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represents two parallel lines.

Find the value of k so that the following equations may represent

pairs of straight lines :

6. 6a;2+lla;?/-10?/2 + a; + 31?/ + ^ = 0.

7. 12a;2-10a;?/ + 2?/2 + lla;-5?/ + A; = 0.

8. 12a;2 + A;a;?/ + 2?/2 + lla;-5?/ + 2:=0.

9. 6a;2 + a;?/ + A-?/2-ll.r + 43?/-35 = 0.

[EXS. XIII.] EXAMPLES. 99

10. kxy-8x + 9y-12 = 0.

11. x'^ + is^-xij + 'if-5x-7y + k=:0.

12. 12x^ + xy-Qif-2dx + 8y + k = 0.

13. 2x^ + oiyy-y'^ + kx + 6y-d = 0.

14. x^ + kxy + y^-5x-ly + Q = 0.

15. Prove that the equations to the straight lines passing through

the origin which make an angle a with the straight line y + x = are

given by the equation

x^ + 2xy sec 2a + y^ = Q.

16. What relations must hold between the coordinates of the

equations

(i) ax^ + hy'^ â– ^cx-\-cy = 0,

and (ii) ay'^ + lxy + dy -{-ex=zQ^

so that each of them may represent a pair of straight lines ?

17. The equations to a pair of opposite sides of a parallelogram

are

a;2-7a; + 6 = and 2/^- 14?/ +40 = ;

find the equations to its diagonals.

120. To prove that a homogeneous equation of the nth

degree represents n straight li7ies, real or imaginary, which

all pass through the orighi.

Let the equation be

y'^ + A^xy''-^ + A^xY'-^ + A.^x'y''-^ 4- . . . + A.^x'' = - 0.

On division by x^, it may be written

This is an equation of the nth degree in - , and hence

JO

must have n roots.

Let these roots be -m^, m^, mg, ... mâ€ž. Then (C. Smith's

Algebra, Art. 89) the equation (1) must be equivalent to

the equation

t

The equation (2) is satisfied by all the points which

satisfy the separate equations

'^)6-"'^)(l-â„¢0- (!-'"Â«)=â€¢'â– â– â€¢<'>â€¢

^ _ mi = 0, ^ - m. = 0, . . . ^ - m,, - 0,

/v. ^ 'T* Â« 5

tLf *^ tKf

7â€”2

100 COORDINATE GEOMETRY.

i.e. by all the points which lie on the n straight lines

y â€” nijX = 0, y â€” m^x = 0, ... y â€” in^x = 0,

all of which pass through the origin. Conversely, the

coordinates of all the points which satisfy these n equa-

tions satisfy equation (1). Hence the proposition.

121. Ex. 1. The equation

which is equivalent to

{y-x){y-'lx){y-^x) = Q,

represents the three straight lines

y-x=.^, ?/-2a: = 0, and y-3a; = 0,

all of which pass through the origin.

Ex. 2. The equation y^ - 5y^ + 6y = 0,

i.e. y{y-2){y-S) = 0,

similarly represents the three straight lines

y=0, y = 2, and y = S,

all of which are parallel to the axis of x.

122. To find the equation to the two straight lines

joining the origin to the points in which the straight line

Ix + my = n (1)

meets the locus whose equation is

ax^ + 2hxy + hy"^ + "Igx + 2fy + c = (2).

The equation (1) may be written

Ix + my

n

= 1 (3).

The coordinates of the points in which the straight line

meets the locus satisfy both equation (2) and equation (3),

and hence satisfy the equation

ax' + Vum) + by^ + 2 {gx +/y) '-^^ + c ('-^^)' =

......(4).

[For at the points where (3) and (4) are true it is clear

that (2) is true.]

Hence (4) represents so7ne locus which passes through

the intersections of (2) and (3).

STRAIGHT LINES THROUGH THE ORIGIN. 101

But, since the equation (4) is homogeneous and of the

second degree, it represents two straight lines passing

through the origin (Art. 108).

It therefore must represent the two straight lines join-

ing the origin to the intersections of (2) and (3).

123. The preceding article may be illustrated geo-

metrically if we assume that the equation (2) represents

some such curve as PQRS in the figure.

â– "'/O

Let the given straight line cut the curve in the points

P and Q.

The equation (2) holds for all points on the curve PQRS.

The equation (3) holds for all points on the line PQ.

Both equations are therefore true at the points of

intersection P and Q.

The equation (4), which is derived from (2) and (3),

holds therefore at P and Q.

But the equation (4) represents two straight lines, each

of which passes through the point 0.

It must therefore represent the two straight lines OP

and OQ.

124. Ex. Prove that the straight lines joining the origin to the

points of intersection of the straight line x-y=:2 and the curve

5a;2 + 12a;i/ - 82/2 + 8a; - 4y + 12 =

make equal angles with the axes.

As in Art. 122 the equation to the required straight lines is

5a;2 + 12a;?/-8i/2+(8a;-4i/)^^ + 12fc^Y = (1),

102 â€¢ COORDINATE GEOMETRY.

For this equation is homogeneous and therefore represents two

straight lines through the origin; also it is satisfied at the points

where the two given equations are satisfied.

Now (1) is, on reduction,

so that the equations to the two lines are

y = 2x and y= -2x.

These lines are equally inclined to the axes.

125. It was stated in Art. 115 that, in general^ an

equation of the second degree represents a curve- line,

including (Art. 116) as a particular case two straight lines.

In some cases however it will be found that such

equations only represent isolated points. Some examples

are appended.

EjX. 1. What is represented hy the locus

{x-y + cf+{x + y-cY = 01 (1).

We know that the sum of the squares of two real

quantities cannot be zero unless each of the squares is

separately zero.

The only real points that satisfy the equation (1)

therefore satisfy both of the equations

cc â€” 2/ + c = and x + yâ€”G = 0.

But the only solution of these two equations is

re = Oj and y = c.

The only real point represented by equation (1) is therefore

(0,c).

The same result may be obtained in a different manner.

The equation (1) gives

{x-y-VGY^-(x + y-cf,

i.e. x â€” y + c = ^ Vâ€” 1 {x + y â€” c).

It therefore represents the two imaginary straight lines

x{l- J'^) -y(l + J^) + c (1 + J^) = 0,

and X (1 + J^l)-y (I - J^) + c(l - J~l) = 0.

EQUATIONS REPRESENTING ISOLATED POINTS. 103

Each of these two straight lines passes through the

real point (0, c). We may therefore say that (1) represents

two imaginary straight lines passing through the point

(0, 0).

Â£jX. 2. What is represented hy the equation

As in the last example, the only real points on the locus

are those that satisfy both of the equations

oc^ â€” a^ â€” and y^ â€” h^ = 0,

i.e. x = =i= a, and y = d=h.

The points represented are therefore

(a, h), {a, â€”h), (â€”a, b), and (â€”a, â€”6).

Ex. 3. WTiat is represented by the equation

The only real points on the locus are those that satisfy

all three of the equations

x = 0, 2/=0, and a = 0.

Hence, unless a vanishes, there are no such points, and

the given equation represents nothing real.

The equation may be written

a? + y^ = â€” a^y

so that it represents points whose distance from the origin

is aslâ€”\. It therefore represents the imaginary circle

whose radius is asJâ€”1 and whose centre is the origin.

126. Es. 1. Obtain the condition that one of the straight lines

given hy the equation

ax^-\-2hxy + by^ = (1)

may coincide with one of those given by the equation

a'x^ + 2h'xy + bY=0 (2).

Let the equation to the common straight line be

y-m^x = (3).

The quantity y -m^x must therefore be a factor of the left-hand of

both (1) and (2), and therefore the value y = m^x must satisfy both (1)

and (2).

104 COORDINATE GEOMETRY.

"We therefore have

bmi^ + 2hm^ + a=:0 (4),

and b'm^^ + 2h'm^ + a' = (5).

Solving (4) and (5), we have

wij^ _ '^h _ ^

2 {ha' - h'a) ~ aV -a'b~2 {bh' - b'h) '

^ ha'-h'a _ ^_ f ab'-a'b ]^

â– *â€¢ bh' -b'h~^~\2 {bh' - b'h)( '

so that we must have

{ab' - a'&)2 = 4 {ha' - h'a) {bh' - b'h) .

Ex. 2. Prove that the equation

m{x^-Sxy^) + y^-3x^y=0

represents three straight lines equally inclined to one another.

Transforming to polar coordinates (Art. 35) the equation gives

m {GQS^d- 3 cos d sin2^) + sin3^ - 3 cos^^ sin ^ = 0,

i.e. m(l-3tan2^) + tan3^-3tan^ = 0,

3tan^-tan3^ ^ â€ž^

If m=tan a, this equation gives

tan 3^ = tan a,

the solutions of which are

3^ = a, or 180Â° + a, or 360Â° + a,

i.e. ^ = |, or 60Â° + ^, or 120Â° + ^.

The locus is therefore three straight lines through the origin

inclined at angles

^, 60Â° + |, and 120Â° + ^

to the axis of x.

They are therefore equally inclined to one another.

Ex. 3. Prove that two of the straight lines represented by the

equation

ax'^ + bx^y + cxy^ + dy^ = (1)

will be at right angles if

a^ + ac + hd + d^ = 0.

Let the separate equations to the three lines be

y-miX = 0, y-m2X=0, and y-m^x=0,

EXAMPLES. 105

so that the equation (1) must be equivalent to

d{y - m-^x) (y - m^) (y - m^x) = 0,

c

and therefore mj^+m2+m^= - (2),

Wom3 + m3?7ij + 77ijm2 = -^ (3)i

and m^m^m^ = - -^ (4).

If the first two of these straight lines be at right angles we have,

in addition,

711^111.2= -1 (5).

From (4) and (5), we have

a

and therefore, from (2),

c a c + a

The equation (3) then becomes

a f c + a\ _b

i.e. a^ + ac + bd + d^ = 0.

EXAMPLES. XIV.

1. Prove that the equation

y^-x^ + 3xy (y -x) =

represents three straight lines equally inclined to one another.

2. Prove that the equation

y^ (cos a + fJ3 sin a) cos a-xy (sin 2a - ^^3 cos 2a)

+ x^ (sin a - ^3 cos a) sin a =

represents two straight lines inclined at 60Â° to each other.

Prove also that the area of the triangle formed with them by the

straight line

(cosa-/sy3 sin a) 2/ -(sin a + ;^3cos a)a; + a-=0

a2

V3'

and that this triangle is equilateral.

3. Shew that the straight lines

(^2 _ 3^2) ^2 + sABxy + [B'^ - 3A^) y^=0

form with the Hne Ax + By + C = an equilateral triangle whose area

^^ J3{A^ + B^)'

106 COORDINATE GEOMETRY. [ExS.

_ 4. Find the equation to the pair of straight lines joining the

origin to the intersections of the straight line y = mx + c and the curve

Prove that they are at right angles if

2c2 = a2(l + m2).

5. Prove that the straight lines joining the origin to the points

of intersection of the straight line

kx + hy = 2hk

with the curve {^-h)^+{y- k)^ = c^

are at right angles if h^+k^=c^.

6. Prove that the angle between the straight lines joining the

origin to the intersection of the straight line y = 3x + 2 with the curve

x'^ + 2xy + Sy^ + 'ix+8y-ll = istan-i?^.

3

7. Shew that the straight lines joining the origin to the other two

points of intersection of the curves whose equations are

ax^ + 2hxy + by^ + 2gx=0

and a'x'^ + 2h'xy + hY + ^g'x =

will be at right angles if

g{a' + b')-g'{a + b) = 0.

What loci are represented by the equations

8. x^-y^=0. 9. x'^-xy = 0. 10. xy-ay = 0.

11. x^-x^-x + l = 0. 12. x^-xy^ = 0. 13. x^ + y^ = 0.

14. x^ + y^=0. 15. x^y = 0. 16. {x' - l){y^-^)=0.

17. {x^-lf + {y^-4y=0. 18. {y-mx-cY + {y-m'x-c')^=0,

19. {a;2-a3)2(^2_52)2 + c'*(?/2-a2) = 0. 20. {x-a)^-y^=0.

21, (x + y)^-c^=0. 22. r=a sec (<?- a).

23. Shew tliat the equation

bx^-2hxy + ay^=0

represents a pair of straight lines which are at right angles to the pair

given by the equation

ax^ + 2 Jixy + by^ = 0.

24. If pairs of straight lines

x^-2pxy -y^=0 and x^-2qxy-y^=0

be such that each pair bisects the angles between the other pair, prove

thatjpgr= -1.

25. Prove that the pair of lines

a^x'^ + 2h{a + b)xy + b^y^=zO

is equally inclined to the pair

ax^ + 2hxy + by^=0.

XIV.] EXAMPLES. 107

26. Shew also that the pair

is equally inclined to the same pair.

27. If one of the straight lines given by the equation

ax^ + Ihxy + hy^ =

coincide with one of those given by

a'x^ + 2h'xy + bY=0,

and the other lines represented by them be perpendicular, prove that

ha'b' h'ah , â€” -

-a b-a ^

28. Prove that the equation to the bisectors of the angle between

the straight lines ax^ + 2hxy + by'^=0 is

h {x^ - y^) + {b - a) xy = {ax^ - by^) cos w,

the axes being inclined at an angle w.

29. Prove that the straight lines

ax^-\-1'kxy + by'^=zQ

make equal angles with the axis of a; if 7t = acosw, the axes being

inclined at an angle w.

30. If the axes be inclined at an angle w, shew that the equation

x^ + 2ocy cos w + 2/^cos 2a; =

represents a pair of perpendicular straight lines.

31. Shew that the equation

cos 3a {x^ - dxy^) + sin 3a (y^ - Sx'^y) + 3a (x^ + y^) -^a^ =

represents three straight lines forming an equilateral triangle.

Prove also that its area is 3 sJSa^.

32. Prove that the general equation

ax^ + 2Jixy + by^ + 2gx + 2fy + c=zO

represents two parallel straight lines if

h^ = ab and bg^=^af^.

Prove also that the distance between them is

/ 9'-

V a(a-

ac

{a + b)'

33. If the equation

ax'^ + 2hxy + by^ + 2gx + 2fy + c =

represent a pair of straight lines, prove that the equation to the third

pair of straight lines passing through the points where these meet the

axis is

ax^ - 2hxy + by^ + 2gx + 2fy + c + -^^xy = 0.

108 COORDINATE GEOMETRY. [Exs. XIV.]

34. If the equation

as(^ + 2hxy + by^ + 2gx + 2fy + c =

represent two straight lines, prove that the square of the distance of

their point of intersection from the origin is

35. Shew that the orthocentre of the triangle formed by the

straight lines

ax^ + 2kxy + by^=0 and lx + viy = l

is a point (x\ y') such that

^ _y' _ a + i>

I ~ m~ aw? - 2hlm + bP '

36. Hence find the locus of the orthocentre of a triangle of which

two sides are given in position and whose third side goes through a

fixed point.

37. Shew that the distance between the points of intersection of

the straight Hne

X cos a + y sin a-p =

with the straight lines ax^ + 2 hxy + dy^=0

2pjh^-ab

& cos^a - 2/i cos a sin a + a sin^ a '

Deduce the area of the triangle formed by them.

38. Prove that the product of the perpendiculars let fall from the

point {x', y') upon the pair of straight lines

ax^ + 2hxy + by"^ â€”

ax''^ + 21tx'y' + by'^

^^ J{a-bf + 4:h^ '

39. Shew that two of the straight lines represented by the

equation

ay"^ + bocy^ + cx^y^ -}-dx^y + ex'^ =

will be at right angles if

(& + d) {ad + be) + {e~ a)^ {a + c + e) = 0.

40. Prove that two of the lines represented by the equation

ax^ + bx^ y + cx^ y^ + dxy^ + ay^ =

will bisect the angles between the other two if

c + Qa=0 and b + d = 0.

41. Prove that one of the lines represented by the equation

ax^ + bx"^ y + cxy^ + dy^ =

will bisect the angle between the other two if

{3a + c)^{bc + 2cd - Sad) = (6+ Sd)^{bc + 2ab - 3ad).

CHAPTER yil.

TRANSFORMATION OF COORDINATES.

127. It is sometimes found desirable in the discussion

of problems to alter the origin and axes of coordinates,

either by altering the origin without alteration of the

direction of the axes, or by altering the directions of the

axes and keeping the origin unchanged, or by altering the

origin and also the directions of the axes. The latter case

is merely a combination of the first two. Either of these

processes is called a transformation of coordinates.

We proceed to establish the fundamental formulae for

such transformation of coordinates.

128. To alter the origin of coordinates without altering

the directions of the axes.

Let OX and Z be the original axes and let the new

axes, parallel to the original, be

O'X' and O'Y'.

Let the coordinates of the new

origin 0\ referred to the original

axes be h and k, so that, if O'L be

perpendicular to OX, we have

OL = h and LO' = h.

Let P be any point in the plane

of the paper, and let its coordinates, referred to the original

axes, be x and y, and referred to the new axes let them be

x' and y'.

Draw PN perpendicular to OX to meet OX' in N'.

Y'

p

N'

X'

N

110

COORDINATE GEOMETRY.

Then

ON^x, NF = y, 0'N' = x, and N'F^y'.

We therefore have

X ^ 0N= OL + O'N' = h + x\

and y = FP = LO' + N'P = k + y'.

The origin is therefore transferred to the point Qi, k) when

we substitute for the coordinates x and y the quantities

X + h and y' + k.

The above article is true whether the axes be oblique

or rectangular.

129. To change the direction of the axes of coordinates,

without changing the origin, both systems of coordinates being

rectangular.

Let OX and OF be the original system of axes and OX'

and OY' the new system, and let

the angle, XOX' , through which

the axes are turned be called 0. Y'

Take any point F in the plane

of the paper.

Draw FN and FN' perpen-

dicular to OX and OX', and also

N'L and N'M. perpendicular to OX and FN.

If the coordinates of F, referred to the original axes,

be X and y, and, referred to the new axes, be xi and y', we

have

ON^x, NF = y, ON'^x', and N'F = y.

The angle

MFN' - 90Â° - z MN'F^ l MN'O = z XOX' = 6.

We then have

X = 0N= OL -MN'= ON' cobO-N'F sine

= x' cos â€” y sin (1),

and y = NF = LN' + MF= ON' sin 6 + N'F cos $

= oj' sin ^ + 2/' cos ^ (2).

CHANGE OF AXES. Ill

If therefore in any equation we wish to turn the axes,

being rectangular, through an angle we must substitute

X' cos ^ â€” y' sin and x' sin ^ + y' cos

for X and y.

"When we have both to change the origin, and also the

direction of the axes, the transformation is clearly obtained

by combining the results of the previous articles.

If the origin is to be transformed to the point (Ji, k)

and the axes to be turned through an angle 6, we have to

substitute

h + X cos â€” y sin 6 and k + x sin + y cos 6

for X and y respectively.

The student, who is acquainted with the theory of projection of

straight lines, will see that equations (1) and (2) express the fact that

the projections of OP on OX and OY are respectively equal to the

sum of the projections of ON' and N'P on the same two lines.

130. Ex. 1. Transform to parallel axes through the 'point ( - 2, 3)

the equation

2a;2 + 4a;2/ + 5^/2 - 4^ - 22?/ + 7 =: 0.

We substitute x=x' -2 and y=y' + S, and the equation becomes

2 (x' - 2)3 + 4 (a;' - 2) (?/' + 3) + 5 (?/' + 3)2 - 4 (x' - 2) - 22 (i/' + 3) + 7 = 0,

i.e. 2x'^ + 4xy + 5y'^-22 = 0.

Ex. 2. Transform to axes inclined at 30Â° to the original axes the

equation

x'^ + 2fj3xy-y'^=2a^.

For X and y we have to substitute

flj' cos 30Â° - ?/ sin 30Â° and a;' sin 30Â° + ?/' cos 30Â°,

t.e. â€” ^^r â€” - and ^ - ^â€” .

The equation then becomes

(a:V3 -2/T + 2 V3 (^V3 -2/') (^' + 2/V3) - (^' + 2/V3)'=8a2,

i.e, x'^-y'^ = a^.

112 COORDINATE GEOMETRY.

EXAMPLES. XV.

1. Transform to parallel axes through the point (1, -2) the

equations

(1) y'^-4:X + 4y + 8 = 0,

and (2) 2x^ + y^-4:X + 4y = 0.

2. What does the equation

{x-a)^+{y-b)^=c^

become when it is transferred to parallel axes through

(1) the point {a-c, b),

(2) the point {a, b-c)?

3. What does the equation

{a-b){x^ + y^-)-2abx=0

become if the origin be moved to the point ( , ) ?

4. Transform to axes inclined at 45Â° to the original axes the

equations

(1) x^-y^ = a\

(2) nx^-l&xy + ny^ = 225,

and (3) y^ + x^ + 6x^y^=2.

5. Transform to axes inclined at an angle a to the original axes

the equations

(1) x^+y^=r^, .

and (2) a^ + 2xy tan 2a- y^=aK

6. If the axes be turned through an angle tan~i 2, what does the

equation Axy - 3x^ = a^ become ?

7. By transforming to parallel axes through a properly chosen

point {h, k), prove that the equation

12ciP-10xy + 2y^ + llx-5y + 2 = 0-

can be reduced to one containing only terms of the second degree.

8. Find the angle through which the axes may be turned so that

the equation Ax + By + 0=0

may be reduced to the form a; = constant, and determine the value of

this constant.

131. The general proposition, which is given in the

next article, on the transformation from one set of oblique

axes to any other set of oblique axes is of very little

importance and is hardly ever required.

CHANGE OF AXES. 113

â– *132. To change from one set of axes, inclined at an

angle w, to another set, inclined at an angle w', the origin

remaining unaltered.

\ __ â€” \ /""1M

O M N L X

Let OX and (9Fbe the original axes, OX' and OY' the

new axes, and let the angle XOX' be 0.

Take any point P in the plane of the paper.

Draw PN and PN' parallel to 07 and OY' to meet OX

and OX' respectively in N and iV", PL perpendicular to OX,

and N'M and X'M' perpendicular to OL and LP.

Now

z PNL = L YOX = 0), and PN'M' = Y'OX = oi' + B.

Hence if

OX^x, NP^y, ON'^x, s.rvAN'P^y',

we have y s>\n.oi = NP ^irna^ LP = MX' + M'P

= OX' sin + X'P sin (w' + 6),

so that y sin w = cc' sin 6 + y' sin (w' + ^) (1).

Also

x-ryco&oi^OX+XL=^OL=^OM-\-X'M'

= x cosO + y'cos{oy' + e) (2).

Multiplying (2) by sinw, (1) by cosw, and subtracting,

we have

X sin (o = x sin (w â€” ^) + y' sin (oi â€” w â€”6) (3).

[This equation (3) may also be obtained by drawing a perpen-

dicular from P upon OT and proceeding as for equation (1).]

The equations (1) and (3) give the proper substitutions

for the change of axes in the general case.

As in Art. 130 the equations (1) and (2) may be obtained by

equating the projections of OP and of ON' and N'P on OX and a

straight line perpendicular to OX.

L. , 8

114 COORDINATE GEOMETKY.

'^133. Particular' cases of the preceding article.

(1) Suppose we wish to transfer our axes from a

rectangular pair to one inclined at an angle w'. In this

case (0 is 90Â°, and the formulse of the preceding article

become

x = x' cos 6 + y' cos (w' + 6),

and y = ^' sin 6 + y sin (w' + 6).

(2) Suppose the transference is to be from oblique

axes, inclined at w, to rectangular axes. In this case co' is

90Â°, and our formulse become

X sin isi = X sin (w â€” ^) â€” y' cos (w â€” 0)^

and y sin o> = a:;' sin 6 + y' cos ^.

These particular formulse may easily be proved in-

dependently, by drawing the corresponding figures.

Ex. Transform the equation -2 - f^ = l from rectangular axes to

axes inclined at an angle 2a, the new axis of x being inclined at an angle

â€” a to the old axes and sin a being equal to â€” , .

Here 6= -a and w' = 2a, so that the formulse of transformation

(1) become

a; = (a;' + y) cos a and y = [y' - x') sin a.

Since sin a = , , we have cos a = , , and hence the

x/a2+&2 Ja'^ + b'^

given equation becomes

i.e. x'y'=i{a'^ + b^).

â– *134. The degree of an equation is unchanged hy any

transformation of coordinates.

For the most general form of transformation is found

by combining together Arts. 128 and 132, Hence the

most general formulse of transformation are

, sin (oi â€” 6) , sin (o> â€” w' â€” 6)

x = h-Â¥x ^ + y ^>-7 ,

sm 0) sm w

- _ , sin , sin (w' + &)

and y = k-vx -. â€” + y ^ .

smo) sinw

CHANGE OF AXES. 115

For X and y we have therefore to substitute expressions

in X and y' of the first degree, so that by this substitution

the degree of the equation cannot be raised.

Neither can, by this substitution, the degree be lowered.

For, if it could, then, by transforming back again, the

degree would be raised and this we have just shewn to be

impossible.

*135. If by any change of axes, without change of origin, the

quantity ax^ + Ihxy + &i/^ become

a'x'^ + 2h'xY + by%

the axes in each case being rectangular, to prove that

a + b = a' + b', and ab-h^ = a'b' -h'^.

By Art. 129, the new axis of x being inclined at an angle 6 to the

old axis, we have to substitute

a;'cos^-2/'sin^ and x' sin + y' cos. 6

for X and y respectively.

Hence ax"^ + 2hxy + by'^

= a{x'cosd-y' sin ef + 2h {x' cose~y' sin d) {x' sin d + y' cos d)

+ b{x'smd + y'cos6)^

= x'^ [a cos2 e + 2h cos Osind + b sin^ d]

+ 2x'y' [-acosdsind + h (cos^ d - sin^i?) + & cos ^ sin 9]

+ y'^ \a sin2 d-2h cos ^ sin ^ + & cos^ d].

We then have

tt' = a cos^ ^ + 2/i cos ^ sin ^ + 6 sin^ (9

= 1 [(a + &) + (Â« -6) cos 2^ + 2/1 sin 2^] (1),

b' = aBin^d- 2h cos ^ sin ^ + 6 cos^^

= i[(a + 6)-(a-&)cos2^-27isin2^] (2),

and /i'= -acos^sin^ + /i(cos2^-sin2^) + &cos^sin^

= i[2;icos2^-(a-&)sin2^j (3).

By adding ( 1) and (2) , we have Â«' + &'=Â« + &.

Also, by multiplying them, we have

4a'&' = (a + 6)2 - { (a - h) cos 2d +-2h sin 26}^.

Hence 4a'&' - 47i'2

= (a + 6)2 _ i{2h sin 2d + {a- b) cos 20]^+ {2h cos 20 -{a- b) sin 20}^

= (a + &)2 _ [(a - 6)2 + 4/i2] = 4a6 - 4h%

so that a'b'-h'^ = ab-h^.

136. To find the angle through which the axes must be turned so

that the expression ax^ + 2hxy + by^ may become an expression in which

there is no term involving x'y'.

8â€”2

116 COORDINATE GEOMETRY.

Assuming the work of the previous article the coefficient of x'y'

vanishes if h' be zero, or, from equation (3), if

27icos2^ = (a-Z>)sin2^,

2h

i.e. if tan 2(9= -.

a~h

The required angle is therefore

i tan~i

â– â– (S)

*137. The proposition of Art. 135 is a particular

case, when the axes are rectangular, of the folloM^ing more

general proposition.

If hy any change of axes, without change of origin, the

quantity ax- + ^hxy + 6?/" becomes a'x'^ + 2h'xy + 6'^/^, then

a + b â€” 2h cos w a' + b' â€” 2h' cos w'

Find the value of k so that the following equations may represent

pairs of straight lines :

6. 6a;2+lla;?/-10?/2 + a; + 31?/ + ^ = 0.

7. 12a;2-10a;?/ + 2?/2 + lla;-5?/ + A; = 0.

8. 12a;2 + A;a;?/ + 2?/2 + lla;-5?/ + 2:=0.

9. 6a;2 + a;?/ + A-?/2-ll.r + 43?/-35 = 0.

[EXS. XIII.] EXAMPLES. 99

10. kxy-8x + 9y-12 = 0.

11. x'^ + is^-xij + 'if-5x-7y + k=:0.

12. 12x^ + xy-Qif-2dx + 8y + k = 0.

13. 2x^ + oiyy-y'^ + kx + 6y-d = 0.

14. x^ + kxy + y^-5x-ly + Q = 0.

15. Prove that the equations to the straight lines passing through

the origin which make an angle a with the straight line y + x = are

given by the equation

x^ + 2xy sec 2a + y^ = Q.

16. What relations must hold between the coordinates of the

equations

(i) ax^ + hy'^ â– ^cx-\-cy = 0,

and (ii) ay'^ + lxy + dy -{-ex=zQ^

so that each of them may represent a pair of straight lines ?

17. The equations to a pair of opposite sides of a parallelogram

are

a;2-7a; + 6 = and 2/^- 14?/ +40 = ;

find the equations to its diagonals.

120. To prove that a homogeneous equation of the nth

degree represents n straight li7ies, real or imaginary, which

all pass through the orighi.

Let the equation be

y'^ + A^xy''-^ + A^xY'-^ + A.^x'y''-^ 4- . . . + A.^x'' = - 0.

On division by x^, it may be written

This is an equation of the nth degree in - , and hence

JO

must have n roots.

Let these roots be -m^, m^, mg, ... mâ€ž. Then (C. Smith's

Algebra, Art. 89) the equation (1) must be equivalent to

the equation

t

The equation (2) is satisfied by all the points which

satisfy the separate equations

'^)6-"'^)(l-â„¢0- (!-'"Â«)=â€¢'â– â– â€¢<'>â€¢

^ _ mi = 0, ^ - m. = 0, . . . ^ - m,, - 0,

/v. ^ 'T* Â« 5

tLf *^ tKf

7â€”2

100 COORDINATE GEOMETRY.

i.e. by all the points which lie on the n straight lines

y â€” nijX = 0, y â€” m^x = 0, ... y â€” in^x = 0,

all of which pass through the origin. Conversely, the

coordinates of all the points which satisfy these n equa-

tions satisfy equation (1). Hence the proposition.

121. Ex. 1. The equation

which is equivalent to

{y-x){y-'lx){y-^x) = Q,

represents the three straight lines

y-x=.^, ?/-2a: = 0, and y-3a; = 0,

all of which pass through the origin.

Ex. 2. The equation y^ - 5y^ + 6y = 0,

i.e. y{y-2){y-S) = 0,

similarly represents the three straight lines

y=0, y = 2, and y = S,

all of which are parallel to the axis of x.

122. To find the equation to the two straight lines

joining the origin to the points in which the straight line

Ix + my = n (1)

meets the locus whose equation is

ax^ + 2hxy + hy"^ + "Igx + 2fy + c = (2).

The equation (1) may be written

Ix + my

n

= 1 (3).

The coordinates of the points in which the straight line

meets the locus satisfy both equation (2) and equation (3),

and hence satisfy the equation

ax' + Vum) + by^ + 2 {gx +/y) '-^^ + c ('-^^)' =

......(4).

[For at the points where (3) and (4) are true it is clear

that (2) is true.]

Hence (4) represents so7ne locus which passes through

the intersections of (2) and (3).

STRAIGHT LINES THROUGH THE ORIGIN. 101

But, since the equation (4) is homogeneous and of the

second degree, it represents two straight lines passing

through the origin (Art. 108).

It therefore must represent the two straight lines join-

ing the origin to the intersections of (2) and (3).

123. The preceding article may be illustrated geo-

metrically if we assume that the equation (2) represents

some such curve as PQRS in the figure.

â– "'/O

Let the given straight line cut the curve in the points

P and Q.

The equation (2) holds for all points on the curve PQRS.

The equation (3) holds for all points on the line PQ.

Both equations are therefore true at the points of

intersection P and Q.

The equation (4), which is derived from (2) and (3),

holds therefore at P and Q.

But the equation (4) represents two straight lines, each

of which passes through the point 0.

It must therefore represent the two straight lines OP

and OQ.

124. Ex. Prove that the straight lines joining the origin to the

points of intersection of the straight line x-y=:2 and the curve

5a;2 + 12a;i/ - 82/2 + 8a; - 4y + 12 =

make equal angles with the axes.

As in Art. 122 the equation to the required straight lines is

5a;2 + 12a;?/-8i/2+(8a;-4i/)^^ + 12fc^Y = (1),

102 â€¢ COORDINATE GEOMETRY.

For this equation is homogeneous and therefore represents two

straight lines through the origin; also it is satisfied at the points

where the two given equations are satisfied.

Now (1) is, on reduction,

so that the equations to the two lines are

y = 2x and y= -2x.

These lines are equally inclined to the axes.

125. It was stated in Art. 115 that, in general^ an

equation of the second degree represents a curve- line,

including (Art. 116) as a particular case two straight lines.

In some cases however it will be found that such

equations only represent isolated points. Some examples

are appended.

EjX. 1. What is represented hy the locus

{x-y + cf+{x + y-cY = 01 (1).

We know that the sum of the squares of two real

quantities cannot be zero unless each of the squares is

separately zero.

The only real points that satisfy the equation (1)

therefore satisfy both of the equations

cc â€” 2/ + c = and x + yâ€”G = 0.

But the only solution of these two equations is

re = Oj and y = c.

The only real point represented by equation (1) is therefore

(0,c).

The same result may be obtained in a different manner.

The equation (1) gives

{x-y-VGY^-(x + y-cf,

i.e. x â€” y + c = ^ Vâ€” 1 {x + y â€” c).

It therefore represents the two imaginary straight lines

x{l- J'^) -y(l + J^) + c (1 + J^) = 0,

and X (1 + J^l)-y (I - J^) + c(l - J~l) = 0.

EQUATIONS REPRESENTING ISOLATED POINTS. 103

Each of these two straight lines passes through the

real point (0, c). We may therefore say that (1) represents

two imaginary straight lines passing through the point

(0, 0).

Â£jX. 2. What is represented hy the equation

As in the last example, the only real points on the locus

are those that satisfy both of the equations

oc^ â€” a^ â€” and y^ â€” h^ = 0,

i.e. x = =i= a, and y = d=h.

The points represented are therefore

(a, h), {a, â€”h), (â€”a, b), and (â€”a, â€”6).

Ex. 3. WTiat is represented by the equation

The only real points on the locus are those that satisfy

all three of the equations

x = 0, 2/=0, and a = 0.

Hence, unless a vanishes, there are no such points, and

the given equation represents nothing real.

The equation may be written

a? + y^ = â€” a^y

so that it represents points whose distance from the origin

is aslâ€”\. It therefore represents the imaginary circle

whose radius is asJâ€”1 and whose centre is the origin.

126. Es. 1. Obtain the condition that one of the straight lines

given hy the equation

ax^-\-2hxy + by^ = (1)

may coincide with one of those given by the equation

a'x^ + 2h'xy + bY=0 (2).

Let the equation to the common straight line be

y-m^x = (3).

The quantity y -m^x must therefore be a factor of the left-hand of

both (1) and (2), and therefore the value y = m^x must satisfy both (1)

and (2).

104 COORDINATE GEOMETRY.

"We therefore have

bmi^ + 2hm^ + a=:0 (4),

and b'm^^ + 2h'm^ + a' = (5).

Solving (4) and (5), we have

wij^ _ '^h _ ^

2 {ha' - h'a) ~ aV -a'b~2 {bh' - b'h) '

^ ha'-h'a _ ^_ f ab'-a'b ]^

â– *â€¢ bh' -b'h~^~\2 {bh' - b'h)( '

so that we must have

{ab' - a'&)2 = 4 {ha' - h'a) {bh' - b'h) .

Ex. 2. Prove that the equation

m{x^-Sxy^) + y^-3x^y=0

represents three straight lines equally inclined to one another.

Transforming to polar coordinates (Art. 35) the equation gives

m {GQS^d- 3 cos d sin2^) + sin3^ - 3 cos^^ sin ^ = 0,

i.e. m(l-3tan2^) + tan3^-3tan^ = 0,

3tan^-tan3^ ^ â€ž^

If m=tan a, this equation gives

tan 3^ = tan a,

the solutions of which are

3^ = a, or 180Â° + a, or 360Â° + a,

i.e. ^ = |, or 60Â° + ^, or 120Â° + ^.

The locus is therefore three straight lines through the origin

inclined at angles

^, 60Â° + |, and 120Â° + ^

to the axis of x.

They are therefore equally inclined to one another.

Ex. 3. Prove that two of the straight lines represented by the

equation

ax'^ + bx^y + cxy^ + dy^ = (1)

will be at right angles if

a^ + ac + hd + d^ = 0.

Let the separate equations to the three lines be

y-miX = 0, y-m2X=0, and y-m^x=0,

EXAMPLES. 105

so that the equation (1) must be equivalent to

d{y - m-^x) (y - m^) (y - m^x) = 0,

c

and therefore mj^+m2+m^= - (2),

Wom3 + m3?7ij + 77ijm2 = -^ (3)i

and m^m^m^ = - -^ (4).

If the first two of these straight lines be at right angles we have,

in addition,

711^111.2= -1 (5).

From (4) and (5), we have

a

and therefore, from (2),

c a c + a

The equation (3) then becomes

a f c + a\ _b

i.e. a^ + ac + bd + d^ = 0.

EXAMPLES. XIV.

1. Prove that the equation

y^-x^ + 3xy (y -x) =

represents three straight lines equally inclined to one another.

2. Prove that the equation

y^ (cos a + fJ3 sin a) cos a-xy (sin 2a - ^^3 cos 2a)

+ x^ (sin a - ^3 cos a) sin a =

represents two straight lines inclined at 60Â° to each other.

Prove also that the area of the triangle formed with them by the

straight line

(cosa-/sy3 sin a) 2/ -(sin a + ;^3cos a)a; + a-=0

a2

V3'

and that this triangle is equilateral.

3. Shew that the straight lines

(^2 _ 3^2) ^2 + sABxy + [B'^ - 3A^) y^=0

form with the Hne Ax + By + C = an equilateral triangle whose area

^^ J3{A^ + B^)'

106 COORDINATE GEOMETRY. [ExS.

_ 4. Find the equation to the pair of straight lines joining the

origin to the intersections of the straight line y = mx + c and the curve

Prove that they are at right angles if

2c2 = a2(l + m2).

5. Prove that the straight lines joining the origin to the points

of intersection of the straight line

kx + hy = 2hk

with the curve {^-h)^+{y- k)^ = c^

are at right angles if h^+k^=c^.

6. Prove that the angle between the straight lines joining the

origin to the intersection of the straight line y = 3x + 2 with the curve

x'^ + 2xy + Sy^ + 'ix+8y-ll = istan-i?^.

3

7. Shew that the straight lines joining the origin to the other two

points of intersection of the curves whose equations are

ax^ + 2hxy + by^ + 2gx=0

and a'x'^ + 2h'xy + hY + ^g'x =

will be at right angles if

g{a' + b')-g'{a + b) = 0.

What loci are represented by the equations

8. x^-y^=0. 9. x'^-xy = 0. 10. xy-ay = 0.

11. x^-x^-x + l = 0. 12. x^-xy^ = 0. 13. x^ + y^ = 0.

14. x^ + y^=0. 15. x^y = 0. 16. {x' - l){y^-^)=0.

17. {x^-lf + {y^-4y=0. 18. {y-mx-cY + {y-m'x-c')^=0,

19. {a;2-a3)2(^2_52)2 + c'*(?/2-a2) = 0. 20. {x-a)^-y^=0.

21, (x + y)^-c^=0. 22. r=a sec (<?- a).

23. Shew tliat the equation

bx^-2hxy + ay^=0

represents a pair of straight lines which are at right angles to the pair

given by the equation

ax^ + 2 Jixy + by^ = 0.

24. If pairs of straight lines

x^-2pxy -y^=0 and x^-2qxy-y^=0

be such that each pair bisects the angles between the other pair, prove

thatjpgr= -1.

25. Prove that the pair of lines

a^x'^ + 2h{a + b)xy + b^y^=zO

is equally inclined to the pair

ax^ + 2hxy + by^=0.

XIV.] EXAMPLES. 107

26. Shew also that the pair

is equally inclined to the same pair.

27. If one of the straight lines given by the equation

ax^ + Ihxy + hy^ =

coincide with one of those given by

a'x^ + 2h'xy + bY=0,

and the other lines represented by them be perpendicular, prove that

ha'b' h'ah , â€” -

-a b-a ^

28. Prove that the equation to the bisectors of the angle between

the straight lines ax^ + 2hxy + by'^=0 is

h {x^ - y^) + {b - a) xy = {ax^ - by^) cos w,

the axes being inclined at an angle w.

29. Prove that the straight lines

ax^-\-1'kxy + by'^=zQ

make equal angles with the axis of a; if 7t = acosw, the axes being

inclined at an angle w.

30. If the axes be inclined at an angle w, shew that the equation

x^ + 2ocy cos w + 2/^cos 2a; =

represents a pair of perpendicular straight lines.

31. Shew that the equation

cos 3a {x^ - dxy^) + sin 3a (y^ - Sx'^y) + 3a (x^ + y^) -^a^ =

represents three straight lines forming an equilateral triangle.

Prove also that its area is 3 sJSa^.

32. Prove that the general equation

ax^ + 2Jixy + by^ + 2gx + 2fy + c=zO

represents two parallel straight lines if

h^ = ab and bg^=^af^.

Prove also that the distance between them is

/ 9'-

V a(a-

ac

{a + b)'

33. If the equation

ax'^ + 2hxy + by^ + 2gx + 2fy + c =

represent a pair of straight lines, prove that the equation to the third

pair of straight lines passing through the points where these meet the

axis is

ax^ - 2hxy + by^ + 2gx + 2fy + c + -^^xy = 0.

108 COORDINATE GEOMETRY. [Exs. XIV.]

34. If the equation

as(^ + 2hxy + by^ + 2gx + 2fy + c =

represent two straight lines, prove that the square of the distance of

their point of intersection from the origin is

35. Shew that the orthocentre of the triangle formed by the

straight lines

ax^ + 2kxy + by^=0 and lx + viy = l

is a point (x\ y') such that

^ _y' _ a + i>

I ~ m~ aw? - 2hlm + bP '

36. Hence find the locus of the orthocentre of a triangle of which

two sides are given in position and whose third side goes through a

fixed point.

37. Shew that the distance between the points of intersection of

the straight Hne

X cos a + y sin a-p =

with the straight lines ax^ + 2 hxy + dy^=0

2pjh^-ab

& cos^a - 2/i cos a sin a + a sin^ a '

Deduce the area of the triangle formed by them.

38. Prove that the product of the perpendiculars let fall from the

point {x', y') upon the pair of straight lines

ax^ + 2hxy + by"^ â€”

ax''^ + 21tx'y' + by'^

^^ J{a-bf + 4:h^ '

39. Shew that two of the straight lines represented by the

equation

ay"^ + bocy^ + cx^y^ -}-dx^y + ex'^ =

will be at right angles if

(& + d) {ad + be) + {e~ a)^ {a + c + e) = 0.

40. Prove that two of the lines represented by the equation

ax^ + bx^ y + cx^ y^ + dxy^ + ay^ =

will bisect the angles between the other two if

c + Qa=0 and b + d = 0.

41. Prove that one of the lines represented by the equation

ax^ + bx"^ y + cxy^ + dy^ =

will bisect the angle between the other two if

{3a + c)^{bc + 2cd - Sad) = (6+ Sd)^{bc + 2ab - 3ad).

CHAPTER yil.

TRANSFORMATION OF COORDINATES.

127. It is sometimes found desirable in the discussion

of problems to alter the origin and axes of coordinates,

either by altering the origin without alteration of the

direction of the axes, or by altering the directions of the

axes and keeping the origin unchanged, or by altering the

origin and also the directions of the axes. The latter case

is merely a combination of the first two. Either of these

processes is called a transformation of coordinates.

We proceed to establish the fundamental formulae for

such transformation of coordinates.

128. To alter the origin of coordinates without altering

the directions of the axes.

Let OX and Z be the original axes and let the new

axes, parallel to the original, be

O'X' and O'Y'.

Let the coordinates of the new

origin 0\ referred to the original

axes be h and k, so that, if O'L be

perpendicular to OX, we have

OL = h and LO' = h.

Let P be any point in the plane

of the paper, and let its coordinates, referred to the original

axes, be x and y, and referred to the new axes let them be

x' and y'.

Draw PN perpendicular to OX to meet OX' in N'.

Y'

p

N'

X'

N

110

COORDINATE GEOMETRY.

Then

ON^x, NF = y, 0'N' = x, and N'F^y'.

We therefore have

X ^ 0N= OL + O'N' = h + x\

and y = FP = LO' + N'P = k + y'.

The origin is therefore transferred to the point Qi, k) when

we substitute for the coordinates x and y the quantities

X + h and y' + k.

The above article is true whether the axes be oblique

or rectangular.

129. To change the direction of the axes of coordinates,

without changing the origin, both systems of coordinates being

rectangular.

Let OX and OF be the original system of axes and OX'

and OY' the new system, and let

the angle, XOX' , through which

the axes are turned be called 0. Y'

Take any point F in the plane

of the paper.

Draw FN and FN' perpen-

dicular to OX and OX', and also

N'L and N'M. perpendicular to OX and FN.

If the coordinates of F, referred to the original axes,

be X and y, and, referred to the new axes, be xi and y', we

have

ON^x, NF = y, ON'^x', and N'F = y.

The angle

MFN' - 90Â° - z MN'F^ l MN'O = z XOX' = 6.

We then have

X = 0N= OL -MN'= ON' cobO-N'F sine

= x' cos â€” y sin (1),

and y = NF = LN' + MF= ON' sin 6 + N'F cos $

= oj' sin ^ + 2/' cos ^ (2).

CHANGE OF AXES. Ill

If therefore in any equation we wish to turn the axes,

being rectangular, through an angle we must substitute

X' cos ^ â€” y' sin and x' sin ^ + y' cos

for X and y.

"When we have both to change the origin, and also the

direction of the axes, the transformation is clearly obtained

by combining the results of the previous articles.

If the origin is to be transformed to the point (Ji, k)

and the axes to be turned through an angle 6, we have to

substitute

h + X cos â€” y sin 6 and k + x sin + y cos 6

for X and y respectively.

The student, who is acquainted with the theory of projection of

straight lines, will see that equations (1) and (2) express the fact that

the projections of OP on OX and OY are respectively equal to the

sum of the projections of ON' and N'P on the same two lines.

130. Ex. 1. Transform to parallel axes through the 'point ( - 2, 3)

the equation

2a;2 + 4a;2/ + 5^/2 - 4^ - 22?/ + 7 =: 0.

We substitute x=x' -2 and y=y' + S, and the equation becomes

2 (x' - 2)3 + 4 (a;' - 2) (?/' + 3) + 5 (?/' + 3)2 - 4 (x' - 2) - 22 (i/' + 3) + 7 = 0,

i.e. 2x'^ + 4xy + 5y'^-22 = 0.

Ex. 2. Transform to axes inclined at 30Â° to the original axes the

equation

x'^ + 2fj3xy-y'^=2a^.

For X and y we have to substitute

flj' cos 30Â° - ?/ sin 30Â° and a;' sin 30Â° + ?/' cos 30Â°,

t.e. â€” ^^r â€” - and ^ - ^â€” .

The equation then becomes

(a:V3 -2/T + 2 V3 (^V3 -2/') (^' + 2/V3) - (^' + 2/V3)'=8a2,

i.e, x'^-y'^ = a^.

112 COORDINATE GEOMETRY.

EXAMPLES. XV.

1. Transform to parallel axes through the point (1, -2) the

equations

(1) y'^-4:X + 4y + 8 = 0,

and (2) 2x^ + y^-4:X + 4y = 0.

2. What does the equation

{x-a)^+{y-b)^=c^

become when it is transferred to parallel axes through

(1) the point {a-c, b),

(2) the point {a, b-c)?

3. What does the equation

{a-b){x^ + y^-)-2abx=0

become if the origin be moved to the point ( , ) ?

4. Transform to axes inclined at 45Â° to the original axes the

equations

(1) x^-y^ = a\

(2) nx^-l&xy + ny^ = 225,

and (3) y^ + x^ + 6x^y^=2.

5. Transform to axes inclined at an angle a to the original axes

the equations

(1) x^+y^=r^, .

and (2) a^ + 2xy tan 2a- y^=aK

6. If the axes be turned through an angle tan~i 2, what does the

equation Axy - 3x^ = a^ become ?

7. By transforming to parallel axes through a properly chosen

point {h, k), prove that the equation

12ciP-10xy + 2y^ + llx-5y + 2 = 0-

can be reduced to one containing only terms of the second degree.

8. Find the angle through which the axes may be turned so that

the equation Ax + By + 0=0

may be reduced to the form a; = constant, and determine the value of

this constant.

131. The general proposition, which is given in the

next article, on the transformation from one set of oblique

axes to any other set of oblique axes is of very little

importance and is hardly ever required.

CHANGE OF AXES. 113

â– *132. To change from one set of axes, inclined at an

angle w, to another set, inclined at an angle w', the origin

remaining unaltered.

\ __ â€” \ /""1M

O M N L X

Let OX and (9Fbe the original axes, OX' and OY' the

new axes, and let the angle XOX' be 0.

Take any point P in the plane of the paper.

Draw PN and PN' parallel to 07 and OY' to meet OX

and OX' respectively in N and iV", PL perpendicular to OX,

and N'M and X'M' perpendicular to OL and LP.

Now

z PNL = L YOX = 0), and PN'M' = Y'OX = oi' + B.

Hence if

OX^x, NP^y, ON'^x, s.rvAN'P^y',

we have y s>\n.oi = NP ^irna^ LP = MX' + M'P

= OX' sin + X'P sin (w' + 6),

so that y sin w = cc' sin 6 + y' sin (w' + ^) (1).

Also

x-ryco&oi^OX+XL=^OL=^OM-\-X'M'

= x cosO + y'cos{oy' + e) (2).

Multiplying (2) by sinw, (1) by cosw, and subtracting,

we have

X sin (o = x sin (w â€” ^) + y' sin (oi â€” w â€”6) (3).

[This equation (3) may also be obtained by drawing a perpen-

dicular from P upon OT and proceeding as for equation (1).]

The equations (1) and (3) give the proper substitutions

for the change of axes in the general case.

As in Art. 130 the equations (1) and (2) may be obtained by

equating the projections of OP and of ON' and N'P on OX and a

straight line perpendicular to OX.

L. , 8

114 COORDINATE GEOMETKY.

'^133. Particular' cases of the preceding article.

(1) Suppose we wish to transfer our axes from a

rectangular pair to one inclined at an angle w'. In this

case (0 is 90Â°, and the formulse of the preceding article

become

x = x' cos 6 + y' cos (w' + 6),

and y = ^' sin 6 + y sin (w' + 6).

(2) Suppose the transference is to be from oblique

axes, inclined at w, to rectangular axes. In this case co' is

90Â°, and our formulse become

X sin isi = X sin (w â€” ^) â€” y' cos (w â€” 0)^

and y sin o> = a:;' sin 6 + y' cos ^.

These particular formulse may easily be proved in-

dependently, by drawing the corresponding figures.

Ex. Transform the equation -2 - f^ = l from rectangular axes to

axes inclined at an angle 2a, the new axis of x being inclined at an angle

â€” a to the old axes and sin a being equal to â€” , .

Here 6= -a and w' = 2a, so that the formulse of transformation

(1) become

a; = (a;' + y) cos a and y = [y' - x') sin a.

Since sin a = , , we have cos a = , , and hence the

x/a2+&2 Ja'^ + b'^

given equation becomes

i.e. x'y'=i{a'^ + b^).

â– *134. The degree of an equation is unchanged hy any

transformation of coordinates.

For the most general form of transformation is found

by combining together Arts. 128 and 132, Hence the

most general formulse of transformation are

, sin (oi â€” 6) , sin (o> â€” w' â€” 6)

x = h-Â¥x ^ + y ^>-7 ,

sm 0) sm w

- _ , sin , sin (w' + &)

and y = k-vx -. â€” + y ^ .

smo) sinw

CHANGE OF AXES. 115

For X and y we have therefore to substitute expressions

in X and y' of the first degree, so that by this substitution

the degree of the equation cannot be raised.

Neither can, by this substitution, the degree be lowered.

For, if it could, then, by transforming back again, the

degree would be raised and this we have just shewn to be

impossible.

*135. If by any change of axes, without change of origin, the

quantity ax^ + Ihxy + &i/^ become

a'x'^ + 2h'xY + by%

the axes in each case being rectangular, to prove that

a + b = a' + b', and ab-h^ = a'b' -h'^.

By Art. 129, the new axis of x being inclined at an angle 6 to the

old axis, we have to substitute

a;'cos^-2/'sin^ and x' sin + y' cos. 6

for X and y respectively.

Hence ax"^ + 2hxy + by'^

= a{x'cosd-y' sin ef + 2h {x' cose~y' sin d) {x' sin d + y' cos d)

+ b{x'smd + y'cos6)^

= x'^ [a cos2 e + 2h cos Osind + b sin^ d]

+ 2x'y' [-acosdsind + h (cos^ d - sin^i?) + & cos ^ sin 9]

+ y'^ \a sin2 d-2h cos ^ sin ^ + & cos^ d].

We then have

tt' = a cos^ ^ + 2/i cos ^ sin ^ + 6 sin^ (9

= 1 [(a + &) + (Â« -6) cos 2^ + 2/1 sin 2^] (1),

b' = aBin^d- 2h cos ^ sin ^ + 6 cos^^

= i[(a + 6)-(a-&)cos2^-27isin2^] (2),

and /i'= -acos^sin^ + /i(cos2^-sin2^) + &cos^sin^

= i[2;icos2^-(a-&)sin2^j (3).

By adding ( 1) and (2) , we have Â«' + &'=Â« + &.

Also, by multiplying them, we have

4a'&' = (a + 6)2 - { (a - h) cos 2d +-2h sin 26}^.

Hence 4a'&' - 47i'2

= (a + 6)2 _ i{2h sin 2d + {a- b) cos 20]^+ {2h cos 20 -{a- b) sin 20}^

= (a + &)2 _ [(a - 6)2 + 4/i2] = 4a6 - 4h%

so that a'b'-h'^ = ab-h^.

136. To find the angle through which the axes must be turned so

that the expression ax^ + 2hxy + by^ may become an expression in which

there is no term involving x'y'.

8â€”2

116 COORDINATE GEOMETRY.

Assuming the work of the previous article the coefficient of x'y'

vanishes if h' be zero, or, from equation (3), if

27icos2^ = (a-Z>)sin2^,

2h

i.e. if tan 2(9= -.

a~h

The required angle is therefore

i tan~i

â– â– (S)

*137. The proposition of Art. 135 is a particular

case, when the axes are rectangular, of the folloM^ing more

general proposition.

If hy any change of axes, without change of origin, the

quantity ax- + ^hxy + 6?/" becomes a'x'^ + 2h'xy + 6'^/^, then

a + b â€” 2h cos w a' + b' â€” 2h' cos w'

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