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the same ratio.

Hence prove that the medians of a triangle meet in a point.

Let the coordinates of the vertices A, J5, and G be (x-^, y-^), (arg, 2/2),

and (ajg, y^) respectively.

The coordinates of D are therefore - ^ ^ and - ^^ .

Let G be the point that divides internally AD in the ratio 2 : 1,

and let its coordinates be x and y.

By the last article

2Xiy "T Xq

^_ 2 ; ^ ^1 + 3^2 + ^3

2+1 3

So ^^2/rt^Â±^3.

3

14 COORDINATE GEOMETRY.

In the same manner we could shew that these are th^ coordinates

of the points that divide BE and CF in the ratio 2 : 1.

Since the point whose coordinates are

x-^ + x^ + x^ and ^LÂ±^2+l_3

3 3

lies on each of the lines AD, BE, and CF, it follows that these three

lines meet in a point.

This point is called the Centroid of the triangle.

EXAMPLES. I.

Find the distances between the following pairs of points.

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

3_ ( _ 3, _ 2) and ( - 6, 7), the axes being inclined at 60Â°.

4. (a, o) and (o, 6). 5. {b + c, c + a) and {c + a, a + b).

6. {a cos a, a sin a) and {a cos |S, a sin /3).

7. {am^^, 2ami) and (am^^ 2am^.

8. Lay down in a figure the positions of the points (1, - 3) and

( - 2,1), and prove that the distance between them is 5.

9. Find the value of x^ if the distance between the points [x^^, 2)

and (3, 4) be 8.

10. A line is of length 10 and one end is at the point (2, - 3) ;

if the abscissa of the other end be 10, prove that its ordinate must be

3 or - 9.

11. Prove that the points (2a, 4a), (2a, 6a), and (2a + s/3a, oa)

are the vertices of an equilateral triangle whose side is 2a.

12. Prove that the points (-2, -1), (1, 0), (4, 3), and (1, 2) are

at the vertices of a parallelogram.

13. Prove that the points (2, -2), (8, 4), (5, 7), and (-1, 1) are

at the angular points of a rectangle.

14. Prove that the point ( - xV. f I) is the centre of the circle

circumscribing the triangle whose angular points are (1, 1), (2, 3),

and ( - 2, 2).

Find the coordinates of the point which

15. divides the line joining the points (1, 3) and (2, 7) in the

ratio 3 : 4.

16. divides the same line in the ratio 3 : - 4.

17. divides, internally and externally, the line joining ( - 1, 2)

to (4, - 5) in the ratio 2 : 3.

[EXS. I.] EXAMPLES. 15

18. divides, internally and externally, the line joining ( - 3, - 4)

to ( - 8, 7) in the ratio 7 : 5.

19. The line joining the points (1, - 2) and ( - 3, 4) is trisected ;

find the coordinates of the points of trisection.

20. The line joining the points ( - 6, 8) and (8, - 6) is divided

into four equal parts ; find the coordinates of the points of section.

21. rind the coordinates of the points which divide, internally

and externally, the line joining the point {a + b, a-h) to the point

(a-&, a + 6) in the ratio a : h.

22. The coordinates of the vertices of a triangle are [x-^, 2/i)Â»

{x^, y^ and (xg, y^. The line joining the first two is divided in the

ratio I : h, and the line joining this point of division to the opposite

angular point is then divided in the ratio m : Jfc + Z. Find the

coordinates of the latter point of section.

23. Prove that the coordinates, x and y, of the middle point of

the line joining the point (2,3) to the point (3, 4) satisfy the equation

x-y + l=:0.

24. If G be the centroid of a triangle ABC and O be any other

point, prove that

^{GA^-^GB'^+GCr-) = BG^+GA^ + AB\

and OA^ + OB^-\-OG'^=GA^ + GB'^+GG^- + ^Ga\

25. Prove that the lines joining the middle points of opposite

sides of a quadrilateral and the Une joining the middle points of its

diagonals meet in a point and bisect one another.

26. -4, B, G, D... are n points in a plane whose coordinates are

(^iÂ» 2/i)> (^2' 2/2)' (^3> 2/3)j - -* -^-S is bisected in the point G-^; G^G is

divided at G^ in the ratio 1:2; G^D is divided at G^ in the ratio

1:3; GgE at G^ in the ratio 1 : 4, and so on until all the points are

exhausted. Shew that the coordinates of the final point so obtained are

^1 + ^2 + 3^3+ â€¢â€¢â€¢+^n ^j^^ yi + y^ + Vz+'-'-^Vn

n n

[This point is called the Centre of Mean Position of the n given

points.]

27. Prove that a point can be found which is at the same

distance from each of the four points

(am,, ^) , (aÂ»â€ž ^) , {am,, ^J , and {am^m^, ^^) .

24. To prove that the area of a trapeziitm, i. e. a quad-

rilateral having two sides parallel, is one half the sum of the

two parallel sides multiplied by the perpendicular distance

between them.

16

COOEDINATE GEOMETRY.

B

Let ABGD be the trapezium having the sides AD and

BC parallel.

Join AC and draw AL perpen-

dicular to BG and ON perpendicular

to AD^ produced if necessary.

Since the area of a triangle is one

half the product of any side and the

perpendicular drawn from the opposite angle, we have

area ABGD = ^ABG â– Â¥ ^AGD

= l.BG ,AL + l,AD.GN

=^i{BG + AD) X AL.

25. To find the area of the triangle^ the coordinates of

whose angular 'points are given^ the axes being rectangular.

Let ABG be the triangle

and let the coordinates of its

angular points A, B and G be

{x^, 2/1), (a?2, 2/2), and {x^, y^).

Draw AL, BM, and Ci\^ per-

pendicular to the axis of x, and

let A denote the required area.

Then

A == trapezium A LNG + trapezium GNMB â€” trapezium A LMB

= \LN {LA + NG) + \NM {NG + MB) - \LM (LA + MB),

by the last article,

= i [(^3 - ^1) (2/1 + ys) + {002 - ^z) (2/2 + 2/3) - (^2 - a^i) {Vi + 2/2)]-

On simplifying we easily have

^ ^ I (^172 - XaYi + y^zYz - ^372 + XgYi - x^yg),

or the equivalent form

^ = J [^1 (2/2 - Vz) + ^2 (2/3 - 2/1) + ^3 (2/1 - 2/2)].

If we use the determinant notation this may be written

(as in Art. 5)

^1) 2/1) ^

^2> 2/2? â– '-

, ^3j 2/35 ^

Cor. The area of the triangle whose vertices are the

origin (0, 0) and the points {x^, y-^, {x^, 2/2) is J (^12/2 â€” ^'22/1)*

AREA OF A QUADRILATERAL,

17

isin w {x^y^

I.e.

fsm wx

26. In the preceding article, if the axes be oblique, the perpen-

diculars AL, BM, and CN, are not equal to the ordinates y^ , y^ , and

2/3, but are equal respectively to yi sin w, 2/2 sin w, and y^ sin w.

The area of the triangle in this case becomes

xu yi, 1

27. In order that the expression for the area in Art. 25 may be

a positive quantity (as all areas necessarily are) the points A, B, and

G must be taken in the order in which they would be met by a

person starting from A and walking round the triangle in such a

manner that the area of the triangle is always on his left hand.

Otherwise the expressions of Art. 25 would be found to be negative.

28. To find the area of a quadrilateral the coordinates

of whose angular foints are given.

Let the angular points of the quadrilateral, taken in

order, be A^ B, C, and D, and let their coordinates be

respectively {x^, y^\ (x.^,, y^), (x^, y^\ and {x^, y^).

Draw ALy BM, CJV, and DH perpendicular to the axis

of X.

Then the area of the quadrilateral

= trapezium ALRD + trapezium BRNO + trapezium GNMB

â€” trapezium ALMB

= ILR {LA + RD) + IRN (RD + NG) + ^NM {JVC + MB)

-lLM{LA-\-MB)

{(^â– 4 - ^1) (2/1 + 2/4) + (^3 - ^4) (2/3 + 2/4) + (^2 - ^^) (2/3 + 2/2)

- (a?2 - x^) (2/1 + 2/2)}

{(^12/2 - Â«^22/i) + fez^s - ^zvi) + {xsy4. - ^42/3) + (^42/i - ^y^)}'

L. 2

_ 1

_ 1

18 COORDINATE GEOMETRY.

29. The above formula may also be obtained by

drawing the lines OA, OB, OC and OD. For the quadri-

lateral ABCn

= AOBG+ AOCD- aOBA- AOAD.

But the coordinates of the vertices of the triangle OBG

are (0, 0), (ajg, 2/2) ^^^ (^35 2/3) ^ hence, by Art. 25, its

area is ^ i^^y-^ â€” ^zV^)-

So for the other triangles.

The required area therefore

^ h [(Â«^22/3 - a^32/2) + {^zVa, - ^m) - (^Wi - ^12/2) - {^i2/4 - ^'42/l)l

= i [{^^2 - ^22/1) + (^22/3 - ^32/2) + {^3y4 - ^m) + (^42/l - Â«^l2/4)]-

In a similar manner it may be shewn that the area

of a polygon of n sides the coordinates of whose angular

points, taken in order, are

(^IJ 2/1/5 V^2) 2/2/3 (^35 2/3)? â€¢â€¢â€¢(.'^)i3 2/ra/

is i [(a?i2/2 - a?22/i) + (^22/3 - ^32/2) + â€¢ â€¢ â€¢ + (^n2/i - a'l^/ri)]-

EXAMPLES. II.

Find the areas of the triangles the coordinates of whose angular

points are respectively

1. (1, 3), ( - 7, 6) and (5, - 1). 2. (0, 4), (3, 6) and ( - 8, - 2).

3. (5,2), (-9, -3) and (-3, -5).

4. {a, l> + c), (a, h-c) and {-a, c).

5. {a,c + a), {a, c) and {-a, c-a).

6. {a cos (pi, b sin <p-^), {a cos ^^, b sin ^g) and (a cos ^3, b sin ^g).

7. (a7?ij2^ 2a7?i;^), [arn^^ lam^ and [am^, 2avi^.

8. {awiimg, a(??ii + ??i2)}, {aWaWg, a (7?i2 + %)} ^-^cl

9. lam-,. â€” I , \am^, â€” y and -Jawo, â€” [ .

Prove (by shewing that the area of the triangle formed by them is

zero) that the following sets of three points are in a straight line :

10. (1,4), (3, -2), and (-3,16).

11. (-i, 3), (-5,6), and (-8,8).

12. (Â«. b + c), (6, c + a), and (c, a + b).

[EXS. II.] POLAR COORDINATES. 19

Find the areas of the quadrilaterals the coordinates of whose

angular points, taken in order, are

13. (1,1), (3,4), (5, -2), and (4, -7).

14. (-1, 6), (-3, -9), (5, -8), and (3, 9).

15. If be the origin, and if the coordinates of any two points

Pj and Pg ^6 respectively (%, y^ and [x^, y.^, prove that

OP^ . OP2 . cos P1OP2 = x-^Xc^ + y-^y^ â€¢

30. Polar Coordinates. There is another method,

which is often used, for determining the position of a point

in a plane.

Suppose to be a fixed point, called the origin or

pole, and OX a fixed line, called the initial line.

Take any other point P in the plane of the paper and

join OP. The position of P is clearly known when the

angle XOP and the length OP are given.

[For giving the angle XOP shews the direction in which OP is

drawn, and giving the distance OP tells the distance of P along this

direction.]

The angle XOP which would be traced out by the line

OP in revolving from the initial line OX is called the

vectorial angle of P and the length OP is called its radius

vector. The two taken together are called the polar co-

ordinates of P.

If the vectorial angle be and the radius vector be r, the

position of P is denoted by the symbol (r, 0).

The radius vector is positive if it be measured from the

origin along the line bounding the vectorial angle; if

measured in the opposite direction it is negative.

31. Ex. Construct the positions of the 'points (i) (2, 80Â°),

(ii) (3, 150Â°), (iii) (-2, 45Â°), (iv) â–

(-3, 330Â°), (v) (3, -210Â°) and (vi) ff\ /^

(-3, -30Â°). ^\^ ,->^^

(i) To construct the first point, ^^^^^^^^^"^^

let the radius vector revolve from '/^^^ yC

OX through an angle of 30Â°, and >/ ""-.,

then mark off along it a distance y ''-â€¢.,

equal to two units of length. We 'p M"

thus obtain the point P^. ^

(ii) For the second point, the radius vector revolves from OX

through 150Â° and is then in the position OP^ ; measuring a distance 3

along it we arrive at Pg .

2â€”2

20 COORDINATE GEOMETKY.

(iii) For the third point, let the radius vector revolve from OX

through 45Â° into the position OL. We have now to measure along

OL a distance - 2, i.e. we have to measure a distance 2 not along OL

but in the opposite direction. Producing iO to Pg, so that OP3 is

2 units of length, we have the required point P3.

(iv) To get the fourth point, we let the radius vector rotate from

OX through 330Â° into the position OM and measure on it a distance

-3, i.e. 3 in the direction MO produced. We thus have the point P^y

which is the same as the point given by (ii).

(v) If the radius vector rotate through - 210Â°, it will be in the

position OP2, and the point required is Pg.

(vi)^ For the sixth point, the radius vector, after rotating through

- 30Â°, is in the position OM: We then measure - 3 along it, i.e. 3 in

the direction MO produced, and once more arrive at the point Pg.

32. It will be observed that in the previous example

the same point P^ is denoted by each of the four sets of

polar coordinates

(3, 150Â°), (-3, 330Â°), (3, -210Â°) and (-3, -30Â°).

In general it v^ill be found that the same point is given

by each of the polar coordinates

(r, 0), (- r, 180Â° + 6), {r, - (360Â° - 6)] and {- r, - (180Â° - 6%

or, expressing the angles in radians, by each of the co-

ordinates

(r, e\ {-r,7r + 6), {r, - (27r - 0)} and {- r, - (tt - $)}.

It is also clear that adding 360Â° (or any multiple of

360Â°) to the vectorial angle does not alter the final position

of the revolving line, so that {r, 6) is always the same point

as (r, ^ + ?i . 360Â°), where n is an integer.

So, adding 180Â° or any odd multiple of 180Â° to the

vectorial angle and changing the sign of the radius vector

gives the same point as before. Thus the point

[-r, ^ + (2n + 1)180Â°]

is the same point as [â€” r, 6 + 180Â°], i.e. is the point [r, 6\

33. To find the length of the straight line joining two

points whose polar coordinates are given.

Let A and B be the two points and let their polar

coordinates be (r^, 6y) and (r^, 6^ respectively, so that

OA^r^, OB = r^, lXOA^O^, and lX0B = 6^,

POLAR COORDINATES.

21

Then (Trigonometry, Art. 164)

AB" - OA'' + OB'' -20 A. OB cos AOB

= r-^ + r^ - 2r-^r^ cos {0^ - 6^.

34. To find the area of a triangle the coordinates of

whose angular points are given.

Let ABC be the triangle and let (r-^, 0^), (r^, 62), and

(rg, ^3) be the polar coordinates of

its angular points.

We have

AABO=AOBC+aOCA

-AOBA (1).

Now

A0BC = i0B,0C sin BOC

[Trigonometry, Art. 198]

= ^r^r^ sin (^3 - $^).

' So A OCA = \0G . OA sin CO A = ^r^r, sin (6, - 6,),

and AOAB^^OA. OB sin AOB = ^r^r^ sin {6^ - 6.^

= - Jn^2 sin (^2 - ^1).

Hence (1) gives

A ABC = J \r<^r^ sin (^3 - 6^ + r^r^ (sin 0-^ - 0^)

+ r^r^ sin {Oo - 0^)].

35. To change from Cartesian Coordinates to Polar

Coordinates, and conversely.

Let P be any point whose Cartesian coordinates, referred

to rectangular axes, are x and y,

and whose polar coordinates, re-

ferred to as pole and OX as

initial line, are (r, 6).

Draw Pit/'perpendicular to OX

so that we have

OM=x, MP = y, LMOP = e,

and OP = r.

From the triangle MOP we

have

x = OM=OPcosMOP = rcosO (1),

y = MP= OPsinMOP ^r smO (2),

r=OP= sJOM^ + MP^^ s/x" + y' (3),

X' O:

22 COORDINATE GEOMETRY,

and

Equations (1) and (2) express the Cartesian coordinates

in terms of the polar coordinates.

Equations (3) and (4) express the polar in terms of the

Cartesian coordinates.

The same relations will be found to hold if P be in any-

other of the quadrants into which the plane is divided by

XOX' and YOT.

Ex. Change to Cartesian coordinates the equations

{!) r = asind, and (2)r=a^cos-.

a

(1) Multiplying the equation by r, it becomes r^rrar sin Q,

i.e. by equations (2) and (3), x^-\-y^ = ay.

(2) Squaring the equation (2), it becomes

r=acos2- = - (1 + cos^),

i. e. 2^2 = ar + ar cos 6,

i.e. 2{x^ + y^) = a sjx^ + y^-\- ax,

i.e. {2x^ + 2y^-ax)^ = a^{x^ + y^).

EXAMPLES. III.

Lay down the positions of the points whose polar coordinates are

L (3,45Â°). 2. (-2, -60Â°). 3. (4,135Â°). 4. (2,330Â°).

5. (-1, -180Â°). 6. (1, -210Â°). 7. (5, -675Â°). 8. (Â«> |) â€¢

9. (2a.-^). 10.{-a,l). n.(-2a.4').

Find the lengths of the straight lines joining the pairs of points

whose polar coordinates are

12. (2, 30Â°) and (4, 120Â°). 13. (-3, 45Â°) and (7, 105Â°).

14. ( Â«> I ) and

(Â«"'!) â€¢

[EXS. III.] EXAMPLES. 23

15. Prove that the points (0, 0), (3, |) , and ( 3, ^ J form an equi-

lateral triangle.

Find the areas of the triangles the coordinates of whose angular

points are

16. (1, 30Â°), (2, 60Â°), and (3, 90Â°).

17. (_3, -30Â°), (5, 150Â°), and (7, 210Â°).

18. (-^. i)' (Â«'i)'^^*l(-2Â«. -y)-

Find the polar coordinates (drawing the figure in each case) of the

points

19. x = J3, y = l. 20. x=-^B, y = l. 21. x=-l, y = l.

Find the Cartesian coordinates (drawing a figure in each case) of

the points whose polar coordinates are

22. (5, I). 23. (-6, I). 24.(5,-^).

Change to polar coordinates the equations

25. x^+y^=a\ 26. y = xtana. 27. x^ + y^ = 2ax.

28. x^-y^=2ay. 29. x^=y^{2a-x). 30. {x^ + y^)^ = a^{x^-y^).

Transform to Cartesian coordinates the equations

31. r=a. 32. ^ = tan-i??i. 33. r = acos^.

34. r = asin2^. 35. ?-2 = a2cos2^. 36. o'^ sin 2d =2a^.

e

2

40. 'T (cos 3^ + sin 3^) = 5h sin 6 cos d

37. r'^G0s2d = a^. 38. r^cos- = a^, 39^ r^=ai sin-.

2i a

CHAPTER III.

LOCUS. EQUATION TO A LOCUS.

36. When a point moves so as always to satisfy a

given condition, or conditions, the path it traces out is

called its Locus under these conditions.

For example, suppose to be a given point in the plane

of the paper and that a point F is to move on the paper so

that its distance from shall be constant and equal to a.

It is clear that all the positions of the moving point must

lie on the circumference of a circle whose centre is and

whose radius is a. The circumference of this circle is

therefore the " Locus" of P when it moves subject to the

condition that its distance from shall be equal to the

constant distance a.

37. Again, suppose A and B to be two fixed points in

the plane of the paper and that a point P is to move in

the plane of the paper so that its distances from A and B

are to be always equal. If we bisect AB in G and through

it draw a straight line (of infinite length in both directions)

perpendicular to AB, then any point on this straight line

is at equal distances from A and B. Also there is no

point, whose distances from A and B are the same, which

does not lie on this straight line. This straight line is

therefore the "Locus" of P subject to the assumed con-

dition.

38. Again, suppose A and B to be two fixed points

and that the point P is to move in the plane of the paper

so that the angle APB is always a right angle. If we

describe a circle on AB as diameter then P may be any

EQUATION TO A LOCUS.

25

point on the circumference of this circle, since the angle

in a semi-circle is a right angle; also it could easily be

shewn that APB is not a right angle except when P lies

on this circumference. The "Locus" of P under the

assumed condition is therefore a circle on -4^ as diameter.

39. One single equation between two unknown quan-

tities x and y, e.g.

x + y = l (1),

cannot completely determine the values of x and y.

*\P6

\Q

vPs

^1?

\1

M

OM

Vs

\ P

Such an equation has an infinite number of solutions.

Amongst them are the followins: :

a: = 0,1

a; =1,1

2/ = 0/'

x= 2,1

y=-l/'

x= 3,1

y=-2/'-

x = â€”\,\ X â€” â€” 2

2/= 2/' y= 3

Let us mark down on paper a number of points whose

coordinates (as defined in the last chapter) satisfy equation

Let OX and OF be the axes of coordinates.

If we mark off a distance OPi (â€”1) along OY^ we have

a point Pi whose coordinates (0, 1) clearly satisfy equation

If we mark off a distance OP^ (=1) along OX, we have

a point Pg whose coordinates (1, 0) satisfy (1).

26 COORDINATE GEOMETRY.

Similarly the point Pg, (2, â€” 1), and P4, (3, â€” 2), satisfy

the equation (1).

Again, the coordinates (â€”1, 2) of Pg and the coordinates

(â€”2, 3) of Pg satisfy equation (1).

On making the measurements carefully we should find

that all the points we obtain lie on the line P^P^ (produced

both ways).

Again, if we took any point Q, lying on P^P^, and draw

a perpendicular QM to OX, we should find on measurement

that the sum of its x and y (each taken with its proper

sign) would be equal to unity, so that the coordinates of Q

would satisfy (1).

Also we should find no point, whose coordinates satisfy

(1), which does not lie on P1P2.

All the points, lying on the straight line P1P2, and no

others are therefore such that their coordinates satisfy the

equation (1).

This result is expressed in the language of Analytical

Geometry by saying that (1) is the Equation to the Straight

Line P^P^.

40. Consider again the equation

Â£c2 + 2/2 = 4 (1).

Amongst an infinite number of solutions of this equa-

tion are the following :

x = 2A x= J?>\ x = J\ x^l \

V3| x = J'I\ x=^i \

2/=2r 2/=x/3r 2/=V2 r y=i r

x = -'2,\ x^-.^?>,\ x = -J2A x = -l, \

2/^0 r 2/^ - i /' 3/=-v2r y=-si^)'

53 = 0, \ x=l, \ x=J2, } x=J3)

2/ = -2j ' y

L and

V3/' 2/ = -./2/'-'"% = -l/-

EQUATION TO A LOCUS.

27

1 â€¢

,5i

â– i|f

-,..

M ;f^ X

All these points are respectively represented by the

points P^, P^^ F^y ... P^Qj and they

will all be found to lie on the

dotted circle whose centre is

and radius is 2.

Also, if we take any other

point Q on this circle and its

ordinate QM, it follows, since

0M^- + MQ^^0Q' = 4:, that the x

and y of the point Q satisfies (1).

The dotted circle therefore

passes through all the points whose

coordinates satisfy (1).

In the language of Analytical Geometry the equation

(1) is therefore the equation to the above circle.

41. As another example let us trace the locus of the

point whose coordinates satisfy the equation

y'=4^'^ (1);

If we give x a negative value we see that y is im-

possible; for the square of a

real quantity cannot be nega-

tive.

We see therefore that there

are no points lying to the left

of OF.

If we give x any positive

value we see that y has two

real corresponding values which

are equal and of opposite signs.

The following values,

amongst an infinite number of

others, satisfy (1), viz.

x = 0,] x=l, "I x = 2,

y = 0}' y = + 2or-2}' y = 2J2ov -2J2

x=4: I cc=16, I aj = + oo, )

y = + 4: or â€”4:) ^ '" y â€” Sor-Sj' '" y = + (X)Ovâ€”ao)'

The origin is the first of these points and P^ and Qi,

Pg and Q^, P^and Q^, ... represent the next pairs of points.

h

28 COORDINATE GEOMETRY.

If we took a large number of values of x and the

corresponding values of ?/, the points thus obtained would

be found all to lie on the curve in the figure.

Both of its branches would be found to stretch away to

infinity towards the right of the figure.

Also, if we took any point on this curve and measured

with sufficient accuracy its x and y the values thus obtained

would be found to satisfy equation (1).

Also we should not be able to find any point, not lying

on the curve, whose coordinates would satisfy (1).

In the language of Analytical Geometry the equation

(1) is the equation to the above curve. This curve is called

a Parabola and will be fully discussed in Chapter X.

42. If a point move so as to satisfy any given condition

it will describe some definite curve, or locus, and there can

always be found an equation between the x and y of any

point on the path.

This equation is called the equation to the locus or

curve. Hence

Def. Equation to a curve. The equation to a

curve is the relation which exists between the coordinates of

any foint on the curve^ and which holds for no other points

except those lying on the curve.

43. Conversely to every equation between x and y it

will be found that there is, in general, a definite geometrical

locus.

Thus in Art. 39 the equation is x + y=\, and the

definite path, or locus, is the straight line P^P^ (produced

indefinitely both ways).

In Art. 40 the equation is x'^ + y'^^ 4, and the definite

path, or locus, is the dotted circle.

Again the equation 2/ = 1 states that the moving point

is such that its ordinate is always unity, i.e. that it is

always at a distance 1 from the axis of x. The definite

path, or locus, is therefore a straight line parallel to OX

and at a distance unity from it.

EQUATION TO A LOCUS. 29

44. In the next chapter it will be found that if the

equation be of the first degree {i.e. if it contain no

products, squares, or higher powers of x and y) the locus

corresponding is always a straight line.

If the equation be of the second or higher degree, the

corresponding locus is, in general, a curved line.

45. We append a few simple examples of the forma-

tion of the equation to a locus.

Ex. 1. A point moves so that the algebraic sum of its distances

from tioo given perpendicular axes is equal to a constant quantity a;

find the equation to its locus.

Take the two straight lines as the axes of coordinates. Let {x, y)

be any point satisfying the given condition. We then ha,wex + y = a.

This being the relation connecting the coordinates of any point

on the locus is the equation to the locus.

It will be found in the next chapter that this equation represents

a straight line.

Ex. 2. The sum of the squares of the distances of a moving point

from the tioo fixed points {a, 0) and {-a, 0) is equal to a constant

quantity 2c^. Find the equation to its locus.

Let (a;, y) be any position of the moving point. Then, by Art. 20,

the condition of the question gives

Hence prove that the medians of a triangle meet in a point.

Let the coordinates of the vertices A, J5, and G be (x-^, y-^), (arg, 2/2),

and (ajg, y^) respectively.

The coordinates of D are therefore - ^ ^ and - ^^ .

Let G be the point that divides internally AD in the ratio 2 : 1,

and let its coordinates be x and y.

By the last article

2Xiy "T Xq

^_ 2 ; ^ ^1 + 3^2 + ^3

2+1 3

So ^^2/rt^Â±^3.

3

14 COORDINATE GEOMETRY.

In the same manner we could shew that these are th^ coordinates

of the points that divide BE and CF in the ratio 2 : 1.

Since the point whose coordinates are

x-^ + x^ + x^ and ^LÂ±^2+l_3

3 3

lies on each of the lines AD, BE, and CF, it follows that these three

lines meet in a point.

This point is called the Centroid of the triangle.

EXAMPLES. I.

Find the distances between the following pairs of points.

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

3_ ( _ 3, _ 2) and ( - 6, 7), the axes being inclined at 60Â°.

4. (a, o) and (o, 6). 5. {b + c, c + a) and {c + a, a + b).

6. {a cos a, a sin a) and {a cos |S, a sin /3).

7. {am^^, 2ami) and (am^^ 2am^.

8. Lay down in a figure the positions of the points (1, - 3) and

( - 2,1), and prove that the distance between them is 5.

9. Find the value of x^ if the distance between the points [x^^, 2)

and (3, 4) be 8.

10. A line is of length 10 and one end is at the point (2, - 3) ;

if the abscissa of the other end be 10, prove that its ordinate must be

3 or - 9.

11. Prove that the points (2a, 4a), (2a, 6a), and (2a + s/3a, oa)

are the vertices of an equilateral triangle whose side is 2a.

12. Prove that the points (-2, -1), (1, 0), (4, 3), and (1, 2) are

at the vertices of a parallelogram.

13. Prove that the points (2, -2), (8, 4), (5, 7), and (-1, 1) are

at the angular points of a rectangle.

14. Prove that the point ( - xV. f I) is the centre of the circle

circumscribing the triangle whose angular points are (1, 1), (2, 3),

and ( - 2, 2).

Find the coordinates of the point which

15. divides the line joining the points (1, 3) and (2, 7) in the

ratio 3 : 4.

16. divides the same line in the ratio 3 : - 4.

17. divides, internally and externally, the line joining ( - 1, 2)

to (4, - 5) in the ratio 2 : 3.

[EXS. I.] EXAMPLES. 15

18. divides, internally and externally, the line joining ( - 3, - 4)

to ( - 8, 7) in the ratio 7 : 5.

19. The line joining the points (1, - 2) and ( - 3, 4) is trisected ;

find the coordinates of the points of trisection.

20. The line joining the points ( - 6, 8) and (8, - 6) is divided

into four equal parts ; find the coordinates of the points of section.

21. rind the coordinates of the points which divide, internally

and externally, the line joining the point {a + b, a-h) to the point

(a-&, a + 6) in the ratio a : h.

22. The coordinates of the vertices of a triangle are [x-^, 2/i)Â»

{x^, y^ and (xg, y^. The line joining the first two is divided in the

ratio I : h, and the line joining this point of division to the opposite

angular point is then divided in the ratio m : Jfc + Z. Find the

coordinates of the latter point of section.

23. Prove that the coordinates, x and y, of the middle point of

the line joining the point (2,3) to the point (3, 4) satisfy the equation

x-y + l=:0.

24. If G be the centroid of a triangle ABC and O be any other

point, prove that

^{GA^-^GB'^+GCr-) = BG^+GA^ + AB\

and OA^ + OB^-\-OG'^=GA^ + GB'^+GG^- + ^Ga\

25. Prove that the lines joining the middle points of opposite

sides of a quadrilateral and the Une joining the middle points of its

diagonals meet in a point and bisect one another.

26. -4, B, G, D... are n points in a plane whose coordinates are

(^iÂ» 2/i)> (^2' 2/2)' (^3> 2/3)j - -* -^-S is bisected in the point G-^; G^G is

divided at G^ in the ratio 1:2; G^D is divided at G^ in the ratio

1:3; GgE at G^ in the ratio 1 : 4, and so on until all the points are

exhausted. Shew that the coordinates of the final point so obtained are

^1 + ^2 + 3^3+ â€¢â€¢â€¢+^n ^j^^ yi + y^ + Vz+'-'-^Vn

n n

[This point is called the Centre of Mean Position of the n given

points.]

27. Prove that a point can be found which is at the same

distance from each of the four points

(am,, ^) , (aÂ»â€ž ^) , {am,, ^J , and {am^m^, ^^) .

24. To prove that the area of a trapeziitm, i. e. a quad-

rilateral having two sides parallel, is one half the sum of the

two parallel sides multiplied by the perpendicular distance

between them.

16

COOEDINATE GEOMETRY.

B

Let ABGD be the trapezium having the sides AD and

BC parallel.

Join AC and draw AL perpen-

dicular to BG and ON perpendicular

to AD^ produced if necessary.

Since the area of a triangle is one

half the product of any side and the

perpendicular drawn from the opposite angle, we have

area ABGD = ^ABG â– Â¥ ^AGD

= l.BG ,AL + l,AD.GN

=^i{BG + AD) X AL.

25. To find the area of the triangle^ the coordinates of

whose angular 'points are given^ the axes being rectangular.

Let ABG be the triangle

and let the coordinates of its

angular points A, B and G be

{x^, 2/1), (a?2, 2/2), and {x^, y^).

Draw AL, BM, and Ci\^ per-

pendicular to the axis of x, and

let A denote the required area.

Then

A == trapezium A LNG + trapezium GNMB â€” trapezium A LMB

= \LN {LA + NG) + \NM {NG + MB) - \LM (LA + MB),

by the last article,

= i [(^3 - ^1) (2/1 + ys) + {002 - ^z) (2/2 + 2/3) - (^2 - a^i) {Vi + 2/2)]-

On simplifying we easily have

^ ^ I (^172 - XaYi + y^zYz - ^372 + XgYi - x^yg),

or the equivalent form

^ = J [^1 (2/2 - Vz) + ^2 (2/3 - 2/1) + ^3 (2/1 - 2/2)].

If we use the determinant notation this may be written

(as in Art. 5)

^1) 2/1) ^

^2> 2/2? â– '-

, ^3j 2/35 ^

Cor. The area of the triangle whose vertices are the

origin (0, 0) and the points {x^, y-^, {x^, 2/2) is J (^12/2 â€” ^'22/1)*

AREA OF A QUADRILATERAL,

17

isin w {x^y^

I.e.

fsm wx

26. In the preceding article, if the axes be oblique, the perpen-

diculars AL, BM, and CN, are not equal to the ordinates y^ , y^ , and

2/3, but are equal respectively to yi sin w, 2/2 sin w, and y^ sin w.

The area of the triangle in this case becomes

xu yi, 1

27. In order that the expression for the area in Art. 25 may be

a positive quantity (as all areas necessarily are) the points A, B, and

G must be taken in the order in which they would be met by a

person starting from A and walking round the triangle in such a

manner that the area of the triangle is always on his left hand.

Otherwise the expressions of Art. 25 would be found to be negative.

28. To find the area of a quadrilateral the coordinates

of whose angular foints are given.

Let the angular points of the quadrilateral, taken in

order, be A^ B, C, and D, and let their coordinates be

respectively {x^, y^\ (x.^,, y^), (x^, y^\ and {x^, y^).

Draw ALy BM, CJV, and DH perpendicular to the axis

of X.

Then the area of the quadrilateral

= trapezium ALRD + trapezium BRNO + trapezium GNMB

â€” trapezium ALMB

= ILR {LA + RD) + IRN (RD + NG) + ^NM {JVC + MB)

-lLM{LA-\-MB)

{(^â– 4 - ^1) (2/1 + 2/4) + (^3 - ^4) (2/3 + 2/4) + (^2 - ^^) (2/3 + 2/2)

- (a?2 - x^) (2/1 + 2/2)}

{(^12/2 - Â«^22/i) + fez^s - ^zvi) + {xsy4. - ^42/3) + (^42/i - ^y^)}'

L. 2

_ 1

_ 1

18 COORDINATE GEOMETRY.

29. The above formula may also be obtained by

drawing the lines OA, OB, OC and OD. For the quadri-

lateral ABCn

= AOBG+ AOCD- aOBA- AOAD.

But the coordinates of the vertices of the triangle OBG

are (0, 0), (ajg, 2/2) ^^^ (^35 2/3) ^ hence, by Art. 25, its

area is ^ i^^y-^ â€” ^zV^)-

So for the other triangles.

The required area therefore

^ h [(Â«^22/3 - a^32/2) + {^zVa, - ^m) - (^Wi - ^12/2) - {^i2/4 - ^'42/l)l

= i [{^^2 - ^22/1) + (^22/3 - ^32/2) + {^3y4 - ^m) + (^42/l - Â«^l2/4)]-

In a similar manner it may be shewn that the area

of a polygon of n sides the coordinates of whose angular

points, taken in order, are

(^IJ 2/1/5 V^2) 2/2/3 (^35 2/3)? â€¢â€¢â€¢(.'^)i3 2/ra/

is i [(a?i2/2 - a?22/i) + (^22/3 - ^32/2) + â€¢ â€¢ â€¢ + (^n2/i - a'l^/ri)]-

EXAMPLES. II.

Find the areas of the triangles the coordinates of whose angular

points are respectively

1. (1, 3), ( - 7, 6) and (5, - 1). 2. (0, 4), (3, 6) and ( - 8, - 2).

3. (5,2), (-9, -3) and (-3, -5).

4. {a, l> + c), (a, h-c) and {-a, c).

5. {a,c + a), {a, c) and {-a, c-a).

6. {a cos (pi, b sin <p-^), {a cos ^^, b sin ^g) and (a cos ^3, b sin ^g).

7. (a7?ij2^ 2a7?i;^), [arn^^ lam^ and [am^, 2avi^.

8. {awiimg, a(??ii + ??i2)}, {aWaWg, a (7?i2 + %)} ^-^cl

9. lam-,. â€” I , \am^, â€” y and -Jawo, â€” [ .

Prove (by shewing that the area of the triangle formed by them is

zero) that the following sets of three points are in a straight line :

10. (1,4), (3, -2), and (-3,16).

11. (-i, 3), (-5,6), and (-8,8).

12. (Â«. b + c), (6, c + a), and (c, a + b).

[EXS. II.] POLAR COORDINATES. 19

Find the areas of the quadrilaterals the coordinates of whose

angular points, taken in order, are

13. (1,1), (3,4), (5, -2), and (4, -7).

14. (-1, 6), (-3, -9), (5, -8), and (3, 9).

15. If be the origin, and if the coordinates of any two points

Pj and Pg ^6 respectively (%, y^ and [x^, y.^, prove that

OP^ . OP2 . cos P1OP2 = x-^Xc^ + y-^y^ â€¢

30. Polar Coordinates. There is another method,

which is often used, for determining the position of a point

in a plane.

Suppose to be a fixed point, called the origin or

pole, and OX a fixed line, called the initial line.

Take any other point P in the plane of the paper and

join OP. The position of P is clearly known when the

angle XOP and the length OP are given.

[For giving the angle XOP shews the direction in which OP is

drawn, and giving the distance OP tells the distance of P along this

direction.]

The angle XOP which would be traced out by the line

OP in revolving from the initial line OX is called the

vectorial angle of P and the length OP is called its radius

vector. The two taken together are called the polar co-

ordinates of P.

If the vectorial angle be and the radius vector be r, the

position of P is denoted by the symbol (r, 0).

The radius vector is positive if it be measured from the

origin along the line bounding the vectorial angle; if

measured in the opposite direction it is negative.

31. Ex. Construct the positions of the 'points (i) (2, 80Â°),

(ii) (3, 150Â°), (iii) (-2, 45Â°), (iv) â–

(-3, 330Â°), (v) (3, -210Â°) and (vi) ff\ /^

(-3, -30Â°). ^\^ ,->^^

(i) To construct the first point, ^^^^^^^^^"^^

let the radius vector revolve from '/^^^ yC

OX through an angle of 30Â°, and >/ ""-.,

then mark off along it a distance y ''-â€¢.,

equal to two units of length. We 'p M"

thus obtain the point P^. ^

(ii) For the second point, the radius vector revolves from OX

through 150Â° and is then in the position OP^ ; measuring a distance 3

along it we arrive at Pg .

2â€”2

20 COORDINATE GEOMETKY.

(iii) For the third point, let the radius vector revolve from OX

through 45Â° into the position OL. We have now to measure along

OL a distance - 2, i.e. we have to measure a distance 2 not along OL

but in the opposite direction. Producing iO to Pg, so that OP3 is

2 units of length, we have the required point P3.

(iv) To get the fourth point, we let the radius vector rotate from

OX through 330Â° into the position OM and measure on it a distance

-3, i.e. 3 in the direction MO produced. We thus have the point P^y

which is the same as the point given by (ii).

(v) If the radius vector rotate through - 210Â°, it will be in the

position OP2, and the point required is Pg.

(vi)^ For the sixth point, the radius vector, after rotating through

- 30Â°, is in the position OM: We then measure - 3 along it, i.e. 3 in

the direction MO produced, and once more arrive at the point Pg.

32. It will be observed that in the previous example

the same point P^ is denoted by each of the four sets of

polar coordinates

(3, 150Â°), (-3, 330Â°), (3, -210Â°) and (-3, -30Â°).

In general it v^ill be found that the same point is given

by each of the polar coordinates

(r, 0), (- r, 180Â° + 6), {r, - (360Â° - 6)] and {- r, - (180Â° - 6%

or, expressing the angles in radians, by each of the co-

ordinates

(r, e\ {-r,7r + 6), {r, - (27r - 0)} and {- r, - (tt - $)}.

It is also clear that adding 360Â° (or any multiple of

360Â°) to the vectorial angle does not alter the final position

of the revolving line, so that {r, 6) is always the same point

as (r, ^ + ?i . 360Â°), where n is an integer.

So, adding 180Â° or any odd multiple of 180Â° to the

vectorial angle and changing the sign of the radius vector

gives the same point as before. Thus the point

[-r, ^ + (2n + 1)180Â°]

is the same point as [â€” r, 6 + 180Â°], i.e. is the point [r, 6\

33. To find the length of the straight line joining two

points whose polar coordinates are given.

Let A and B be the two points and let their polar

coordinates be (r^, 6y) and (r^, 6^ respectively, so that

OA^r^, OB = r^, lXOA^O^, and lX0B = 6^,

POLAR COORDINATES.

21

Then (Trigonometry, Art. 164)

AB" - OA'' + OB'' -20 A. OB cos AOB

= r-^ + r^ - 2r-^r^ cos {0^ - 6^.

34. To find the area of a triangle the coordinates of

whose angular points are given.

Let ABC be the triangle and let (r-^, 0^), (r^, 62), and

(rg, ^3) be the polar coordinates of

its angular points.

We have

AABO=AOBC+aOCA

-AOBA (1).

Now

A0BC = i0B,0C sin BOC

[Trigonometry, Art. 198]

= ^r^r^ sin (^3 - $^).

' So A OCA = \0G . OA sin CO A = ^r^r, sin (6, - 6,),

and AOAB^^OA. OB sin AOB = ^r^r^ sin {6^ - 6.^

= - Jn^2 sin (^2 - ^1).

Hence (1) gives

A ABC = J \r<^r^ sin (^3 - 6^ + r^r^ (sin 0-^ - 0^)

+ r^r^ sin {Oo - 0^)].

35. To change from Cartesian Coordinates to Polar

Coordinates, and conversely.

Let P be any point whose Cartesian coordinates, referred

to rectangular axes, are x and y,

and whose polar coordinates, re-

ferred to as pole and OX as

initial line, are (r, 6).

Draw Pit/'perpendicular to OX

so that we have

OM=x, MP = y, LMOP = e,

and OP = r.

From the triangle MOP we

have

x = OM=OPcosMOP = rcosO (1),

y = MP= OPsinMOP ^r smO (2),

r=OP= sJOM^ + MP^^ s/x" + y' (3),

X' O:

22 COORDINATE GEOMETRY,

and

Equations (1) and (2) express the Cartesian coordinates

in terms of the polar coordinates.

Equations (3) and (4) express the polar in terms of the

Cartesian coordinates.

The same relations will be found to hold if P be in any-

other of the quadrants into which the plane is divided by

XOX' and YOT.

Ex. Change to Cartesian coordinates the equations

{!) r = asind, and (2)r=a^cos-.

a

(1) Multiplying the equation by r, it becomes r^rrar sin Q,

i.e. by equations (2) and (3), x^-\-y^ = ay.

(2) Squaring the equation (2), it becomes

r=acos2- = - (1 + cos^),

i. e. 2^2 = ar + ar cos 6,

i.e. 2{x^ + y^) = a sjx^ + y^-\- ax,

i.e. {2x^ + 2y^-ax)^ = a^{x^ + y^).

EXAMPLES. III.

Lay down the positions of the points whose polar coordinates are

L (3,45Â°). 2. (-2, -60Â°). 3. (4,135Â°). 4. (2,330Â°).

5. (-1, -180Â°). 6. (1, -210Â°). 7. (5, -675Â°). 8. (Â«> |) â€¢

9. (2a.-^). 10.{-a,l). n.(-2a.4').

Find the lengths of the straight lines joining the pairs of points

whose polar coordinates are

12. (2, 30Â°) and (4, 120Â°). 13. (-3, 45Â°) and (7, 105Â°).

14. ( Â«> I ) and

(Â«"'!) â€¢

[EXS. III.] EXAMPLES. 23

15. Prove that the points (0, 0), (3, |) , and ( 3, ^ J form an equi-

lateral triangle.

Find the areas of the triangles the coordinates of whose angular

points are

16. (1, 30Â°), (2, 60Â°), and (3, 90Â°).

17. (_3, -30Â°), (5, 150Â°), and (7, 210Â°).

18. (-^. i)' (Â«'i)'^^*l(-2Â«. -y)-

Find the polar coordinates (drawing the figure in each case) of the

points

19. x = J3, y = l. 20. x=-^B, y = l. 21. x=-l, y = l.

Find the Cartesian coordinates (drawing a figure in each case) of

the points whose polar coordinates are

22. (5, I). 23. (-6, I). 24.(5,-^).

Change to polar coordinates the equations

25. x^+y^=a\ 26. y = xtana. 27. x^ + y^ = 2ax.

28. x^-y^=2ay. 29. x^=y^{2a-x). 30. {x^ + y^)^ = a^{x^-y^).

Transform to Cartesian coordinates the equations

31. r=a. 32. ^ = tan-i??i. 33. r = acos^.

34. r = asin2^. 35. ?-2 = a2cos2^. 36. o'^ sin 2d =2a^.

e

2

40. 'T (cos 3^ + sin 3^) = 5h sin 6 cos d

37. r'^G0s2d = a^. 38. r^cos- = a^, 39^ r^=ai sin-.

2i a

CHAPTER III.

LOCUS. EQUATION TO A LOCUS.

36. When a point moves so as always to satisfy a

given condition, or conditions, the path it traces out is

called its Locus under these conditions.

For example, suppose to be a given point in the plane

of the paper and that a point F is to move on the paper so

that its distance from shall be constant and equal to a.

It is clear that all the positions of the moving point must

lie on the circumference of a circle whose centre is and

whose radius is a. The circumference of this circle is

therefore the " Locus" of P when it moves subject to the

condition that its distance from shall be equal to the

constant distance a.

37. Again, suppose A and B to be two fixed points in

the plane of the paper and that a point P is to move in

the plane of the paper so that its distances from A and B

are to be always equal. If we bisect AB in G and through

it draw a straight line (of infinite length in both directions)

perpendicular to AB, then any point on this straight line

is at equal distances from A and B. Also there is no

point, whose distances from A and B are the same, which

does not lie on this straight line. This straight line is

therefore the "Locus" of P subject to the assumed con-

dition.

38. Again, suppose A and B to be two fixed points

and that the point P is to move in the plane of the paper

so that the angle APB is always a right angle. If we

describe a circle on AB as diameter then P may be any

EQUATION TO A LOCUS.

25

point on the circumference of this circle, since the angle

in a semi-circle is a right angle; also it could easily be

shewn that APB is not a right angle except when P lies

on this circumference. The "Locus" of P under the

assumed condition is therefore a circle on -4^ as diameter.

39. One single equation between two unknown quan-

tities x and y, e.g.

x + y = l (1),

cannot completely determine the values of x and y.

*\P6

\Q

vPs

^1?

\1

M

OM

Vs

\ P

Such an equation has an infinite number of solutions.

Amongst them are the followins: :

a: = 0,1

a; =1,1

2/ = 0/'

x= 2,1

y=-l/'

x= 3,1

y=-2/'-

x = â€”\,\ X â€” â€” 2

2/= 2/' y= 3

Let us mark down on paper a number of points whose

coordinates (as defined in the last chapter) satisfy equation

Let OX and OF be the axes of coordinates.

If we mark off a distance OPi (â€”1) along OY^ we have

a point Pi whose coordinates (0, 1) clearly satisfy equation

If we mark off a distance OP^ (=1) along OX, we have

a point Pg whose coordinates (1, 0) satisfy (1).

26 COORDINATE GEOMETRY.

Similarly the point Pg, (2, â€” 1), and P4, (3, â€” 2), satisfy

the equation (1).

Again, the coordinates (â€”1, 2) of Pg and the coordinates

(â€”2, 3) of Pg satisfy equation (1).

On making the measurements carefully we should find

that all the points we obtain lie on the line P^P^ (produced

both ways).

Again, if we took any point Q, lying on P^P^, and draw

a perpendicular QM to OX, we should find on measurement

that the sum of its x and y (each taken with its proper

sign) would be equal to unity, so that the coordinates of Q

would satisfy (1).

Also we should find no point, whose coordinates satisfy

(1), which does not lie on P1P2.

All the points, lying on the straight line P1P2, and no

others are therefore such that their coordinates satisfy the

equation (1).

This result is expressed in the language of Analytical

Geometry by saying that (1) is the Equation to the Straight

Line P^P^.

40. Consider again the equation

Â£c2 + 2/2 = 4 (1).

Amongst an infinite number of solutions of this equa-

tion are the following :

x = 2A x= J?>\ x = J\ x^l \

V3| x = J'I\ x=^i \

2/=2r 2/=x/3r 2/=V2 r y=i r

x = -'2,\ x^-.^?>,\ x = -J2A x = -l, \

2/^0 r 2/^ - i /' 3/=-v2r y=-si^)'

53 = 0, \ x=l, \ x=J2, } x=J3)

2/ = -2j ' y

L and

V3/' 2/ = -./2/'-'"% = -l/-

EQUATION TO A LOCUS.

27

1 â€¢

,5i

â– i|f

-,..

M ;f^ X

All these points are respectively represented by the

points P^, P^^ F^y ... P^Qj and they

will all be found to lie on the

dotted circle whose centre is

and radius is 2.

Also, if we take any other

point Q on this circle and its

ordinate QM, it follows, since

0M^- + MQ^^0Q' = 4:, that the x

and y of the point Q satisfies (1).

The dotted circle therefore

passes through all the points whose

coordinates satisfy (1).

In the language of Analytical Geometry the equation

(1) is therefore the equation to the above circle.

41. As another example let us trace the locus of the

point whose coordinates satisfy the equation

y'=4^'^ (1);

If we give x a negative value we see that y is im-

possible; for the square of a

real quantity cannot be nega-

tive.

We see therefore that there

are no points lying to the left

of OF.

If we give x any positive

value we see that y has two

real corresponding values which

are equal and of opposite signs.

The following values,

amongst an infinite number of

others, satisfy (1), viz.

x = 0,] x=l, "I x = 2,

y = 0}' y = + 2or-2}' y = 2J2ov -2J2

x=4: I cc=16, I aj = + oo, )

y = + 4: or â€”4:) ^ '" y â€” Sor-Sj' '" y = + (X)Ovâ€”ao)'

The origin is the first of these points and P^ and Qi,

Pg and Q^, P^and Q^, ... represent the next pairs of points.

h

28 COORDINATE GEOMETRY.

If we took a large number of values of x and the

corresponding values of ?/, the points thus obtained would

be found all to lie on the curve in the figure.

Both of its branches would be found to stretch away to

infinity towards the right of the figure.

Also, if we took any point on this curve and measured

with sufficient accuracy its x and y the values thus obtained

would be found to satisfy equation (1).

Also we should not be able to find any point, not lying

on the curve, whose coordinates would satisfy (1).

In the language of Analytical Geometry the equation

(1) is the equation to the above curve. This curve is called

a Parabola and will be fully discussed in Chapter X.

42. If a point move so as to satisfy any given condition

it will describe some definite curve, or locus, and there can

always be found an equation between the x and y of any

point on the path.

This equation is called the equation to the locus or

curve. Hence

Def. Equation to a curve. The equation to a

curve is the relation which exists between the coordinates of

any foint on the curve^ and which holds for no other points

except those lying on the curve.

43. Conversely to every equation between x and y it

will be found that there is, in general, a definite geometrical

locus.

Thus in Art. 39 the equation is x + y=\, and the

definite path, or locus, is the straight line P^P^ (produced

indefinitely both ways).

In Art. 40 the equation is x'^ + y'^^ 4, and the definite

path, or locus, is the dotted circle.

Again the equation 2/ = 1 states that the moving point

is such that its ordinate is always unity, i.e. that it is

always at a distance 1 from the axis of x. The definite

path, or locus, is therefore a straight line parallel to OX

and at a distance unity from it.

EQUATION TO A LOCUS. 29

44. In the next chapter it will be found that if the

equation be of the first degree {i.e. if it contain no

products, squares, or higher powers of x and y) the locus

corresponding is always a straight line.

If the equation be of the second or higher degree, the

corresponding locus is, in general, a curved line.

45. We append a few simple examples of the forma-

tion of the equation to a locus.

Ex. 1. A point moves so that the algebraic sum of its distances

from tioo given perpendicular axes is equal to a constant quantity a;

find the equation to its locus.

Take the two straight lines as the axes of coordinates. Let {x, y)

be any point satisfying the given condition. We then ha,wex + y = a.

This being the relation connecting the coordinates of any point

on the locus is the equation to the locus.

It will be found in the next chapter that this equation represents

a straight line.

Ex. 2. The sum of the squares of the distances of a moving point

from the tioo fixed points {a, 0) and {-a, 0) is equal to a constant

quantity 2c^. Find the equation to its locus.

Let (a;, y) be any position of the moving point. Then, by Art. 20,

the condition of the question gives

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