intermediate maths 1B

7/24/2019 intermediate maths 1B

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First year Mathematics IB AIMS TUTORIAL

Aims tutorial Page 1

1. If Q (h, k) is the foot of perpendicular from

P (, ) on the straight line + + =. Then P.T

= = (++)+ .findthe foot of perpendicular of (4, 1) w.r.t the

line

+

=

.

Sol: Q (h, k) is the foot of perpendicular fromP (1 , 1) on the straight line + + = 0 ..(1)1 1 = 2 = 2121= 1 1 (1) 1. 2 = 1( ). ( 1 1) = 1

1

= 1

(& )1 = 1 = ( 1)+( 1) .+ . 1 = 1 = ( 1+1)2+ 2 1 = 1 = (11+ + )2+ 2 (2), 1

+

+

= 0

+

=

(2)1 = 1 = (11)2+ 2 = = (++)+ ., (4, 1) w.r.t the line 3 4 + 12 = 0.4

3=

14 = (3441+12)32+(4)2 43

=14 = (20)25

4

3 = 4

5 1

4 = 4

5 4 = 125

& 1 = 165

= 125

+ 4 & = 165

+ 1

= 12+205

=8

5 , k=

16+5

5=

21

5

, = (85

,21

5)

2.

If Q (h, k) is image of P (, ) on the straightline

+ + = .Then P.T

= = (++)+ .find theimage of (1, 2) w.r.t the line

+

=

.

Sol: Q (h, k) is image of P (1, 1) on thestraight line + + = 0 ..(1)1 1 = 2 = 2121= 1 1 (1) 1. 2 = 1( ). ( 1 1) = 11 = 1 ( & )1 = 1 = ( 1)+( 1) .+ . 1 = 1 = ( 1+1)2+ 2 1 = 1 = ( 1 1+ + )2+ 2 (2) ,

+2

,+

2=+

2,

+2

+

2

,+

2

1

1+2 + 1+2 + = 0 1 + + 1 + + 2 = 0 + = 1 1 2 (2)1 = 1 = ( 1 11 12)2+ 2 = = (++)+ ., (1, 2) w.r.t the line3 + 4 1 = 0.

13 = 24 = 231+42132+(4)2 43

=1

4= 2(10)

25 1

3= 4

5 2

4= 4

5

1 = 125

& 2 = 165

= 125

+ 1 & = 165

+ 2

= 12+55

=75

, k=16+10

5=

65

, = 7

5 , 6

5.


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3. Find the circumcentre of the triangle whose

vertices are (1, 3), (-3, 5) and (5, -1).

Sol:Let the given vertices are

A(1, 3), B(-3, 5) and C(5, -1).

Let S(x, y) be the circumcentre of

= =

() = . . . 2 = 2 S(x, y)

A (1, 3) B (-3, 5)

12 + 32 = + 32 + 52 2 + 1 2 + 2 + 9 6= 2 + 9 + 6 + 2 + 25 108 8 1 6 + 4 = 08 + 4 24 = 0 ( 4)2 + 6 = 0 . (1)(ii) =

.

.

.

2 = 2S(x, y)A (-3, 5) B (5, -1)

+ 32 + 52 = 52 + + 12 2 + 9 + 6 + 2 + 25 10= 2 + 25 10 + 2 + 1 + 216+16 + 24 12 = 016

12

+ 8 = 0 ( 4)

4 3 + 2 = 0 . (2)Solving (1) and (2)

2 1 6 24 3 2 4, = 2+186+4 , 2446+4 = 162 , 202

8,

10

.

[2,3, 2, 1, 4, 0& (1,3), (0, 2),(3,1)]

4. Find the circumcentre of the triangle whose

sides are + + = , = + = .: Let + + 2 = 0 .. 15

2 = 0

. 2

2 + 5 = 0 . (3) 1& (2) 1 1 2 15 -1 -2 5

, = 2 + 21 5 , 1 0 + 21 5 = (0, 2) 2& (3) 5 -1 -2 5

1 -2 5 1

,

=

5 4

1 0 + 1,

2 251 0 + 1

= (1, 3)

1& (3) 1 1 2 11 -2 5 1, = 5 + 42 1 , 2 52 1 = (3, 1)

A(0, -2), B(1, 3) and C(-3, 1).

Let S(x, y) be the circumcentre of = =

() = . . . 2 = 2S(x, y)

A (0, -2) B (1, 3)

02 + + 22 = 12 + 32 2 + 0 + 0 + 2 + 4 + 4=

2 + 1

2

+

2 + 9

6

1 + 2 5 + 1 0 = 02 + 10 6 = 0 ( 2)1 + 5 3 = 0 . (4)(ii) = . . . 2 = 2

S(x, y)

A (1, 3) B (-3, 1)

12 + 32 = + 32 + 12

2=(2 1)2 + (2 1)2

2=(2 1)2 + (2 1)2


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2 + 1 2 + 2 + 9 6= 2 + 9 + 6 + 2 + 1 2

8 8 + 8 4 = 08 4 = 0 ( 4)2 + = 0 .(5)

Solving (1) and (2)

1 5 3 12 1 0 2, = 0+3

110 , 60110 = 39 , 69 1

3,

2

3.

5.

Find the Orthocenter of the triangle whose

vertices are (-5, -7), (13, 2) and (-5, 6).

Sol: Given vertices are (-5, -7), (13, 2) and (-

5, 6)

Slope of B (13, 2), C (-5, 6) = 2121 = 62513 = 418 = 29 , = 92 1 = ( 1)

5,

7

=

9

2

+ 7 = 92 ( + 5)2 + 14 = 9 + 459 2 + 31 = 0 . .(1)Slope of A (-5, -7), C (-5, 6) = 2121 = 6+75+5 = 130 , = 0 1 = ( 1)13, 2 = 0 2 = 0( 13)

2 = 0

y=2.. (2) = 2 (1)9 2(2) + 31 = 09 = 27 = 279

= 3(3, 2)

6. If the eqns of the sides of the triangle are + = , + = + + = ..sol:Let 7 + 10 = 0 .. 1

2

+ 5 = 0

. 2

+ + 2 = 0 . (3) 1& (2) 7 1 -10 71 -2 5 1

, = 5 2014 1 , 10 3514 1 = (1, 3) 2& (3) 1 -2 5 1

1 1 2 1

,

=

4 51 + 2

,5 21 + 2

= (

3,1)

1& (3) 7 1 -10 71 1 2 1, = 2 + 1 0

7 1 , 10 147 1 = (2, 4)A(1, 3), B(-3, 1) and C(2, -4).

Now

Slope of B (-3,1), C (2, -4) = 2121 = 412+3 = 55 = 1

,

= 1

1 = ( 1)1, 3 = 1 3 = 1( 1) 3 = 1 + 2 = 0 . .(4)Slope of A (1, 3), C (2, -4) = 2121 = 4321 = 71 , = 1

7 1 = ( 1)

3, 1

=

1

7

1 = 17 ( + 3)7 7 = + 3 7y+10=0.. (5) 4& (5) 1 -1 2 1

1 -7 10 1

, = 10+147+1 , 2107+1 = 46 , 86

( 23 , 43)


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7.

If P and q are the lengths of perpendiculars

from the origin to the st lines + = = ,

.

+

=

.

: + = + = + = + = 0 0, 0is = 2+ 2 = 2+ 2 = 2 = 2

2

=

2

S.O.B.S

42 = 222 . .(1) 0, 0 2 = 0 = 2 2+ 2 = 2 S.O.B.S 2 = 222 . .(2) 1& (2)

4

2 +

2 =

2

22

+

2

22

= 2(22 + 22)= 2(1)

+ =

8.

ind the eqn of the st lines passing through

the point of intersection of the lines + + = , + = whosedistance from (2, -1) is 2.

Sol: Given eqns3 + 2 + 4 = 0 . (1)2 + 5 1 = 0 (2) 1& (2)

3 2 4 3

2 5 -1 2, = 220154 , 8+3154 = 2211 , 1111

= (2, 1)Let m be the slope of the line passingthrough

P (-2, 1) is 1 = 1 + 2 + 1 = 0 + 2 + 1 = 0 . (3)Since distance from (2, -1) to (3) is 2 = (++)+ 2 = +++ 2 =

(4

+2)

2+1 2 = 2(2 +1) 2+1 S.O.B.S 2 + 1 = ( 2 + 1)2 2 + 1 = 42 + 4 + 1 42 + 4 + 1 -2 1 = 0 32 + 4 = 0 (3 + 4) = 0 = 0 = 43

= 0

=

4

3

1 = 0 + 2; 1 = 43 ( + 2) 1 = 0. 3 3 =4 + 8 4 + 3 11 = 0.

1 = + 2


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9.

If is the angle between the pair of lines + + = ,then P.T = +()+.Sol: let2 + 2 + 2 = 0 1 + 1 = 0 . .(1)2 + 2 = 0 . .(2) 2 + 2 + 2 1 + 12 + 2

= 0 12 + 2 + 12 + 2= 0 122 + 12 + 21 + 122

= 0

1

2

2 + (

1

2 +

2

1)

+

1

2

2

= 0Comparing both sides2, 2& , 12 = , 12 = & 12 + 21

= 2 = 12+1 212+1222+22 = 12+12(12)2+(12)2+(21)2+( 12)2= 12+12(12)2+(12)221212+(12)2+(21)2+21212=

12+12(12 12)2+(12+2 1)2=

+ 2+22=

+( )2+42.

10.

Prove that product of perpendiculars from a

point (

, ) to the pair of st lines + + = is ++()+ Sol: : let

2

+ 2 + 2

=0 1 + 1 = 0 . .(1)2 + 2 = 0 . .(2) 2 + 2 + 2 1 + 12 +2 = 0 12 + 2 + 12 + 2 =0 122 + 12 + 21 +122 = 0 122 + (12 + 21) +

1

2

2 = 0

Comparing both sides 2, 2 & ,12 = , 12 = & 12 + 21 = 2. . (1, 1)to the line + + = 0 (++)+ , to thelines (1) and (2) is = 1+112+12 2+222+ 22

= 12+2 +1 2+2 (12)2+(12)2+(21)2+(12)212 2+12 +21 +12 2(12)2+(12)221212+(12)2+(21)2+21212

=12 2+(12+21) +1 2 2(1212)2+(12+21)2

= 2+2 + 22+22

= 2+2 + 2( )2+42 .


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11.

If the eqn + + = representa pair of lines,P.T the combined eqn of the

pair of bisectors bisecting the angle b/w these

lines is = ( ).Sol: let

2 + 2

+

2 =

0 1 + 1 = 0 . .(1)2 + 2 = 0 . .(2) 2 + 2 + 2 1 + 12 + 2= 0 12 + 2 + 12 + 2

= 0 122 + 12 + 21 + 122= 0

1

2

2 + (

1

2 +

2

1)

+

1

2

2

= 0Comparing both sides2, 2& , 12 = , 12 = & 12 + 21

= 2Now eqns of bisectors of angle b/w 1 & 2are

1 + 11

2 +

1

2

= 2 + 22

2 +

2

2

. . . , 1 + 12(22 + 22)=(2 + 2)2(12 + 12) (122 + 122 + 211)(22 + 22)=222 + 222 + 222(12 + 12)(12)22 + (12)22 + (21)22 +(12)22+21122 + 21122

=(

1

2)

2

2 + (

2

1)

2

2 + (

1

2)

2

2 +

(12)22+22212 + 222122(12)2 (21)22(12)2 (21)2

= 21212 21 12(12 21

12+

2112 212

2

= 212 2112 12

12 + 212 2 = 212 1222 2 = 2( )

2

2

=

(

).

12.

S.T the area of the triangle formed by the lines + + = and + + = + . .Sol: let2 + 2 + 2 = 0 1 + 1 = 0 . .(1)2 + 2 = 0 . .(2) + + = 0 . (3) 2 + 2 + 2 1 + 12 +2 = 0

1

2

+

2

+

1

2

+

2

=

0 122 + 12 + 21 +122 = 0 122 + (12 + 21) +122 = 0Comparing both sides 2, 2 & ,12 = , 12 = & 12 + 21 = 2Solving (1) & (2) we get, (0, 0)

Solving (1) & (3)1 1 0 1

A (1, 1) = 101 1 , 0 111Similarly by solving (2) & (3) we get,

B (2, 2) = 22 2 , 222Now area of = 1

212 21

=1

2 1

1 1 22 2 22 2 111

=12

2

12211 12 2=

1

2 21 2+212412 12

12 21221 + 1 22=

1

2 2224 2(12+21) +2

=1

2 2424 22 +2

=22 22 22 +2= 22 22 +2 .


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13.

If the eqn

+ + + + + = represent a pair of lines,P.T() + = , , .:let 2 + 2 + 2 + 2 + 2 +

= 01 + 1 + 1 = 0 . .(1)2 + 2 + 2 = 0 . .(2)

2 + 2

+

2 + 2

+ 2

+

= 0

1 + 1 + 12 + 2 + 2 = 0 122 + 12 + 12 + 21+ 122

+12 + 21 + 21 + 12 = 0 122 + (12 + 21) + 122 +(12 + 21)+(12 + 21) + 12 =0

Comparing both sides

2, , 2 , , ,12 = , 12 = , 12 + 21 = 212 + 21 = 2, 12 + 21 = 2 And12 = 222= (12 + 21) (12 + 21)(12 + 21)

= 122212 + 21

intermediate maths 1B

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