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S. L. (Sidney Luxton) Loney.

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{ [x - af + /} + { (a; + af + if] = 2c\
i.Ci x^ + y'^ = c^- a~.

This being the relation between the coordinates of any, and every,
point that satisfies the given condition is, by Art. 42, the equation to
the required locus.

This equation tells us that the square of the distance of the point
{x, y) from the origin is constant and equal to c^ - a^, and therefore
the locus of the point is a circle whose centre is the origin.

Ex. 3. A point moves so that its distance from the point (-1,0)
is always three times its distance from the point (0, 2).

Let {x, y) be any point which satisfies the given condition. We

then have

J{x + iy' + {y-0)^=Bj{x - 0)2+ {y - 2)2,
so that, on squaring,

x'^ + 2x + l + y'^=9{x'^ + y'^-4:y + 4),
i.e. 8(a;2 + y2)_2a;-36?/ + 35 = 0.

This being the relation between the coordinates of each, and
every, point that satisfies the given relation is, by Art. 42, the
required equation.

It will be found, in a later chapter, that this equation represents
a circle.



30 COOEDINATE GEOMETRY.

EXAMPLES. IV.

By taking a number of solutions, as in Arts. 39 — 41, sketch
the loci of the following equations :

1. 2x + dy = l0. 2. ^x-y = l. 3. x'^-2ax-Vy'^ = Q.

4. a;2-4aa; + ?/2 + 3a2 = 0. 5. y'^ = x. 6. ^x = y^-^.

7 ^' + ^'=1.
'■4^9

A and B being the fixed points (a, 0) and ( - a, 0) respectively,
obtain the equations giving the locus of P, when

8. PA"^ - P52 _ a constant quantity = 2fc2.

9. PA = nPB, n being constant.

10. P^+PjB = c, a constant quantity.

11. PB^ + PC^=2PA^, C being the point (c, 0).

12. Find the locus of a point whose distance from the point (1, 2)
is equal to its distance from the axis of y.

Find the equation to the locus of a point which is always equi-
distant from the points whose coordinates are

13. (1, 0) and (0, -2). 14. (2, 3) and (4, 5).

15. {a + b, a-h) and {a-b, a + b).

Find the equation to the locus of a point which moves so that

16. its distance from the axis of x is three times its distance from
the axis of y.

17. its distance from the point (a, 0) is always four times its dis-
tance from the axis of y.

18. the sum of the squares of its distances from the axes is equal
to 3.

19. the square of its distance from the point (0, 2) is equal to 4.

20. its distance from the point (3, 0) is three times its distance
from (0, 2).

21. its distance from the axis of x is always one half its distance
from the origin.

22. A fixed point is at a perpendicular distance a from a fixed
straight line and a point moves so that its distance from the fixed
point is always equal to its distance from the fixed line. Find the
equation to its locus, the axes of coordinates being drawn through
the fixed point and being parallel and perpendicular to the given
line.

23. In the previous question if the first distance be (1), always half,
and (2), always twice, the second distance, find the equations to the
respective loci.



CHAPTER IV.

THE STRAIGHT LINE. RECTANGULAR COORDINATES.

46. To find the equation to a straight line which is
parallel to one of the coordinate axes.

Let CL be any line parallel to the axis of y and passing
through a point C on the axis of x such that OG = c.

Let F be any point on this line whose coordinates are
X and y.

Then the abscissa of the point F is
always c, so that

x = c (1).

This being true for every point on
the line CL (produced indefinitely both
ways), and for no other point, is, by
Art. 42, the equation to the line.

It will be noted that the equation does not contain the
coordinate y.

Similarly the equation to a straight line parallel to the
axis oi X is y — d.

Cor. The equation to the axis of a? is 2/ = 0.
The equation to the axis oi y is x — 0.

47. To find the equation to a st7'aight line which cuts
off a given intercept on the axis of y and is inclined at a
given angle to the axis of x.

Let the given intercept be c and let the given angle be a.



X



32 COORDINATE GEOMETRY.

Let C be a point on the axis of y such that OC is c.
Through C draw a straight
line Z(7Z' inclined at an angle
a (= tan~^ m) to the axis of x^
so that tan a — m.

The straight line LCL' is ^^

therefore the straight line ^^
required, and we have to -'l O MX

find the relation between the
coordinates of any point P lying on it.

Draw PM perpendicular to OX to meet in ^ a line
through G parallel to OX.

Let the coordinates of P be cu and ?/» so that OM=x
and MP = y.

Then MP = NP + MN =C]Srt^iia + 00 = m.x + c,
i.e. y = mx+c.

This relation being true for any point on the given
straight line is, by Art. 42, the equation to the straight
line.

[In this, and other similar cases, it could be shewn,
conversely, that the equation is only true for points lying
on the given straight line.]

Cor. The equation to any straight line passing through
the origin, i.e. which cuts off a zero intercept from the axis
of 2/, is found by putting c — O and hence is 3/ = mx.

48. The angle a which is used in the previous article is the
angle through which a straight line, originally parallel to OZ, would
have to turn in order to coincide with the given direction, the rotation
being always in the positive direction. Also m is always the tangent
of this angle. In the case of such a straight line as AB, in the figure
of Art. 50, m is equal to the tangent of the angle PAX (not of the
angle PAO). In this case therefore wi, being the tangent of an obtuse
angle, is a negative quantity.

The student should verify the truth of the equation of the last
article for all points on the straight line LCL', and also for straight
Hnes in other positions, e.g. for such a straight line as A^B^ in the
figure of Art. 59. In this latter case both m and c are negative
quantities.

A careful consideration of all the possible cases of a few proposi-
tions will soon satisfy him that this verification is not always
necessary, but that it is sufficient to consider the standard figure.



THE STRAIGHT LINE.



33



49. Ex. The equation to the straight line cutting off an
intercept 3 from the negative direction of the axis of y, and inclined
at 120° to the axis of a;, is

?/ = a;tanl20° + (-3),

i.e. y= -x^S-S,

i.e. y + x^S + S = 0.

50. 1^0 find the equation to the straight line which cuts
off given i7itercepts a and h from the axes.

Let A and B be on OX and OY respectively, and be
such that OA = a and OB = h.

Join AB and produce it in-
definitely both ways. Let P be
any point (x, y) on this straight
line, and draw PM perpendicular
to OX.

We require the relation that
always holds between x and 3/, so
long as P lies on AB.

By Euc. YI. 4, we have

OM_PB MP _AP

OA~AB' ^"""^ 'OB~AB




OM MP PB + AP



+



OA OB



AB



= 1,



^.e.



X y ^
a D



This is therefore the required equation ; for it is the
relation that holds between the coordinates of any point
lying on the given straight line.

51. The equation in the preceding article may he also obtained
by expressing the fact that the sum of the areas of the triangles OP A
and OPB is equal to OAB, so that

\axy + \hy.x = \ax'b,



and hence



a



52. Ex. 1. Find the equation to the straight line passing
through the -point (3, - 4) and cutting off intercepts, equal but of
opposite signs, from the tioo axes.

Let the intercepts cut off from the two axes be of lengths a and



— a.



34



COORDINATE GEOMETRY.



The equation to the straight line is then

a -a

i.e. x-y = a (1).

Since, in addition, the straight line is to go through the point
(3, -4), these coordinates must satisfy (1), so that

3-(-4) = a,
and therefore a = l.

The required equation is therefore

x-y = 7.

Ex. 2. Find the equation to the straight line lohich passes through
the point (-5, 4) and is such that the portion of it between the axes is
divided by the point in the ratio ofl : 2.

Let the required straight line be - + t = 1. This meets the axes

a b

in the points whose coordinates are {a, 0) and (0, &).

The coordinates of the point dividing the line joining these

points in the ratio 1 : 2, are (Art. 22)



2.a+1.0 , 2.0 + 1.&
If this be the point ( - 5, 4) we have



. 2(1 , b
i.e,-^ and -.



„ 2a , , b
-5:=- and 4=-,

so that a= - Y- and b = 12.

The required straight line is therefore

X y



-i^^l2 '



I.e.



oy



8a; = 60.



53. To find the equation to a straight line in tenns of
the perpendicular let fall upon it from the origin and the
angle that this perpendicular makes with the axis of x.

Let OR be the perpendicular from and let its length
be jo.

Let a be the angle that OR makes
with OX.

Let P be any point, whose co-
ordinates are x and y, lying on AB ',
draw the ordinate PM, and also ML
perpendicular to OR and PN perpen-
dicular to ML.




THE STRAIGHT LINE. 85

Then OL = OMco^a (1),

and LR = NP = MF&inNMP.

But lNMP^W - lNMO= iMOL^a.

LR = MP&m.a (2).

Hence, adding (1) and (2), we have

Oil/ cos a + ifPsin a=OL + LR=OR =79,
i.e. X COS a + y sin a = p.

This is the required equation.

54. In Arts. 47 — 53 we have found that the correspond-
ing equations are only of the first degree in x and y. We
shall now prove that

Any equation of the first degree i7i x and y always repre-
sents a straight line.

For the most general form of such an equation is

Ax + By^C = ^ (1),

where A^ B, and C are constants, i.e. quantities which do
not contain x and y and which remain the same for all
points on the locus.

Let (cCi, 2/1), (a?2) 2/2)) ^iicl (rt's, 2/3) be any three points on
the locus of the equation (1).

Since the point {x-^, y^) lies on the locus, its coordinates
when substituted for x and y in (1) must satisfy it.

Hence Ax^ + Ry^+C-=0 (2).

So Ax^ + Ry^ + C^O (3),

and Axs + £ys+C = (4).

Since these three equations hold between the three quanti-
ties A, B, and C, we can, as in Art. 12, eliminate them.

The result is

^35 2/35 -•-

But, by Art. 25, the relation (5) states that the area of the
triangle whose vertices are (x^, y^), (x^, 3/2)5 ^^^ (^3> 2/3) is
zero.

Also these are any three points on the locus.

3—2



= (5).



36 COORDINATE GEOMETRY.

The locus must therefore be a straight line ; for a curved
line could not be such that the triangle obtained by joining
any three points on it should be zero.

55. The proposition of the preceding article may also be deduced

from Art. 47. For the equation

^a; + % + (7=0

A C
may be written y=- — x-^,

and this is the same as the straight line

y = mx + c,

A ^ C

if ?3i=-— and c = - — .

x> is

But in Art. 47 it was shewn that y = mx + c was the equation to
a straight line cutting off an intercept c from the axis of y and
inclined at an angle tan~^m to the axis of x.

The equation Ax + By + C=0

C

therefore represents a straight line cutting off an intercept - — from

x>

the axis of y and inclined at an angle tan~^ ( - — | to the axis of x.

56. We can reduce the general equation of the first
degree Ax + By + C = (1)

to the form of Art. 53.

For, if p be the perpendicular from the origin on (1)
and a the angle it makes with the axis, the equation to the
straight line must be

X cos a 4- 2/ sin a - /» = (2).

This equation must therefore be the same as (1).
cos a sin a —p



Hence



ABC



p cos a sin a \/cos^ a + sin^ a 1



C -A -B Ja^ + B' sJa^' + B^

Hence

-A . -B ^ C

cos a - - , sm a = , , and p =



s/A^ + B^' \fA-' + B'' sfA^ + B^

The equation (1) may therefore be reduced to the form (2)
by dividing it by JA^ + B^ and arranging it so that the
constant term is negative.



THE STRAIGHT LINE. 37

57. Ex. Reduce to tlie perpendicular form the equation

^ + 2/\/3 + 7 = (1).

Here JA'' + B^= ^TTs = sJ4:=2.

Dividing (1) by 2, we have

i.e. ^(-i)+y( - ^)-i=o,

i.e. X cos 240° + y sin 240° - 1 = 0.

58. To trace the straight line given hy an equation of
the first degree.

Let the equation be

Ax + By + G = (1).

(a) This can be written in the form

A B

Comparing this with the result of Art. 50, we see that it

(J

represents a straight Hne which cuts off intercepts — -^ and


— — from the axes. Its position is therefore known.

jO

If G be zero, the equation (1) reduces to the form

A

and thus (by Art. 47, Cor.) represents a straight hne
passing through the origin inchned at an angle tan~^ I ~ r )
to the axis of x. Its position is therefore known.

(^) The straight line may also be traced by firnding
the coordinates of any two points on it.

G
If we put y — in (1) we have x — —-r. The point

JL



i-'i-')



therefore lies on it.



38



COORDINATE GEOMETRY.



G



If we put oj = 0, we have 2/ = — ^ , so that the point

G^



(»-.)



lies on it.



line.



Hence, as before, we have the position of the straight



69. Ex. Trace the straight lines

(1) 3a;-4i/ + 7 = 0; (2) 7a; + 8y + 9 = 0j
(3) %y = x', (4) x = ^i (5) 2/= -2.




(1) Putting 2/ = 0, we have rc= -|,

and putting x = Q, we have y = ^.

Measuring 0A-^{= -^) along the axis of x we have one point on
the Hne.

Measuring OB^ (=t) along the axis of y we have another point.
Hence A-^B^ , produced both ways, is the required line,

(2) Putting in succession y and x equal to zero, we have the
intercepts on the axes equal to - f and - f.

If then 0-42= -f and 0^2= - |, we have A^B^, the required line.

(3) The point (0, 0) satisfies the equation so that the origin is on
the line.

Also the point (3, 1), i.e. C.^, lies on it. The required line is
therefore OC3.

(4) The line ic = 2 is, by Art. 46, parallel to the axis of y and passes
through the point A^ on the axis of x such that 0A^ = 2.

(5) The line y= - 2 is parallel to the axis of x and passes through
the point B^ on the axis of y, such that 0B^= - 2.

60. Straight Line at Infinity. We have seen
that the equation Ax + By + (7 = represents a straight line



STRAIGHT LINE JOINING TWO POINTS. 39

c c

which cuts oiF intercepts — - and — — from the axes of

Ji. Jj

coordinates.

If A vanish, but not B or C, the intercept on the axis
of X is infinitely great. The equation of the straight line
then reduces to the form y = constant, and hence, as in
Art. 46, represents a straight line parallel to Ox.

So if B vanish, but not A or C, the straight line meets
the axis of y at an infinite distance and is therefore parallel
to it.

If A and B both vanish, but not C, these two in-
tercepts are both infinite and therefore the straight line
Q .x + .y + C = is altogether at infinity.

61. The multiplication of an equation by a constant
does not alter it. Thus the equations

2a;-32/+5 = and 10a;- 152/+ 25 -

represent the same straight line.

Conversely, if two equations of the first degree repre-
sent the same straight line, one equation must be equal to
the other multiplied by a constant quantity, so that the
ratios of the corresponding coefficients must be the same.

For example, if the equations

a^x + \y + Ci = and A-^^x + B^y + Cj =
we must have

«! \ CjL

62. To jind the equation to the straight line which
passes through the two given points {x\ y') and (x", y").

By Art. 47, the equation to any straight line is

y - mx -VG (1).

By properly determining the quantities m and c we can
make (1) represent any straight line we please.

If (1) pass through the point (a;', y')^ we have

2/' = mas' + c (2).

Substituting for c from (2), the equation (1) becomes

y-y' = m(x-x') (3).



40 COOKDINATE GEOMETRY.

This is the equation to the line going through (x\ y') making
an angle tan~^ m with OX. If in addition (3) passes through
the point {x", y"), then

y —y=m{x — x),

ti r

y -y



* * X' - X

Substituting this value in (3), we get as the required
equation

V" — v'
*^ X" — x^ '

63. Ex. Find the equation to the straight line which passes
through the points (-1, 3) and (4, -2).

Let the required equation be

y=mx + c (1).

Since (1) goes through the first point, we have
3=-m + c, so that c = m + S.
Hence (1) becomes

y = mx + m + S (2).

If in addition the line goes through the second point, we have

-2 = 47?i + m + 3, so that m= -1.
Hence (2) becomes

y=-x + 2, i.e. x + y = 2.
Or, again, using the result of the last article the equation is

y-B = ^-^^^{x + l)=-x-l,

i.e. y + x-=2.

64. To fix definitely the position of a straight line we
must have always two quantities given. Thus one point
on the straight line and the direction of the straight line
will determine it; or again two points lying on the straight
line will determine it.

Analytically, the general equation to a straight line
will contain two arbitrary constants, which will have to be
determined so that the general equation may represent any
particular straight line.

Thus, in Art. 47, the quantities m and c which remain
the same, so long as we are considering the same straigld
line, are the two constants for the straight line.



EXAMPLES. 41

Similarly, in Art. 50, the quantities a and h are the
constants for the straight line.

65. In any equation to a locus the quantities x and y,
which are the coordinates of any point on the locus, are
called Current Coordinates ; the curve may be conceived as
traced out by a point which " runs " along the locus.

EXAMPLES. V.

Find the equation to the straight line

1. cutting off an intercept unity from the positive direction of the
axis of y and inclined at 45° to the axis of x.

2. cutting off an intercept - 5 from the axis of y and being equally
inclined to the axes.

3. cutting off an intercept 2 from the negative direction of the
axis of y and inclined at 30° to OX.

4. cutting off an intercept - 3 from the axis of y and inclined at
an angle tan~i f to the axis of x.

Find the equation to the straight line

5. cutting off intercepts 3 and 2 from the axes.

6. cutting off intercepts - 5 and 6 from the axes.

7. Find the equation to the straight line which passes through the
point (5, 6) and has intercepts on the axes

(1) equal in magnitude and both positive,

(2) equal in magnitude but opposite in sign.

8. Find the equations to the straight lines which pass through
the point (1, - 2) and cut off equal distances from the two axes.

9. Find the equation to the straight line which passes through
the given point {x\ y') and is such that the given point bisects the
part intercepted between the axes.

10. Find the equation to the straight line which passes through
the point ( - 4, 3) and is such that the portion of it between the axes
is divided by the point in the ratio 5 : 3.

Trace the straight lines whose equations are

11. a; + 2?/+3 = 0. 12. 5a - 7//-9 = 0.
13. 3a; + 7r/ = 0. 14. 2a;-3?/ + 4 = 0.

Find the equations to the straight lines passing through the
following pairs of points.

15. (0, 0) and (2, -2). 16. (3, 4) and (5, 6).

17. (-1, 3) and (6, -7). 18. (0, -a) and (&, 0).



42



COORDINATE GEOMETRY.



[Exs. v.]



19. (a, &) and {a + h, a-h).

20. {at^, 2at-^) and (at^^ 2at;).



21. (a«„^)and(a«„^j.



22. (« cos 01 , a sin <pi) and (a cos (p^, a sin ^a)-

23. (acos0jLJ & sin 0j) and (acos02> ^sin^g)*

24. (* sec 01, 6 tan 0i) and (a sec 02, 6 tan 02).

Find the equations to the sides of the triangles the coordinates of
whose angular points are respectively

25. (1,4), (2,-3), and (-1,-2).

26. (0,1), (2,0), and (-1, -2).

27. Find the equations to the diagonals of the rectangle the
equations of whose sides are x = a, x = a\ y = b, and y = b\

28. Find the equation to the straight line which bisects the
distance between the points {a, b) and {a', b') and also bisects the
distance between the points ( - a, b) and (a', - b').

29. Find the equations to the straight lines which go through the
origin and trisect the portion of the straight line 3a; + 2/ = 12 which
is intercepted between the axes of coordinates.

Angles between straight lines.

66. To find the angle between two given straight lines.

Let the two straight lines be AL^ and AL^j meeting the
axis of X in L^ and L^,




I. Let their equations be

y — m^x^-G-^ and y ~ in.j,x ^r c.^

By Art. 47 we therefore have

tan^ZjA'^mi, and td^Vi. AL.^X^Wj.^,.
Now L L-^AL^^ — L AL^X — L AL.2.X.

tan L^AL^ — tan \AL^X — AL^X\
ta,n AL^X— tan AL^X rn^ — n^



(1).



1 + tan AL^X. tan AL^X



1 +mi«i2



ANGLES BETWEEN STRAIGHT LINES. 43



Hence the required angle — lL^AL



= tan-i "'^•""'^ (2).

l + mim2

[In any numerical example, if the quantity (2) be a positive quan-
tity it is the tangent of the acute angle between the lines ; if negative,
it is the tangent of the obtuse angle.]

II. Let the equations of the straight lines be
^i£c + ^i2/ + Ci = 0,
and A^^x^- B^^y + G^^O.

By dividing the equations by B^ and B^, they may be
written



and









A,


c,
















2/ -


A.


'A'












L




y = -


sr


A'












Comparing


these with the equations of


(I-x


we


see


that






.„ ^1


A




^2











Hence the required angle



- tan-i-,— i == tan ^



B



r(-J)



1 + mi7n.2 , / -41"^^ ^ -^2^



(-!)(-»



= *"" 3STA^ <^>-

III. If the equations be given in the form
X cos a + y sin a — ^^ = and x cos ^ + 2/ sin j^ —p-^ — O,
the perpendiculars from the origin make angles a and p
with the axis of x.

Now that angle between two straight lines, in which
the origin lies, is the supplement of the angle between the
perpendiculars, and the angle between these perpendiculars
is ^ — a.

[For, if OEi and OR2 be the perpendiculars from the origin upon
the two hnes, then the points O, R^, R^, and A lie on a circle, and
hence the angles R^OR^ and R.^AR^ are either equal or supplementary.]



44 COORDINATE GEOMETRY.

67. To find the condition that two straight lines may
he parallel.

Two straight lines are parallel when the angle between
them is zero and therefore the tangent of this angle is zero.

The equation (2) of the last article then gives

Two straight lines whose equations are given in the
"m" form are therefore parallel when their "7?i's" are the
same, or, in other words, if their equations differ only in
the constant term.

The straight line Ax + By + G' = is any straight line which is
parallel to the straight line Ax + By + C = 0. For the "m's" of the
two equations are the same.

Again the equation A {x-x')+B {y-y') = clearly represents the
straight line which passes through the point {x', y') and is parallel to
Ax + By + C=0.

The result (3) of the last article gives, as the condition
for parallel lines,

68. Ex. Find the equation to the straight line, which passes
through the point (4, - 5), and which is parallel to the straight line

3:c + 4r/ + 5 - =0 (1).

Any straight line which is parallel to (1) has its equation of the
form

3a; + 4^/ + (7=0 (2).

[For the "w" of both (1) and (2) is the same.]

This straight line will pass through the point (4, - 5) if
3x4 + 4x(-5) + C = 0,
i.e. if (7=20-12 = 8.

The equation (2) then becomes

3a;+42/ + 8 = 0.

69. To find the condition that two st^'aight lines j whose
equations are given, may he 'perpendicular.

Let the straight lines be

y — m^x -i-Ci,

and y — m.^x -\- G.2_.



CONDITIONS OF PERPENDICULARITY. 45

If be the angle between them we have, by Art. 66,

tan^^ r^""^^ (1).

1 +mim2

If the lines be perpendicular, then ^ = 90°, and therefore
tan = 00 .

The right-hand member of equation (1) must therefore
be infinite, and this can only happen when its denominator
is zero.

The condition of perpendicularity is therefore that

1 + m^TTi^ — O, i.e. Tn-^Tn2 = — I.'

The straight line y — tu^x + c.^ is therefore perpendicular

to y = »...H-.c., if «, = -!.

y/c'-t

It follows that the straight lines

A^x +B^y + C^ = and A^x + B^y + 0^ = 0,



for which m^ = — ^ and m^^ — ^ , are at right angles if

a) V A



AA / A,, _



i.e. a A^A^+B^B^ = 0.

70. From the preceding article it follows that the two
straight lines

A^x + B,y + Ci = Q (1),

and B,x-A,y+C^ = (2),

are at right angles ; for the product of their m's

Also (2) is derived from (1) by interchanging the coefficients
of a; and y, changing the sign of one of them, and changing
the constant into any other constant.

Ex. The straight line through (x', y') perpendicular to (1) is (2)
where B^x' - A-^y' + 62= 0, so that Cg = A^y'- B^x'.

This straight line is therefore

B,{x-x')-A^{y-y') = 0.



46 COORDINATE GEOMETRY.

71. Ex. 1. Find the equation to the straight line which passes
through the point (4, —5) and is perpendicular to the straight line

Sx + 4ij + 5 = (1).

First Method. Any straight line perpendicular to (1) is by the



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