in the equation of Art. 28, viz., y — y' = m (x − x'). The question, however, admits of an easier solution by the help of the following important principle: If S=0, S' = 0, be the equations of any two loci, then the locus represented by the equation S+kS' = 0 (where k is any constant) passes through every point common to the two given loci. For it is plain that any co-ordinates which satisfy the equation S=0, and also satisfy the equation S'= 0, must likewise satisfy the equation S+kS' = 0.

Thus, then, the equation

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

which is obviously the equation of a right line, denotes one passing through the intersection of the right lines

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

for if the co-ordinates of the point common to them both be substituted in the equation (Ax+ By +C) + k (A'x + B'y+C') = 0, they will satisfy it, since they make each member of the equation separately = 0.

Ex. 1. To find the equation of the line joining to the origin the intersection of

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

Multiply the first by C', the second by C, and subtract, and the equation of the required line is (AC' — A'C') x + (BC' – CB') y = 0; for it passes through the origin (Art. 18), and by the present article it passes through the intersection of the given lines.

Ex. 2. To find the equation of the line drawn through the intersection of the same lines, parallel to the axis of x. Ans. (BA' AB') y + CA' — AC′ = 0.

Ex. 3. To find the equation of the line joining the intersection of the same lines to the point x'y'. Writing down by this article the general equation of a line through the intersection of the given lines, we determine k from the consideration that it must be satisfied by the co-ordinates x'y', and find for the required equation

(Ax + By + C) (A'x' + B'y' + C') = (Ax' + By' + C) (A'x + B'y +C'). Ex. 4. Find the equation of the line joining the point (2, 3) to the intersection of 2x + 3y +1 = 0, 3x - 4y = 5.

Ans. 11 (2x + 3y + 1) + 14 (3x4y-5)=0; or 64x-23y = 59.

F

41. The principle established in the last article gives us a test for three lines intersecting in the same point, often more convenient in practice than that given in Art. 38. Three right lines will pass through the same point if their equations being multiplied each by any constant quantity, and added together, the sum is identically =0: that is to say, if the following relation be true, no matter what x and У are:

1 (Ax+By+C) + m (A'x + B'y +C) + n (A′′x + B′′y+C")=0. For then those values of the co-ordinates which make the first two members severally = 0 must also make the third = 0.

Ex. 1. The three bisectors of the sides of a triangle meet in a point. Their equations are (Art. 29, Ex. 4)

1

(y'' + y'"' — 2y' ) x − (x" + x'''

(y'" + y' - 2y") x − (x""' + x′

2x′ ) y + (x'y' - y′′x′ ) + (x""'y' — y′′x′ ) = 0,

- 2x") y + (x""y" — y'"x") + (x'y" — y'x'' ) = 0,

(y' + y" - 2y''') x − (x' + x′′ – 2x"") y + (x'y'" — y'x""') + (x"y""' — y′′x"") = 0. And since the three equations when added together vanish identically, the lines represented by them meet in a point. Its co-ordinates are found by solving between any two, to be } (x' + x" + x'''), } (y' + y'' + y'").

Ex. 2. Prove the same thing, taking for axes two sides of the triangle whose lengths are a and b.

x 2y
a b

x

Ans. +/- 1 = 0, + -1=0, -0.

2x y
a b

a

Ex. 3. The three perpendiculars of a triangle, and the three perpendiculars at middle points of sides respectively meet in a point. For the equations of Ex. 5 and 6, Art. 32, when added together, vanish identically.

Ex. 4. The three bisectors of the angles of a triangle meet in a point. For their equations are

*42. To find the co-ordinates of the intersection of the line joining the points x'y', x"y", with the right line Ax + By +C=0.

We give this example in order to illustrate a method (which we shall frequently have occasion to employ) of determining the point in which the line joining two given points is met by a given locus. We know (Art. 7) that the co-ordinates of any point on the line joining the given points must be of the form