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. 34); if any two of them are parallel, it becomes infinite (Art. 22).

36. 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. 33, and then to substitute these values for x'y' in the equation of Art. 28, viz., y - y = m(xx). 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 + By + 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 + By + 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. 19), 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 + By + 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 (3x - 4y-5)=0; or 64x - 23y = 59.

37. 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. 34. 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 y are

7 (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. 5)—

4

(y" + y" - 2y ) x − (x' + x1" — 2x′ ) y + (x”y — y'x' ) + (x”y → y′′x′ ) = 0.
(y" + y2y") x − (x" + x′ - 2x") y + (x"y"— y"x") + (x'y" — y'x" ) = 0.

(y + y' - 2y") x − (x′ + x' — 2x") y + (x'y"— yx") + (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 (Art. 33)

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

Ans.

2x
y
+

2y

y

1=

0,

α

+
a b

- 1 = 0,

= 0.

α b

* 38. 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 might solve this question by forming the equation of the line joining the two points, and then determining, by Art. 33, its intersection with the given line. There is, however, another method (which we shall frequently 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

m

and we may take as our unknown quantity the ratio, namely,

n

in which the line joining the points is cut by the given locus, and we may 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 'y', x"y" is cut, is equal to the ratio of the perpendiculars from these points upon the given line; but (Art. 27) these perpendiculars are

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. 27).

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

Λ

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

1.

* 39. To find the ratio in which the line joining two points X1Y1, X2Y2, is cut by the line joining two other points x¿Yз9 XY4.

The

equation of this latter line is (Art. 29)

(X3 − X4) Y + X3Y4 — X4Y3 = 0.

(Yз - Y4) X Therefore by the last article

m

n

It is plain (by Art. 31) that this is the ratio of the two triangles

whose vertices are

xıyı, Xзyз, X1y1, and x2y2, xâyз, X1y49

as also is geometrically evident.

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

CD AE BF

= + 1.

Let the assumed point be xy, and the vertices y1, X2Y2, X3Y8

=

− y)
Y)

BF X2 (Y3 — Y1) + X3 (Y4 − Y2) + X1 (Y2 — Y3)'

and the truth of the theorem is evident.

40. To find the angle between two lines, whose equations with regard to rectangular co-ordinates are given.

The angle between the lines is manifestly equal to the angle between the perpendiculars on the lines from the origin; if therefore these perpendiculars make with the axis of x the angles a, a', we have (Art. 25)

COR. 1.-The two lines are parallel to each other when

BA' - AB' = 0 (Art. 22),

since then the angle between them vanishes.

COR. 2.-The two lines are perpendicular to each other when
AA' + BB' = 0,

since then the tangent of the angle between them becomes infinite. If the equations of the lines had been given in the form

m'x + b;

since the angle between the lines is the difference of the angles they make with the axis of x, and since (Art. 22) the tangents of these angles are m and m', it follows that the tangent of the re