Suppose OQ=p, and the angle QOX= a. Let r, ◊ be the polar co-ordinates of P; then

OQ= OP cos POQ;

that is, p=r cos (-a).

This is the polar equation to the line.

29. The polar equation may also be derived from the equation referred to rectangular co-ordinates. Let

Ax+By+C=0

be the equation to a line referred to rectangular co-ordinates. Put rcos for x, and r sin for y, Art. 8; thus

30. The equation to a line passing through the origin is, by Art. 14,

Put rcos for x and rsine for y; the equation then becomes

this is therefore the polar equation to a line passing through the origin.

31. We will collect here the different forms of the equation to a straight line which have been investigated,

x= constant, or, y = constant, Arts. 15 and 25.

CHAPTER III.

PROBLEMS ON THE STRAIGHT LINE.

32. WE proceed to apply the results of the preceding articles to the solution of some problems.

To find the form of the equation to a straight line which passes through a given point.

Let x, y, be the co-ordinates of the given point, and

suppose

y = mx + c........

...(1)

to represent the straight line. Since the point (x,, y1) is on the line, its co-ordinates must satisfy (1); hence

33. The equation (3) of the preceding article obviously represents what is required, namely, a line passing through the point (x, y). For the equation is of the first degree in the variables x, y, and therefore, by Art. 16, must represent some straight line. Also the equation is obviously satisfied by the values x=x1, y=y1; that is, the line which the equation represents does pass through the given point. The constant m is the tangent of the angle which the line makes with the axis of x, and by giving a suitable value to m we may make the equation (3) represent any straight line which passes through the assigned point.

The geometrical meaning of equation (3) is obvious. For let AB be any straight line passing through the given point 2. Let P be any point in the line; x, y its co-ordinates. Draw the ordinates PM, QN; and QR parallel to OX; then

which agrees with equation (3).

34. In Art. 32 we eliminated c between the equations (1) and (2) and retained m; we may if we please eliminate m and retain c. From (2)

This equation obviously represents a straight line passing through the given point, because it is an equation of the first degree and is satisfied by the values xx1, y=y1•

=

35. To find the equation to the straight line which passes through two given points.

Let x, y, be the co-ordinates of one given point; x,, Y2 those of the other; suppose the equation to the straight line to be

y = mx + c.........

Since the line passes through (x,, y1) and (x, y2),

Substitute the value of m in (4) and we have for the

(x, — x ̧) (y — y ̧) = (Y2 — Y1) (x — x ̧)......................... (6).

2

2

Some particular cases may here be noted. Suppose Y2=Y1, then (6) becomes (x,-x) (y-y,) = 0, therefore y=y; the required line is thus parallel to the axis of x. Similarly if we suppose x=x, then (6) becomes (y,—y1) (x − x,) = 0, thereforex=x; thus the required line is parallel to the axis of y. Lastly, suppose the point (x, y) to be the origin; hence x=0 and y=0; thus (6) becomes x,y=y,x. The student should illustrate these particular cases by figures.

36. The equation (6) of Art. 35 becomes by reduction

x1y − xу1+x2у1 − x1у2+xу, − x2Y = 0.

If we compare the expression on the left-hand side of this equation with the expression in brackets in equation (2) of