Substituting these values in (3), there results for the required distance

the upper or lower sign to be used according as the numerator is positive or negative. For as d represents the distance of two points, and is not susceptible of opposition of direction, being an absolute magnitude, it is necessary to reject that of the two signs which will give a negative value for d.

EXERCISES.

1. Find the equations of the lines which pass through the following points, taken two and two:

(3, 4), (2,5), (— 6, 7), and (-1, - 2).

2. Find the equation of the line which passes through the point (5, 3), and makes an angle of 60° with the line whose equation is 7y+6x-8= 0.

3. Determine the equation of the line which meets the axis of x at a distance 4 from the origin, and makes an angle of 45° with the line whose equation is 3y-6x+7 = 0.

=

4. Find the equation of the line which meets the axis of y at a distance 3 from the origin, and is perpendicular to the line which passes through the points (3, 4) and (− 5, 2).

5. Find the angle included by the lines whose equations are 2y-5x-7= 0, and 3y+ 6 x 10 = 0.

6. Determine the distance of the point (2, 3), from the line which passes through the points (1, 2) and (6, 7).

OBLIQUE COORDINATES.

Modifications necessary to adapt the preceding results to oblique axes. 19. To find the equation of a straight line in reference to oblique

line referred to oblique axes, a, the angular coefficient, expresses the ratio of the sines of the angles which the line makes with the axes of x and y; and b, as in rectangular coordinates, is the part of the axis of y between the line and the origin. Moreover, if o be the angle which the line makes with the axis of x, and w the angle of ordination, or the angle contained by the axes, then

The forms (2), (3), (4), (5), (6), and (7), of Art. 12 are the same for oblique axes, except the form (2), in which a is the same as in this Article.

Scholium. This Article contains all the modifications necessary to adapt the results in Arts 12... 16, to oblique axes.

20. To find the conditions that the two lines

y = ax + b, and =

y a' x + b',

which are referred to oblique axes inclined at an angle = w, may be parallel or perpendicular to each other.

The condition of parallelism is obviously the same as for rectangular axes, or a = a'. The condition of perpendicularity must be modified

thus:

Let and ' be the angles which the lines make with the axis of x; then by (2), Art. 19, a' sin w 1 + a' cos w

a sin w

tan =

tan e' =

9 1+ a cos w

But because the lines are perpendicular to each other,

2

and hence (Plane Trig., Art. 15),

1

1

tan o

21. To find the perpendicular distance of a given point from a given line in reference to oblique axes.

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

p = ± √ {(x − h)2 + (y − k)2 + 2 (x − h) (y — k) cos w}.... (3). If now we eliminate x and y from these equations (the point B whose coordinates are x and y being common to all), the resulting equation will contain p and known quantities.

Equating the values of y in (1) and (2), and taking a h from each side of the resulting equation, we get

(a + cos w) (k

ah

b)

the upper or lower sign to be used as in the analogous expression for rectangular coordinates (Art. 18).

22. To find the polar equation of a straight line.

Let O be the pole (Art. 7), O C the prime radius, A B the line whose equation is required, and O P = r, angle

=

OB sin (B

or

=

sin B

•)' a sin (8-0)'

which is a relation between the polar co

ordinates of any point in the line, and hence it is the equation required.

INVESTIGATION OF PROPERTIES FOR ILLUSTRATION RELATIVE TO THE POINT AND STRAIGHT LINE.

Rectangular Axes.

23. The coordinates of two points are given to find the coordinates of that point which divides in a given ratio the line joining the two points.

Denote the given points A and B by (a, b), (c, d), and the required point C by (x, y). Then if m n be the given ratio, we have by Art. 11 and the conditions of the question,

or m (x − a) √ { 1 + ( − b ) } = ±n (x−c) √ {1+

(y-d)2
(x − c)2]
axis of x, then,

(1).

d

(2).

Hence by (2) the equation (1) becomes

m (x − a) = ± n (x − c) .....

By (2) and (3) we readily get

(3).

the required coordinates. The upper or lower sign must be used according as the point (x, y) or C is in the given line A B, or its extension. This property is much used in mechanics (Centre of gravity). Cor. If AB be bisected in C, then m =

n, and x = (a+c), y = (b + d),

are the coordinates of the middle point of the line which joins the points (a, b) and (c, d).

24. Two lines BA and AC are perpendicular to each other; B is a given point in BA; R a variable point

in A D; BRP is a right angle; and BR is to RP in a constant ratio 1 to m. Find the locus of P.

B

Since BRP is a right angle, the sum of the angles B RA and PRC is equal to the sum of the angles PRC and RPM. Hence the triangles BRA and PRM are A similar, consequently

R

...

RM: AB:: RP: RB::m: 1, or RM = m AB (1). Also B PB R2+R P2 = BA+ (AM-RM) +PM +RM2.. (2). Whence putting BA y (CA and A B being the axes), we have by Art. 11, and (1) and (2) of the preceding,

a, AM

x, MP

=

(y a)o + x2 = a2 + (x − m a)2 + y2 + m2 a2, or y = m (x —m a).. (3), for the equation of the locus, which (Art. 13) is a straight line. Let y = 0 in (3); then x = ma the line (3) meets the axis of x. passes through the points B and N,

Hence (Art. 17) the line (4) is perpendicular to (3). The locus in question may hence be constructed thus:

=

Make A N = m. BA ma, and draw NP perpendicular to N B, then is N P the locus required.

25. The following theorem may now be proved :

Upon BA, A C, the base and perpendicular of a right-angled triangle ABC, describe squares A F, A K, and join FC, BK; then will the lines F C, BK, intersect on the perpendicular from A on B C.

Take BA and AC for axes, and put AB=c, AC=b, then the points B, K, F, C will be denoted thus

which (Art. 12) is the equation of a straight line passing through the origin A; and as it is formed by the combination of (1) and (2), it also passes through the intersection of the lines FC and B K. Again, the equation of BC is (Art. 16)

Consequently the equation of a perpendicular to this line from the origin A is (Art. 17)

Hence as (3) and (5) are identical, the theorem is established, viz., FC, BK, and the perpendicular from A on BC, intersect in the same point. Scholium.-When the axes are oblique, the mode of investigation is exactly similar. In this case, however, the angular coefficients of the lines and the conditions of perpendicularity must be interpreted by Arts. 19, 20, 21.

EXERCISES ON THE STRAIGHT LINE.

1. Prove that the sum of the perpendiculars on BC (Art. 25), from the points F and K, is equal to BC, and that if BG and C H be drawn, these lines will be parallel.

2. Prove that the perpendiculars from the angles of a triangle on the opposite sides pass through the same point.

3. In a given triangle ABC a variable straight line M N moves parallel to the side B C, so that the points M and N are always on the sides A B, A C. From B and C straight lines BN and CM are drawn to N and M; it is required to prove that the locus of the intersection P of these lines is a straight line passing through A and the middle of B C.

4. In the figure to Euclid, i. 43, let the diagonals AF, HC, and BK be drawn, then will these lines meet in the same point on BK produced. Prove this by coordinates.

5. Express the area of a triangle in terms of the coordinates of its three angles.

6. Three lines AB, AC, AD, which emanate from the same point A, are given in position. In A D any point P is taken, and lines PCF, PE B, are drawn to A B, A C (meeting A B in B, F, and A C in E, C); then if we draw the lines EF, C B, the locus of their intersection O will be a straight line passing through A, however the point P be taken in A D.

7. Prove that the lines joining the angular points and the middle points of the opposite sides of a triangle divide each other in the ratio of 1: 2.