8. When one equation only is given between two variable quantities x and y, to one of them, as x, innumerable values can Be given, and for each value of x there will be a corresponding value of y (real or imaginary), determined by the given equation (Algebra, Art. 133). Now the values given to x may differ as little as possible, and therefore it will be obvious that in general some straight line or curve will be given by the values of y, or by the extremities of the ordinates, in reference to a system of coordinates as in the preceding articles. This straight line or curve is called the locus of the equation. And conversely, the equation which expresses the relation of the abscissa and ordinate for every point of a curve, is called the equation of the curve.

9. Hence if h and k be quantities which being substituted for x and y in an equation between x and y, satisfy that equation; then (h, k) is a point in the locus of that equation. And conversely, if (h, k) be a point in a line or curve represented by an indeterminate equation between x and y; then if h and k be substituted for x and y, the equation will be satisfied. Thus, since the values x 3, y 4, satisfy the equation 5x6y+9 = 0, the point (3, 4) is a point in the locus of the equation 5 x 6y+9=0; and so on.

10. Equations are said to be of different orders according to the highest degree of either of the coordinates (x or y), or the product of these. Thus ax + bx + c = O, is an equation of the first order, ay2+ bxy + c x2 + dy + ex+ƒ = 0, an equation of the second order, etc.

11. To find the distance between two points, and the angle formed by the axis of x and the line which joins these points, the axes being rectangular.

which express the distance and angle of inclination required.

EXERCISES.

N' N

Q

1. Prove that the distance (d) between two points (h, k) and (h', k'), in reference to axes which make an angle O with each other is

=

d = ± √ { (h− h')2 + (k — k')2 + 2 (h—h') (k-k') cos e}.

2. Prove that the distance (d) between two points (r,0) and (r', 0'), in reference to polar coordinates is

12

d = √ { p2 + p12 - 2 rr' cos (0')}.

3. Find the distance (d) between the points (2, 3) and (- 5, 7), and the angle which this line makes with the line of abscissas.

THE STRAIGHT LINE REFERRED TO RECTANGULAR

COORDINATES.

12. To find the equation of a straight line, in reference to rectangular

axes.

अ

B

M

Let A B be the line referred to the rectangular axes Ox and Oy; and P any point in it, the coordinates ON, NP, of which are x and y, respectively. Draw BM parallel to the axis of x; then whatever be the position of P, the ratio PM: BM = tan PBM (Art. 14, Plane Trigonometry) = tan BAO a, suppose. Hence (putting OB = b), PM = PN - MN PN-BO = y — b, and BM = ON

=

=

=

x, therefore

= a, or y = ax + b .... (1),

N

which is a relation between the variable coordinates (x and y) of any point in the line, and therefore it is the equation required (Art. 8).

In general, therefore, the equation of a straight line is of the first degree (Art. 10), and contains two constants a and b; the former is the tangent of the angle which the line makes with the axis of x, that is, the angle formed by the part of the line above the axis of x, and the axis of x itself taken in the positive direction; the latter (b) is the part of the axis of y, intercepted between the line and the origin.*

The student must render himself familiar by frequent and varied exercise with the geometrical signification of all the algebraic circumstances proper to this fundamental equation, which will represent every straight line in the plane by attributing to the arbitrary constants a and b suitable values. The following are modifications of the general form (1), for particular positions of the line :

=

If the line pass through the origin, then b = 0, and its equation is

If it be parallel to the axis of x, and meet the axis of y at a distance

b from the origin; then a = O, and the equation of the line is

denotes a line parallel to the axis of y, at a distance origin.

Again, if the line coincide with the axis of x, then a = hence the axis of x is denoted by the equation

The following additional form of the equation of the line is frequently used :

* The number of constants (a, b), or parameters as they are sometimes called, corresponds, geometrically, to the number of points through which a line must pass, in order that its position may be completely determined. As a is angular, and b linear, a is frequently called the angular, and b the linear coefficient or parameter.

This is the symmetrical form; c and b, it will be seen, are the portions of the axes of x and y, between the line and the origin. The negative value of c merely implies that the intersection of the line with the axis of x is to the left of the origin.

Scholium.-In Art. 6 it is shown that the equations_x = a and y = b give the point P. These are sometimes said to be the equations of the point P. This point may also be represented by the single equation

=

(x − a)2 + (y - b) = 0, or m (x − a)2 + n (y — b)2 = 0, as no other values of x and y, except x = a, y b, will satisfy these equations. The symbols m and n denote any finite quantities.

13. To prove that the locus of the general equation of the first degree between two variables x and y, is a straight line.

The equation of the first degree between two variables x and y is, in its most general form (Art. 10),

Ax+ By + C = 0 ... (1),

in which A, B, C, are independent of x and y.

Let (h, k), (h', k') be any two points in (1); then h, k and k', k', being put for x and y respectively, must (Art. 9) satisfy this equation. Hence Ah+Bk + C = 0, A h' + Bk' + C

Eliminating C between these, we get

= 0.

But if o be the angle which the line joining the points (h, k) and (h', k'), makes with the axis of x, then (Art. 11),

The same result will be obtained by taking any other two points in (1). It follows, therefore, that whatever two points be taken in (1), the line which joins these points makes the same angle with the axis of x, which could not be the case unless (1) were a straight line.

14. To construct a line from its equation in reference to given coordinate axes.

If in the equation of the line we put x = O, and find the value of y in the resulting equation, and again in the original equation put y = 0, and find x; the values of y and x thus found will be, by (7) of Art. 12, the portions of the axes of y and x between the line and the origin. Hence, as a straight line is determined when any two points in it are known, we can, by means of this property, construct any line whose equation is of the form y = ax + b, or Ax+ By + C 0. The points in the axes found in this way must be taken along the positive or negative directions of the axes, according as the value of x in the one case and that of y in the other are positive or negative.

There is one form of the equation of the line to which this method

does not apply, that is, the form (2), Art. 12, or y = ax; for when x = = 0, y = 0, and therefore this merely shows that the line passes through the origin. A second point, however, may be found, by giving to x some determinate value as 1, and then finding the value of y from the resulting equation. The construction of the line whose equation is of the form (3) or (4), Art. 12, will be obvious.

15. The equations of two lines referred to the same coordinate axes being given, to find their point of intersection.

The variable coordinates of the lines (x and y) are identical at their point of intersection,* and hence it will be sufficient and necessary to resolve the two equations for x and y, for the coordinates of the required point.

EXERCISES.

1. Construct the following equations to the same scale:

(4.) 2x+6y ·

9 = 0,

(5.)

5y + 8

=

(6.)

x = 10.

0,

2 Find the angles which the lines (1) and (2) of the preceding make, respectively, with the axis of x.

3. Find the points of intersection, respectively, of the lines (1) and (2), (3) and (4), and (5) and (6), of Ex. 1.

16. To find the equation of a line subject to the condition of passing through one given point, or two given points.

be the equation of the line to be determined, and (h, k) one of the given points; then, because (h, k) is a point in (1), we have, by Art. 9,

This is the condition that (h, k) is a point in (1). If this condition be combined with (1), the resulting equation will be that of a line passing through the point (h, k). Now this can be done by eliminating either a or b between (1) and (2), and as the elimination of b is effected by a simple subtraction, it is the one which is most easily performed. Hence

This is the equation of a line subject to the condition of passing through a given point (h, k).

Again, let (h', k') be another point in the line; then (3) must be satisfied when h' and k' are written for x and y,

which is the equation of a line passing through the two points (h, k) and (h', k').

* It must be kept in mind that, in reference to the equations of two or more lines, a point (x, y) in one is not the same as a point (x, y) in another, except at the point of intersection of the lines, though the x and y seem to be the same in both,

Cor. If m be any arbitrary quantity (Algebra, Art. 97), the equation y − a x − b = m (y — a' x − b') . .

(5),

denotes any line passing through the intersection of the lines

For (5) is a straight line by Art. 13, and as it is satisfied by the simultaneous equations

y-ax

b

=

0, y

a' x

=

0, or y = ax + b, y = a' x + b', which give (Art. 15) the point of intersection of the equations (6), it obviously represents an indefinite number of lines, all passing through the intersection of the equations (6), as m admits of all possible values. 17. To find the conditions that the two lines

y = ax + b, and y a' x + b', may be parallel or perpendicular to each other.

When the lines are parallel, they make equal angles with the axis of x, and hence, by Art. 12,

Again, if the lines be perpendicular to each other, and e, e' be the angles they make with the axis of x, then 0' = 0 + : hence, Art. 15, (Plane Trig.),

The required conditions are contained in (1) and (2).

18. To find the perpendicular distance of a given point from a given line.

be the equation of the given line A B, and (h, k) the given point P. Draw PB perpendicular to A B; then because PB passes through the point (h, k), and is perpendicular to A B, whose equation is (1); its equation by Arts. 16, 17, is

Also, if the point B be denoted by (x, y), and the distance P B by d, we have, by Art. 11,

d = √ {(x − h)2 + (y — k)2 } . . . . . (3).

M

If now x and y be eliminated from these equations (the point B whose coordinates are x and y being common to all), the resulting equation will contain d and known quantities. Equating, then, the values of y in (1) and (2), and then subtracting a h from both sides of the resulting equation, we have

*It will be obvious that every locus whose equation is formed by the combination of the equations (6), in any way whatever, passes through the intersection of these lines; for it is tacitly assumed that a point (x, y) is common to the two lines, and the equation which results from their combination.