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S. L. (Sidney Luxton) Loney.

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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^±^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 ^L±^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 60°.

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)»
{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
(^i» 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+ •••+^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,, ^) , (a»„ ^) , {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 ■¥ ^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 — 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? ■'-
, ^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 — ^'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

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

-lLM{LA-\-MB)

{(^■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 - «^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-^ — ^zV^)-

So for the other triangles.

The required area therefore

^ h [(«^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 - «^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)? •••(.'^)i3 2/ra/

is i [(a?i2/2 - a?22/i) + (^22/3 - ^32/2) + • • • + (^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-,. — I , \am^, — y and -Jawo, — [ .



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. («. 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^ •

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, 80°),

(ii) (3, 150°), (iii) (-2, 45°), (iv) ■

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

(-3, -30°). ^\^ ,->^^

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

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

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

then mark off along it a distance y ''-•.,

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 150° and is then in the position OP^ ; measuring a distance 3
along it we arrive at Pg .

2—2



20 COORDINATE GEOMETKY.

(iii) For the third point, let the radius vector revolve from OX
through 45° 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 330° 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 - 210°, it will be in the
position OP2, and the point required is Pg.

(vi)^ For the sixth point, the radius vector, after rotating through
- 30°, 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, 150°), (-3, 330°), (3, -210°) and (-3, -30°).
In general it v^ill be found that the same point is given
by each of the polar coordinates
(r, 0), (- r, 180° + 6), {r, - (360° - 6)] and {- r, - (180° - 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 360° (or any multiple of
360°) 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 . 360°), where n is an integer.

So, adding 180° or any odd multiple of 180° to the
vectorial angle and changing the sign of the radius vector
gives the same point as before. Thus the point

[-r, ^ + (2n + 1)180°]

is the same point as [— r, 6 + 180°], 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 (3,45°). 2. (-2, -60°). 3. (4,135°). 4. (2,330°).

5. (-1, -180°). 6. (1, -210°). 7. (5, -675°). 8. («> |) •
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, 30°) and (4, 120°). 13. (-3, 45°) and (7, 105°).



14. ( «> I ) and



(«"'!) •



[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, 30°), (2, 60°), and (3, 90°).

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

18. (-^. i)' («'i)'^^*l(-2«. -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 = —\,\ X — — 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 (—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, — 1), and P4, (3, — 2), satisfy
the equation (1).

Again, the coordinates (—1, 2) of Pg and the coordinates
(—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

£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 •


,5i




■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 —4:) ^ '" y — Sor-Sj' '" y = + (X)Ov—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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