then sin p, x is positive, and sin (x, yp, x) is negative; therefore the straight line lies in the direction AZ'.

50. PROB. 2. To find the equation to a straight line in terms of the perpendicular dropped upon it from the origin, and the angles which it forms with the axes.

Let p be the perpendicular, and suppose the symbols p, x and p, y to denote the angles which p makes with the axes of x and of y.

therefore by substitution, we have

x cos p, x + y cos p, y = P,

which is the equation sought.

H

COR. 1. When the line passes through the origin,

x cosp, x + y cos p, y = 0.

COR. 2. Since cos p, x=- - sin

and cos p,y=

we have y sin

p, x,

sin p, y,

Py-x sin

—x sin p, x=p.

51. PROB. 3. To determine the position of the straight , lines, defined by the equation y=ax+b.

The various combinations of sign, of which the equation is susceptible, may be exhibited by writing the equation thus, y= ±(ax+b).

(1) Let y=+ax+b.

The corresponding line has been already considered in Prob. 1. (2) Let y=+ax-b.

Hence, in YA produced, (fig. 29.), take AB'=b, and in AX

take AC

b

= then C'B'z' is the line required.

E

a

(3) Let y-ax+b,

b

then, as before, AB=b, AC′ = And the line required is

C'B2'.

a

Hence, the lines defined by y = ± (ax+b), are CZ, Cz.

Those defined by y= ± (ax − b), are C'Z', C'z'.

CHAP. III.

ON PROBLEMS RELATING TO THE STRAIGHT LINE.

52. PROB. 1. REQUIRED the equation to a straight line passing through a given point, whose co-ordinates are x', y'.

Any point, of which the co-ordinates are x, y, being assumed in the line, we have

y = ax + b,

but since x', y' are also the co-ordinates of a point in the same line, they will satisfy this equation;

:. y'=ax'+b;

whence, subtracting the lower from the upper equation, we have y—y' = a (x − x′),

which is the equation required.

Note. For the sake of brevity, we shall always designate the point whose co-ordinates are x', y', as the point (x', y'), and the straight line whose equation is y = ax+b, as the straight line y=ax+b.

53. PROB. 2. Required the equation to the line which passes through two given points (x', y′), and (x", y′).

Since (x", y") is a point in the line passing through (x', y'), we have, by the last proposition,

Substituting this value of (a) in the equation just referred to,

Since two points fix the position of a straight line, the coefficient (a) is no longer indeterminate as in the last Problem, and we have therefore expressed it in terms of the co-ordinates of the two given points.

COR. 1. The equation may be thus written,

which is of the general form y = ax + b.

COR. 2. If there be three points (x,y): (x′′, y′′) : (x", y'"' in the same straight line, the equation to the line passing through the first two points, is

and that to the line passing through the first, and third, is

Hence, in order that the three points may be in the same line, we must have

or (y" — y') (x"”— x') = (y′′" — y') (x" — x′).

54. PROB. 3. Required the co-ordinates of the point of intersection of two lines

y = ax + b, and y = ax + b'.

The point of intersection being common to the two lines, the

ON THE STRAIGHT LINE.

co-ordinates of that point will be the same in both equations. Hence, subtracting the second equation from the first, we have

Observation. The above method of determining analytically the intersection of two straight lines, is applicable to all lines whatever, situated in the same plane.

Thus, let f(x, y) =0, and F(x, y)=0 be the equations to any two curves, which are supposed to intersect each other; then since the co-ordinates of the points of intersection must satisfy both equations, they will be obtained by elimination. The principle of elimination, thus employed, is of most extensive use in Analytical Geometry.

55. PROB. 4. Required the distance between two points (x', y',) and (x, y).

Let P, P' be two points, (fig. 30.) draw PM, P'M' parallel to AY, and PQ parallel to AX, meeting P'M' in Q.

Let PP'= r.

Then r2 PQ2+P'Q2-2 PQ. P'Q cos Q

=

= PQ2 + P′Q2 +2PQ. P'Q cos x, y

since cos Q=cos (π — x, y): = cos x,y;

but P'Q=M'P' — MP=y'-y, and PQ=x' − x; therefore by substitution,

r2 = (x' —x)2 + (y′—y)2 + 2 (x' —x) (y' —y) cos x,y;