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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 = â€” 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 â€” 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 â€” x) (1),

and y ~y' ~ t^^ <fi' X (x â€” x') (2).

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

and <^'^L LNS+ L SLN= (180Â° ^a) + e.

Hence

â– , ^. tan a + tan ^ tana + m

tan <^ = tan (a + ^) - .j â€” ^ = ,

i â€” tan a tan ff i â€”m tan a

and tan </>' = tan ( 1 80Â° + ^ - a)

. â€ž . tan d â€” tan a mâ€” tan a

= tan (^ â€” a) = ^â€” -p, â€” = ., .

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 â€” tan a , ,.

and y-y = 1 1 \^ - ^ )â€¢

^ ^ 1 + m tan a

EXAMPLES. VI.

Find the angles between the pairs of straight lines

1. x-ijsj^ â€” ^ 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 â€” 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', &')â€¢

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 45Â° 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 60Â° 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 75Â° 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 (.'Â«', y").

Since Q lies on the given line, we

have

^Â£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" â€” 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-Â¥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 /Â«,

so that _i = _^ w^â€”Ti (3 â€¢

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

Â£CCOSa+ 2/ sin a â€” 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' â€” ^â€¢

Since this passes through the point {x, y\ we have

Â£c' cos a + 2/' sin a.â€”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 â€” 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.

4â€”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 = ,,â€ž , sm a = , - , and â€” Â«

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

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

= x' cos a + 2/" sin aâ€”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

â– 12{x + 6) = 5{y-2).

4. the point (&, a) upon the straight line â€” f =!â€¢

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 ( Â± \/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

Â«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^ â€” h.^c^ - c^a^ â€” c^a^

a^^-ajb^ a-}><^â€”a^^

78. The coordinates of the point of intersection found

in the last article are infinite if

a^^ â€” 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

â€”r -J- and â€”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 â€” Oc\C-\ -t C-\Cto C.-^Cti -

Â«3 X -^r â€” T- + ^3 X -V â€” ~-Y + C3 = 0,

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

i. e. ai (b^c^ â€” b^c^) + b^ {c^a^ â€” c^a^ + Cj (^2^3 â€” ^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 =: â€” - (a-^x + h^y + Cj) â€” - {a^x + h^y + Co) . â€¢ â€¢(!).

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 2Â«-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^ = "^ â€” - {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) (!)â€¢

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)+Â«'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^ , ,

â€” ,_ and â€” . (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 â€” 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 â€” 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-Â¥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 â– Â¥ 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. { â€” ^\ = â€” \,

i. e. if tan ^ = â€” â– ,

so that

and hence

A

cos 6 sin 6 1

â– 4 B JA'- + B'

AcosO + BsinO= 4=^Â£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-Â¥^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 = Â«â€¢

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

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 = â€” 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 â€” 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 â€” x) (1),

and y ~y' ~ t^^ <fi' X (x â€” x') (2).

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

and <^'^L LNS+ L SLN= (180Â° ^a) + e.

Hence

â– , ^. tan a + tan ^ tana + m

tan <^ = tan (a + ^) - .j â€” ^ = ,

i â€” tan a tan ff i â€”m tan a

and tan </>' = tan ( 1 80Â° + ^ - a)

. â€ž . tan d â€” tan a mâ€” tan a

= tan (^ â€” a) = ^â€” -p, â€” = ., .

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 â€” tan a , ,.

and y-y = 1 1 \^ - ^ )â€¢

^ ^ 1 + m tan a

EXAMPLES. VI.

Find the angles between the pairs of straight lines

1. x-ijsj^ â€” ^ 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 â€” 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', &')â€¢

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 45Â° 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 60Â° 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 75Â° 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 (.'Â«', y").

Since Q lies on the given line, we

have

^Â£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" â€” 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-Â¥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 /Â«,

so that _i = _^ w^â€”Ti (3 â€¢

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

Â£CCOSa+ 2/ sin a â€” 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' â€” ^â€¢

Since this passes through the point {x, y\ we have

Â£c' cos a + 2/' sin a.â€”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 â€” 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.

4â€”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 = ,,â€ž , sm a = , - , and â€” Â«

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

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

= x' cos a + 2/" sin aâ€”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

â– 12{x + 6) = 5{y-2).

4. the point (&, a) upon the straight line â€” f =!â€¢

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 ( Â± \/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

Â«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^ â€” h.^c^ - c^a^ â€” c^a^

a^^-ajb^ a-}><^â€”a^^

78. The coordinates of the point of intersection found

in the last article are infinite if

a^^ â€” 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

â€”r -J- and â€”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 â€” Oc\C-\ -t C-\Cto C.-^Cti -

Â«3 X -^r â€” T- + ^3 X -V â€” ~-Y + C3 = 0,

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

i. e. ai (b^c^ â€” b^c^) + b^ {c^a^ â€” c^a^ + Cj (^2^3 â€” ^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 =: â€” - (a-^x + h^y + Cj) â€” - {a^x + h^y + Co) . â€¢ â€¢(!).

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 2Â«-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^ = "^ â€” - {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) (!)â€¢

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)+Â«'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^ , ,

â€” ,_ and â€” . (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 â€” 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 â€” 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-Â¥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 â– Â¥ 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. { â€” ^\ = â€” \,

i. e. if tan ^ = â€” â– ,

so that

and hence

A

cos 6 sin 6 1

â– 4 B JA'- + B'

AcosO + BsinO= 4=^Â£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-Â¥^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 = Â«â€¢

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

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