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S. L. (Sidney Luxton) Loney.

The elements of coordinate geometry online

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4:X-Sij + C=0 (2).

[We should expect an arbitrary constant in (2) because there are
an infinite number of straight lines perpendicular to (1).]
The straight line (2) passes through the point (4, - 5) if
4x4-3x(-5) + C = 0,
i.e. a (7= -16-15= -31.

The required equation is therefore

4:X-Sy = 31.
Second Method. Any straight line passing through the given
point is

y -{-5)=m{x~4:).

This straight line is perpendicular to (1) if the product of their
m's is - 1,
i.e. if m X ( - 1) = - 1,

i.e. if m=|.

The required equation is therefore

y + 5=i{x-4),
i.e. 4:X-'6y = Sl.

Third Method. Any straight line is y=mx + c. It passes through
the point (4, - 5), if

-5 = 4m + c (3).

It is perpendicular to (1) if

mx{-i)=-l (4).

Hence m = f and then (3) gives c = — V.

The required equation is therefore y = '^x-^-^,
i.e. 4x-By = Sl.

[In the first method, we start with any straight line which is
perpendicular to the given straight line and pick out that particular
straight line which goes through the given point.

In the second method, we start with any straight line passing
through the given point and pick out that particular one which is
perpendicular to the given straight hne.

In the third method, we start with any straight line whatever and
determine its constants, so that it may satisfy the two given
conditions.

The student should illustrate by figures. ]

Ex. 2. Find the equation to the straight line which passes through
the point (x', y') and is perpendicular to the given straight line

yy' = 2a {x + x').



THE STRAIGHT LINE. 47

The given straight line is

yy' - 2ax - 2ax' = 0.

Any straight line perpendicular to it is (Art. 70)

2ay + xy'+G=0 (1).

This will pass through the point (x', y') and therefore will be the
straight line required if the coordinates x' and y' satisfy it,

i.eAt 2ai/ + xY+C = 0,

i.e. if G=-2ay' -x'y'.

Substituting in (1) for G the required equation is therefore

2a{y-y') + y'{x-x') = 0.

72. To find the equations to the straight lines which
pass through a given point (x', t/') and make a given angle a
with the given straight line y — nix + c.

Let P be the given point and let the given straight line
be LMJSf, making an angle
with the axis of x such that

tan = m.

In general (i.e. except when
a is a right angle or zero) there
are two straight lines PMR and
FNS making an angle a with
the given line.

Let these lines meet the axis of x in R and S and let
them make angles ^ and <^' with the positive direction of
the axis of x.

The equations to the two required straight lines are
therefore (by Art. 62)

y -y' = tan ^ x (x — x) (1),

and y ~y' ~ t^^ <fi' X (x — x') (2).

Now cf, = L LMR + L RLM = a + 6*,

and <^'^L LNS+ L SLN= (180° ^a) + e.

Hence

■ , ^. tan a + tan ^ tana + m

tan <^ = tan (a + ^) - .j — ^ = ,

i — tan a tan ff i —m tan a

and tan </>' = tan ( 1 80° + ^ - a)

. „ . tan d — tan a m— tan a

= tan (^ — a) = ^— -p, — = ., .

1 + tan tan a 1 +7n tan a




48 COORDINATE GEOMETRY.

On substituting these values in (1) and (2), we have as
the required equations

, m + tan a , ,,

^ ^ 1-mtana^ ^

, m — tan a , ,.

and y-y = 1 1 \^ - ^ )•

^ ^ 1 + m tan a



EXAMPLES. VI.

Find the angles between the pairs of straight lines

1. x-ijsj^ — ^ and ^/3a;+2/ = 7.

2. ic-4?/ = 3 and ^x-y = ll. 3. 2/ = 3a3 + 7 and 3?/-a; = 8.

4. 2/ = (2-V3)a: + 5 and 2/= (2 + ^/3) a;- 7.

5. {m'^-mn)y = (inn^-n^)x + n^ and (?n7H- m^) y = (??i?i - w'^) a; + m^

6. Find the tangent of the angle between the lines whose inter-
cepts on the axes are respectively a, - 6 and 6, - a.

7. Prove that the points (2, - 1), (0, 2), (2, 3), and (4, 0) are the
coordinates of the angular points of a parallelogram and find the
angle between its diagonals.

Find the equation to the straight line

8. passing through the point (2, 3) and perpendicular to the
straight line 4a;-3i/ = 10.

9. passing through the point ( - 6, 10) and perpendicular to the
straight line 7aj + 8?/ = 5.

10. passing through the point (2, -3) and perpendicular to the
straight line joining the points (5, 7) and ( - 6, 3).

11. passing through the point (-4, -3) and perpendicular to the
straight line joining (1, 3) and (2, 7).

12. Find the equation to the straight line drawn at right angles to

the straight line — v = 1 through the point where it meets the axis
a b

of X.

13. Find the equation to the straight line which bisects, and is
perpendicular to, the straight line joining the points (a, b) and
(a', &')•

14. Prove that the equation to the straight line which passes
through the point {a cos^ 6, a sin^ 6) and is perpendicular to the
straight line xsecd + y cosec d = ais x cos d-y sin d = a cos 26.

15. Find the equations to the straight lines passing through {x', y')
and respectively perpendicular to the straight lines

xxf-\-yy'=a\



[Exs. VI.]



EXAMPLES.



49



XX yy



= 1,



a- 62
and x'y + xy' = a-.

16. Find the equations to the straight lines which divide, internally
and externally, the line joining (-3,7) to (5, - 4) in the ratio of 4 : 7
and which are perpendicular to this line.

17. Through the point (3, 4) are drawn two straight lines each
inclined at 45° to the straight line x-y = 2. Find their equations
and find also the area included by the three lines.

18. Shew that the equations to the straight lines passing through
the point (3, - 2) and incHned at 60° to the line

iJ3x + y = l are y + 2 = and y -J3x + 2 + 3^S = 0.

19. Find the equations to the straight lines which pass through
the origin and are inclined at 75° to the straight line

x + y + ^S{y-x) = a.

20. Find the equations to the straight lines which pass through
the point {h, k) and are inclined at an angle tan~'^m to the straight
line y = mx + c.

21. Find the angle between the two straight lines 3a; = 4?/ + 7 and
5y = 12x + 6 and also the equations to the two straight lines which
pass through the point (4, 5) and make equal angles with the two
given lines.

73. To sheiv that the point (x', y') is on one side or the
other of the straight line Ax + By +(7 = according as the
quantity Ax + By' + C is positive or negative.

Let LM be the given straight line and P any point
ix\ y).

Through P draw P^, parallel to
the axis of 3/, to meet the given
straight line in Q^ and let the co-
ordinates of Q be (.'«', y").

Since Q lies on the given line, we
have

^£c' + %" + C=:0,
Ax + C

y




i>x:x



so that



B



.(1).



It is clear from the figure that PQ is drawn parallel to
the positive or negative direction of the axis of y according
as P is on one side, or the other, of the straight line LM^
i.e. according as y" is > or < y\
i.e. according as y" — y is positive or negative.



L.



4



50 COORDINATE GEOMETKY.

Now, by (1),

The point {x ^ y') is therefore on one side or the other of
LM according as the quantity Ax' + By' + C is negative or
positive.

Cor. The point (cc', y') and the origin are on the same
side of the given line if Ax + By + G and AxO-\- B xO + C
have the same signs, i.e. if Ax' + By' + C has the same sign
as C.

If these two quantities have opposite signs, then the
origin and the point (x, y') are on opposite sides of the
given line.

!74. The condition that two points may lie on the
same or opposite sides of a given line may also be obtained
by considering the ratio in which the line joining the two
points is cut by the given line.

For let the equation to the given line be

Ax + By-¥C=0 (1),

and let the coordinates of the two given points be [x-^, y^
and (a?2, 2/2)-

The coordinates of the point which divides in the ratio
7?ii : vu the line joining these points are, by Art. 22,

Tn^ + TYi.j 7)1-^ + m^ ^ ''

If this point lie on the given line we have

^^ m^x^ + m^x^ ^ ^ m^y^ + m^y^ ^ ^^^

mj + mg m^ + m^ '

^1 . ^1 Ax.-\-By. + G /«,

so that _i = _^ w^—Ti (3 •

m^ Ax^ + ^2/2 + ^

If the point (2) be between the two given points {x-^, y^)
and (x^, 2/2), i.e. if these two points be on opposite sides of
the given line, the ratio in-^ : 711^ is positive.

In this case, by (3) the two quantities Ax^ + By^ + C
and Ax^ + By^ + C have opposite signs.

The two points (a?i, y^ and {x^^ y^) therefore lie on the op-



LENGTHS OF PEKPENDICULARS. 51

posite (or the same) sides of the straight line Ax + By + (7 =
according as the quantities Ax^ + By^ + G and Ax.^ + By.^ + G
have opposite (or the same) signs.

Lengths of perpendiculars.

75. To find the length of the perpendicular let fcdl from,
a given point upon a given straight line.

. Y




(i) Let the equation of the straight line be

£CCOSa+ 2/ sin a — p = (1),

so that, if p be the perpendicular on it, we have
ON^p and lXON=^o..

Let the given point F be {x ^ y').

Through P draw PR parallel to the given line to meet
OiV produced in R and draw PQ the required perpendicular.
If OR be 2^'i the equation to PR is, by Art. 53,

X cos a + 2/ ^^^ ^ ~P' — ^•
Since this passes through the point {x, y\ we have

£c' cos a + 2/' sin a.—p' = 0,
so that p = x' cos a + 2/' sin a.

But the required perpendicular

^PQ = NR = OR-ON = p'-p

= X' cos a + y' sin a — p (2).

The length of the required perpendicular is therefore
obtained by substituting x and y for x and y in the given
equation.

(ii) Let the equation to the straight line be

Ax + By + G=^0 (3),

the equation being written so that (7 is a negative quantity.

4—2



52 COORDINATE GEOMETRY.

As in Art. 56 this equation is reduced to the form (1)
by dividing it by V-^^ + B^. It then becomes



^A^-^B" slA^^B" ^|A^\E'
Hence

A , B ^ G

cos a = ,,„ , sm a = , - , and — «



^'A^ + B^' \/J'' + B ^A'^ + B''

The perpendicular from the point {x\ y') therefore

= x' cos a + 2/" sin a—p

_ Ax' + By^ + C

VAZ + B2

The length of the perpendicular from (x, y') on (3) is
therefore obtained by substituting x and y for x and 2/ in
the left-hand member of (3), and dividing the result so
obtained by the square root of the sum of the squares of
the coefficients of x and y.

Cor. 1. The perpendicular from the origin

^G-rs]~AFVB\

Cor. 2. The length of the perpendicular is, by Art. 73,

positive or negative according as {x ^ y) is on one side or
the other of the given line.

76. The length of the perpendicular may also be
obtained as follows :

As in the figure of the last article let the straight line
meet the axes in L and M^ so that

OL^ -% and 0M=-%,.
A B

Let PQ be the perpendicular from F (re', y'^ on the
given line and PS and PT the perpendiculars on the axes
of coordinates.

"We then have

/\PML + /\MOL = /\OLP + AOPM,

i.e., since the area of a triangle is one half the product of
its base and perpendicular height,

PQ,LM+OL. OM^ OL.PS+ OM . PT.



EXAMPLES. 53

since (7 is a negative quantity.
Hence

Ax' + By' +



so that P<?r=



Ja^ + b^



EXAMPLES. VII.

Find the length of the perpendicular drawn from

1. the point (4, 5) upon the straight line 3a; + 4?/ = 10.

2. the origin upon the straight line -^ - j=l.

3. the point ( - 3, - 4) upon the straight line

■12{x + 6) = 5{y-2).

4. the point (&, a) upon the straight line — f =!•

5. Find the length of the perpendicular from the origin upon the
straight line joining the two points whose coordinates are

{a cos a, a sin a) and (a cos j8, a sin j8).

6. Shew that the product of the perpendiculars drawn from the
two points ( ± \/a2 - h^, 0) upon the straight line

- cos ^ + ^ sin ^= 1 is 62.
a

7. If p and p' he the perpendiculars from the origin upon the
straight lines whose equations are x sec ^ + ^ cosec 6 = a and

a; cos ^ - 2/ sin ^ = a cos 2^,

prove that 4Lp^+p'^ = a'^.

8. Find the distance between the two parallel straight lines

y = mx + c and y = mx + d.

9. What are the points on the axis of x whose perpendicular

X 1/

distance from the straight line - + ~ =lis a^

ah

10. Shew that the perpendiculars let fall from any point of the
straight line 2a; + 11?/ = 5 upon the two straight lines 24a; + 7t/ = 20
and 4a; -3?/ = 2 are equal to each other.



54 COORDINATE GEOMETRY. [EXS. VII.l

11. Find the perpendicular distance from the origin of the
perpendicular from the point (1, 2) upon the straight line

77. To find the coordinates of the foint of intersection
of two given straight lines.

Let the equations of the two straight lines be

a-^x + h{y + Ci = (1),

and fta^^ + 522/ + ^2 = {2)5

and let the straight lines be AL^ and AL^ as in the figure
of Art. 66.

Since (1) is the equation of AL^^ the coordinates of any
point on it must satisfy the equation (1). So the coordi-
nates of any point on AL^ satisfy equation (2).

Now the only point which is common to these two
straight lines is their point of intersection A.

The coordinates of this point must therefore satisfy
both (1) and (2).

If therefore A be the point {x^, t/i)? ^^ have

«ia^i + ^i2/i + Ci = (3),

and a2^i + 522/i + <^2 = W-

Solving (3) and (4) we have (as in Art. 3)

^x ^ 2/1 ^ \__

so that the coordinates of the required common point are

h-fi^ — h.^c^ - c^a^ — c^a^
a^^-ajb^ a-}><^—a^^

78. The coordinates of the point of intersection found
in the last article are infinite if

a^^ — a^hi = 0.

But from Art. 67 we know that the two straight lines
are parallel if this condition holds.

Hence parallel lines must be looked upon as lines whose
point of intersection is at an infinite distance.



CONCUKEENCE OF STRAIGHT LINES. 55

79. To find the condition that three straight lines may
ineet in a point.

Let their equations be

a^x + hyy + c^ = (1),

a.jX + h^y + C2 = (2),

and a.p: + h-^y + C3 = (3).

By Art. 77 the coordinates of the point of intersection
of (1) and (2) are

—r -J- and —jf J - (4).

If the three straight lines meet in a point, the point of
intersection of (1) and (2) must lie on (3). Hence the
values (4) must satisfy (3), so that

0-iCn — Oc\C-\ -t C-\Cto C.-^Cti -

«3 X -^r — T- + ^3 X -V — ~-Y + C3 = 0,

i. e. a^ (b^c^ — b^c-^) + 63 (ci^g — 02%) + c^ (a^b^ — a^bj) — 0,

i. e. ai (b^c^ — b^c^) + b^ {c^a^ — c^a^ + Cj (^2^3 — ^3^2) = . . , (5).

Aliter. If, the three straight lines meet in a point let
it be (ccj, 2/1), so that the values x\ and y-^ satisfy the
equations (1), (2), and (3), and hence

a-^x-^ + 6i2/i + Ci = 0,

ftoa^i + b^yi + C2 = 0,
and a^x-^ + b^y^ + Cg = 0.

The condition that these three equations should hold
between the two quantities x^ and y^ is, as in Art. 12,

^3 J ^3 ) <^3

which is the same as equation (5).

80. Another criterion as to whether the three straight
lines of the previous article meet in a point is the following.

If any three quantities ^p, q, and r can be found so
that
p (ajX + b^y + C;^) + q (a^x + b.2y + c^) + r (a^x + b^y + Cg) =

identically, then the three straight lines meet in a point.



56 COORDINATE GEOMETRY.

For in this case we have

a^x + h^y + Cg =: — - (a-^x + h^y + Cj) — - {a^x + h^y + Co) . • •(!).

Now the coordinates of the point of intersection of the
first two of the lines make the right-hand side of (1) vanish.
Hence the same coordinates make the left-hand side vanish.
The point of intersection of the first two therefore satisfies
the equation to the third line and all three therefore meet
in a point.

81. Ex. 1. Shew that the three straight lines 2«-3i/ + 5 = 0,
dx + ^y -1 = 0, and 9a; -5y 4- 8 = meet in a point.

If we multiply these three equations by 6, 2, and - 2 we have
identically

6 (2x - 32/ + 5) + 2 (3x -I- 4t/ - 7) - 2 (9a; - 5?/ -F 8) = 0.

The coordinates of the point of intersection of the first two lines
make the first two brackets of this equation vanish and hence make
the third vanish. The common point of intersection of the first two
therefore satisfies the third equation. The three straight lines
therefore meet in a point.

Ex. 2. Prove that the three 'perpendiculars draicn from the
vertices of a triangle upon the opposite sides all meet in a point.
Let the triangle be ABC and let its angular points be the points

K,2/i). (^2 '2/2)5 and (x^^ys)-
The equation to BG is y-y^ = "^ — - {x - x^).
The equation to the perpendicular from A on this straight line is



x^ x^



i'^' 2/ (2/3 -2/2) + ^(^3 -^2) =2/1 (1/3 -2/2) +^1(^3 -^2) (!)•

So the perpendiculars from B and C on CA and AB are

2/(2/i-2/3)+^(aa-^3) = 2/2(2/i-2/3)+«'2{^i-^3) (2).

and 2/ (2/2 -2/]) + ^(^2-^1) =2/3 (2/2 -2/1) + ^3 (^2-^1) (3).

On adding these three equations their sum identically vanishes.
The straight lines represented by them therefore meet in a point.

This point is called the ortliocentre of the triangle.

82. To Jind the equation to any straight line which
2)asses through the intersection of the two straight lines

a-^x + \y + c-^ = (1)>

and a^x + b^y + Cg = (2).



INTERSECTIONS OF STRAIGHT LINES. 57

If (a?i, 2/1) be the common point of the equations (1)
and (2) we may, as in Art. 77, find the values of x^ and 2/1,
and then the equation to any straight line through it is

where m is any quantity whatever.

Aliter. If A be the common point of the two straight
lines, then both equations (1) and (2) are satisfied by the
coordinates of the point A.

Hence the equation

a-^x + h^y + Ci + X {a,^x + h^y + Co) = (3)

is satisfied by the coordinates of the common point A,
where \ is any arbitrary constant.

But (3), being of the first degree in x and y, always
represents a straight line.

It therefore represents a straight line passing through A .

Also the arbitrary constant X may be so chosen that (3)
may fulfil any other condition. It therefore represents
any straight line passing through A.

83. Ex. Find the equation to the straight line ivhich passes
through the intersection of the straight lines

2x-Sy + 4: = 0, Sx + ^y-5 = (1),

and is perpendicular to the straight line

Gx-7y + 8 = (2).

Solving the equations (1), the coordinates x^^, 7/1 of their common
point are given by

^1 ^ yi _ 1 _ 1

(-3)(-5)-4x4 4x3-2x(-5) 2x4-3x(-3) ^^'

so that x-^= - yY and 2/1 =Tf-

The equation of any straight line through this common point is
therefore

y-H=m{x + ^).

This straight line is, by Art. 69, perpendicular to (2) if

m X f = - 1, i.e. if mr= - |.
The required equation is therefore

i.e. 119a; + 1022/ = 125.



58



COORDINATE GEOMETRY.



Aliter. Any straight line through the intersection of the straight
lines (1) is

i.e. (2 + 3X)a; + ?/(4X-3) + 4-5X = (3).

This straight line is perpendicular to (2), if

6(2 + 3X)-7(4X-3) = 0, (Art. 69)

i.e. a \ = u.

*
The equation (3) is therefore

i.e. 119a; +102?/ -125 = 0.

Bisectors of angles between straight lines.

84. To find the equations of the bisectors of the angles
between the straight lines

ajpc + b-yy + Cj^ = (1),

and a^ + b^ + c^ = (2).




Let the two straight lines be AL^ and AL^, and let the
bisectors of the angles between them be AM^ and A^f^-

Let P be any point on either of these bisectors and
draw PiVi and FJV^ perpendicular to the given lines.

The triangles PAN-^ and PAN<^ are equal in all respects,
so that the perpendiculars PN^ and PN^ ^-re equal in
magnitude.

Let the equations to the straight lines be written
so that Ci and c^ are both negative, and to the quantities

Ja^ + b^ and Ja^ + b^ let the positive sign be prefixed.



EQUATIONS TO BISECTOKS OF ANGLES. 59

If P be the point (/i, k), the numerical values of PN^
and PN^ are (by Art. 75)

aJh + hJc + c. , aji + hjc + c^ , ,

— ,_ and — . (1).

If P lie on AM^^ i.e. on the bisector of the angle
between the two straight lines in which the origin lies, the
point P and the origin lie on the same side of each of the
two lines. Hence (by Art. 73, Cor.) the two quantities (1)
have the same sign as c^ and c^ respectively.

In this case, since c^ and c^ have the same sign, the
quantities (1) have the same sign, and hence
aji + hjc + Ci aji + hjc + c^

But this is the condition that the point (h, k) may lie on
the straight line

a^x + h-{y + Cj a^ + h^y + c^

which is therefore the equation to AM-^.

If, however, P lie on the other bisector AM^, the two
quantities (1) will have opposite signs, so that the equation
to AM^ will be

a-^x + \y + Ci a^x + })c^y + c^

The equations to the original lines being therefore
arranged so that the constant terms are both positive (or
both negative) the equation to the bisectors is

VS7+b7 - Vi7+bp '

the upper sign giving the bisector of the angle in which
the origin lies.

85. Ex. Find the equations to the bisectors of the angles
between the straight lines

3x-4:y + 7 = and 12x-5y-S = 0.
Writing the equations so that their constant terms are both
positive they are

Sx-Ay + 7 = and -12x + 5y + 8 = 0.



60



COORDINATE GEOMETRY.



t.e.
i.e.



The equation to the bisector of the angle in which the origin lies
is therefore

Bx-^y + 7 _ -12x + 5y + 8

i.e. 13{Bx-^y + l) = 5{-12x + 5y + 8),

i.e. 99a; -772/ + 51 = 0.

The equation to the other bisector is

Sx-iy + l _ - 12x + 5?/ + 8
V32 + 42 ~ ~ x/122 + 52 '
13 {3a; - 4?/ + 7) + 5 ( - 12a; + 5?/ + 8) = 0,
21a: + 272/- 131 = 0.

86. It will be found useful in a later chapter to have
the equation to a straight line, which passes through a
given point and makes a given angle 6 with a given line, in
a form different from that of Art. 62.

Let A be the given point {h, k) and L'AL a straight
line through it inclined at an
angle 6 to the axis of x.

Take any point F, whose
coordinates are (x, y), lying on
this line, and let the distance
AP be r.

Draw PM perpendicular
to the axis of x and AN perpendicular to PM.

Then x - h=- AN = AP cosO=r cos 6,

and y — k = NP - AP sin ^ = r sin 0.

x-h y-k




Hence



= r



.(1).



This being the relation holding between the coordinates
of any point P on the line is the equation required.

Cor. From (1) we have

x — h + r cos 6 and y = k + r sin 0.

The coordinates of any point on the given line are

therefore h-\-r cos 6 and k + r sin 0.

87. To find the length of the straight line drawn
through a given point in a given direction to meet a given
straight line.



EXAMPLES. 61

Let the given straight line be

Ax-¥By + C^Q (1).

Let the given point A be (7t, k) and the given direction
one making an angle 6 with the axis of x.

Let the line drawn through A meet the straight line
(1) in P and let AP be r.

By the corollary to the last article the coordinates
of P are

h + r cos 6 and k + r sin 6.

Since these coordinates satisfy (1) we have
^ (A + r cos ^) + ^ (^ + r sin ^) + (7 - 0.
Ah-[-Bk+C



r =



(2),



A cos 6 ■¥ B sin 6
giving the length AP which is required.

Cor. From the preceding may be deduced the length
of the perpendicular drawn from (A, k) upon (1).

For the "m" of the straight line drawn through A is

tan ^ and the "m" of (1) is - -.

This straight line is perpendicular to (1) if

tan y. { — ^\ = — \,
i. e. if tan ^ = — ■ ,



so that
and hence



A

cos 6 sin 6 1



■4 B JA'- + B'



AcosO + BsinO= 4=^£L^s/A^ + B'.
V-i' + ^-
Substituting this value in (2) we have the magnitude
of the required perpendicular.

EXAMPLES. VIII.

Find the coordinates of the points of intersection of the straight
lines whose equations are

1. 2x-Si/ + 5 = and 7x-¥^y=S.



62 COOEDINATE GEOMETRY. [EXS.

2. - + T=1 and -+^ = 1.
a ha

3 a ^ a
. y = vi-,x-\ and y=moX-\ .

4. a; cos 01 + ^ sin 01 = a and a; cos 02+2/ sin 02 = «•

5. Two straight lines cut the axis of x at distances a and - a and
the axis of y at distances & and 6' respectively ; find the coordinates
of their point of intersection.

6. Find the distance of the point of intersection of the two
straight lines

2x-Sy + 5 = and dx + 4:y =

from the straight line

5x-2y = 0.

7. Shew that the perpendicular from the origin upon the
straight line joining the points

(a cos a, a sin a) and {a cos /3, a sin ^)

bisects the distance between them.

8. Find the equations of the two straight lines drawn through
the point (0, a) on which the perpendiculars let fall from the point
(2a, 2a) are each of length a.

Prove also that the equation of the straight line joining the feet
of these perpendiculars is y -r2x = oa.

9. Find the point of intersection and the inclination of the two
lines

Ax + By = A+B and A{x-y)+B{x + y)=2B.

10. Find the coordinates of the point in which the line

2y-Sx + l =

meets the line joining the two points (6, - 2) and ( - 8, 7). Find also
the angle between them.

11. Find the coordinates of the feet of the perpendiculars let fall
from the point (5, 0) upon the sides of the triangle formed by joining



Online LibraryS. L. (Sidney Luxton) LoneyThe elements of coordinate geometry → online text (page 4 of 26)