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Published by tsujit pathak, 2020-08-03 10:48:48

Vedanta Optional Math Teacher's guide

Vedanta Optional Math Teacher's guide

Questions for practice

1. Solve graphically (a) x2 –8x +12=0
(b) x2 –9x – 10=0
(c) x2 +4x +3=0
2. Solve the following equations graphically.
(a) y= x2 , y=2
(b) y=x2 – 6x +9, 3x + 4y=12
(c) y= x2 –2x, y=x+2
(d) y=x2 – 2x –15, 25x – 8y+20 =0

Vedanta Optional Mathematics Teacher's Guide ~ 10 151

UNIT

six Continuity

1. Objectives

S.N. Level Objectives

– To define natural numbers, rational numbers, whole numbers

i) Knowledge (K) – To define continuity by using graphs.

– To define discontinuity by using graphs.

To say meaning of continuity.

To define and check continuity or discontinuity of numbers in
ii) Understanding(U) number line.

To identify continuity or discontinuity of given graphical figures.

To define continuity of a function.

iii) Application (A) To discuss the continuity or discontinuity of given functions
given in graphs or equations.

iv) Higher Ability To examine the continuity or discontinuity of given functions
(HA) at given points calculating functional values and limits.

2. Teaching materials
– Number lines with natural numbers, whole numbers, integers.
– diagrams in graphs to discuss continuity or discontinuity with intervals.

3. Teaching Learning Strategies
– Review the concept of real number system.
– Draw the number lines to show the following.

i) natural numbers ii) integers iii) whole numbers.

– Discuss the continuity or discontinuity of the number drawn in above number lines.
– Give concept intervals with diagrams (open interval, left open interval, right open

intervals, closed intervals )
– Discuss the continuity or discontinuity of given graphical diagrams with intervals.

– Review the meaning of , lim f(x) , lim f(x) , f(a) with an example.

x a+ x a-

– Define continuity of a function at a point.

– Review the concept of existence of limit of a function at a point.
– Explain continuity or discontinuity of a function at a given point calculating

– functional value f(a), at x=a
– right hand limit, lim f(x).

x a+

– left hand limit,lim f(x).

x a–

152 Vedanta Optional Mathematics Teacher's Guide ~ 10

Note :
1) In mathematics, the word "continuous " applies to functions not in sets.

2) The continuity of a simple function can be checked by drawing a curve. If there is no
breakage at any point on the curve, then the function is continuous.

3. If there is a breakage or a hole on the given curve, then it is discontinuous at that point.

4. A function f(x) is said to be continuous at x =0, if the following conditions are
satisfied.

i) f(a) exists or f(a) is finite

ii) lim f(x) exists ie. lim f(x) =lim f(x)

xa x a+ x a-

iii)lim f(x) =f(a)
xa

If any one of the above conditions fails, then the function is said to be discontinuous at
that point.

Some solved problems

1. In the following given curves.
(a) (i) Find the initial and the terminating points of the curve.

(ii) State the continuity or discontinuity of the curve.

Solution
(Graph 1(a) page 117)
i) The initial point is x =1 and terminating point is x=14.
ii) The given curve is discontinuous at x =10

graph 1(b)
(b) i) The initial paint is x=0 and the terminating point is x=6.
ii) The the points of discontinuity are x=2 and x=4.

graph 1(c)
(c) i) The initial point is x=0 and terminating point is x=7
ii) The straight line is continuous.

2. Discuss the continuity and discontinuity of the following curves from point x=–6 and
x=6. (stating the intervals for continuity and points of discontinuity for discontinuity).

(a) Page 118 graph of 2 (a)

interval point of discontinuous
Continuous in interval [–6,–1] x=1
Continuous in interval [–1,6]
(b) Graph of page 118 2(c)

interval point of discontinuous
Continuous in interval (–6,3) x=–3

Vedanta Optional Mathematics Teacher's Guide ~ 10 153

Continuous in interval (–3,2) x=2
Continuous in interval (2,6)

3. Write a sentance for each of the following notation.

(a) lim f(x) or lim f(x)

x a- x a-0

Solution f(x)
Left hand limit of function f(x) at x=a is denoted by lim

x a-

or

It denotes the left hand limit of f(x) at x=a
(b) lim f(x)

xa

Solution f(x)
The limit of function f(x) at x=a is denoted by lim

xa

It denotes the limit of f(x) at x =a.

(c) lim f(x) = lim f(x)

x a+ x a-

Solution
The limit of f(x) at x=a exists.
The left hand limit and the right hand limit of f(x) at x = a are equal.

(d) write conditions for continuity of a function f(x) at x=a, using notations.

Solution
The following are the required conditions for continuity of a function f(x) at x=a.

i) f(a) is finite

ii) limit of f(x) at x=a exists

ie. lim f(x) = lim f(x) , lim f(x) is finite.

x a+ x a- xa

iii) f(a) = lim f(x)
xa

page 121(2) long question

4. Let f:R R be a real valued function defined by f(x)=x+4

(a) For x=3.9, 3.99, 3.999, 3.9999, find the value of f(x).

(b) For x=4.1, 4.01, 4.001,4.0001, find the value of f(x)

(c) Find the value of f(x) at x=4.

(d) Find the values of lim f(x) = lim f(x)

x 4- x 4+

154 Vedanta Optional Mathematics Teacher's Guide ~ 10

(e) Does limit of the function f(x) exists at x=4 ?
(f) Write the notation to show above function is continuous at x=4.

Solution
Here, f(x)= x+4
(a) f(3.9)=3.9+4=7.9
f(3.99)=3.99+4=7.99
f(3.999)=3.999+4=7.999
f(3.9999)=3.9999+4=7.9999
(b) f(4.1) =4.1 +4 =8.1
f(4.01) =4.01 +4 =8.01

f(4.001) =4.001 +4 =8.001

f(4.0001) =4.0001 +4 =8.0001

(c) f(4) =4+4 =8
(d) lim f(x) = 4 +4 =8

x 4–

lim f(x) = 4+4=8
x 4+

(e) (i) f(4)=8

(ii) Now, lim f(x) = 8
x 4+

lim f(x) = 8
x 4-

lim f(x) = lim f(x)

x 4– x 4+

limit of the function exists at x =4

ie. lim f(x) = 8
x4

(iii) f(4)= lim f(x)
x4

Hence f(x) is continuous at x =4.

5. Let f : R R be a real valued function defined by f(x) = x+3, 1≤x<2
4x-3 x≥2. at x=2
(a) Find lim f(x)
x 2–

(b) lim f(x)
x 2+

(c)Is lim f(x) = lim f(x)

x 2- x 2+

(d) Find f(2).

Vedanta Optional Mathematics Teacher's Guide ~ 10 155

(e) Draw your conclusions

Solution
x+3, 1≤x<2

Here, 4x-3, x≥2. at x=2

(a) For left hand limit, we take

lim f(x) = lim (x+3) ( 1≤x<2)

x 2– x 2–

=2+3=5

(b) For right hand limit, we take

lim f(x) = lim (4x–3) ( x≥2)

x 2+ x 2+

=4×2–2= 8–3=5

(c) lim f(x) = lim f(x)

x 2– x 2+

(d) For functional value, we take

f(x) = 4x–3, ( x≥2)

f(2) = 4×2–3 =5

(e) From (a), (b), (c) and (d), we get.

lim f(x) = lim f(x) =5

x 2+ x 2-

i.e lim f(x) = 5
x2

f(2) =5
and lim f(x) =f(2)

x2

Hence f(x) is continuous at x=2.

6. Discuss the continuity of the function f(x) at x=2.
2x - 1, when x<2

f(x) = 3 , when x=2 at x = 2.
x +1 , when x > 2.

Solution 2x - 1, when x<2 at x = 2.
Here, f(x) = 3 , when x=2
x +1 , when x > 2.

156 Vedanta Optional Mathematics Teacher's Guide ~ 10

For x < 2, we take f(x)=2x–1

lim f(x) = lim (2x–1) =2.2 –1 =3

x 2– x 2–

For x=3, we have, f(2) =3

For x>2, we take, f(x) = x +1

lim f(x) = lim x +1 =2 +1 =3

x 2+ x 2+

we have,

lim f(x) = lim f(x) =3

x 2+ x 2-

i.e. lim f(x) =3
x2

From above, we get

f(2) =lim f(x)
x2

Hence the given function f(x) is continuous at x =2.

Some solved problems

1. Examine the continuity or discontinuity of the following functions at the points mentioned.
(a) f(x) =4x +1, at x=3.

Solution
Functional value at x =3, f(3)=4×3 +1=13

Also, lim f(x) = lim (4x+1)

x3 x3

=4×3 +1 f(x)

=12+1

=13
f(3) =lim
x3

Hence f(x) is continuous at x =3.

(b) f(x)= x2– 64
x–8

Solution x2–64
x–8
Here, f(x)=

For x=8, f(8)= x2–64 = 0 which is not finite.
x–8 0

ie. the functional value of f(x) at x =8 does not exists. Hence f(x) is discontinuous at x =8.

Vedanta Optional Mathematics Teacher's Guide ~ 10 157

2. Examine the continuity or discontinuity of the following functions at the points mentioned.

x2 –7x , when x≠7
x –7
at x = 7
(a)f(x) =

3 , When x=7

For x≠7, we take limit of the function when x 7.

lim f(x) = lim x2–7x
x7 x7 x–7

= lim x(x–7)
x7 (x–7)

=lxim x
7

=7

Functional value at x =7 is given as 3.

ie. f(7)=3

f(7) ≠ lxim f(x)
7

Hence the function f(x) is discontinuous at x=7.

Note: v00al,uweeoffaxctdoirriezcetltyh.e

To calculate limit of a function at x=a, if the function take the form of
numerator and denominator if possible. In this case we do not put the

Example Evaluate lim x2-25
x5 x-5

Here, if we put x=5, we geeqtual00tofo5rmbustwnhoitcehxaisctnlyotefqinuiatle.toIn5.sense of limit, x 5 means,
the value of x is slightly

Now, lim x2–25 0 forms)
x5 x–5 ( 0

= lim (x+5)(x–5)
x5 (x–5)

= lim (x+5)
x 5

=5+5
=10

10 is the limit of f(x) at x=5.

158 Vedanta Optional Mathematics Teacher's Guide ~ 10

x2 –2x , when x≠2
x –2
at x = 2
(b) f(x) =

2 , When x=2

Solution x2 –2x
x –2
, when x≠2

Here, f(x) = at x = 2

2 , When x=2

Functional value at x = 2

f(2)=2

Limit of f(x) at x =2

lim f(x) = lim x2–2x 0 forms)
x2 x–2 ( 0
x2

= lim x(x–2)
x2 x–2
lim
= x 2 x

=2
f(2)= lim f(x)
x2

Hence f(x) is continuous at x=2.

x2 –x–6 , when x≠3
x –3
at x = 3
(C) f(x) =

5, When x=3

Solution x2 -x-6 , when x≠3
Here,f(x) = x -3
at x = 3

5, When x=3

Functional value at x = 3

f(3)=5

For limit of f(x) at x =5

lim f(x) = lim x2–x–6 ( 0 forms)
x3 x–3 0
x3

Vedanta Optional Mathematics Teacher's Guide ~ 10 159

= lim x2–3x+2x–6
x3 x–3

= lim x(x-3)+2(x-3)
x3 x-3

= lim (x-3)(x+2)
x3 (x-3)

==3lx+im2 (x+2)
3

=5

f(3)= lim f(x)
x3

Hence f(x) is continuous at x=3.

x2 -3x+2 , when x≠2
x2+x-6
at x = 2
(d) f(x) =
1
5 , When x = 2

Solution x2 -3x+2 , when x≠2
Here,f(x) = x2+x-6
at x = 2

1
5 , When x = 2

when x = 2, f(x)=f(2)= 22-3.2+2
22+2-6
6-6 0
= 6-6 = 0 form

For limit of f(x) at x = 2 we factorize the denominator and numerator of the function.

lim f(x) = lim x2-3x+2
x2 x2+x-6
x2

= lim x2 - 2x -x+2
x2 x2+3x-2x-6

= lim x(x-2)-1(x-2)
x2 x(x+3)-2(x+3)

= lim (x-3)(x-1)
x2 (x+3)(x-2)

160 Vedanta Optional Mathematics Teacher's Guide ~ 10

= lim (x-1)
x2 (x+3)

= 2-1
2+3
1
=5
1
Functional value is 5 at x=2(given)

lim f(x) =f(2)

x2

Hence f(x) is continuous at x=2.

3. Examine the continuity or discontinuity the following functions at the points muntioned
by calculating left hand limit (LHL), right hand limit (RHL) and functioned value.

3 -x , for x ≤ 0

(a) f(x) = at x = 0

3 , for x > 0

Solution

3 -x, for x ≤ 0

Here, f(x) = at x = 0

3, for x > 0

For functional value at x = 0, we take
f(x) = 3 – x (⸪ x ≤ 0)
f(0) = 3 – 0 =3

For right hand limit, we take

lim f(x) = lim 3 = 3

x 0+ x 0+

=2+3=5

For left hand limit, we take

lim f(x) = lim 3–x

x 0- x 0-

=3–0 = lim f(x)
x 0-
=3
lim f(x)
x 0+

Hence the limit of f(x) at x = 0 exists.

Now, we take f(0) =lim f(x)
x0

Vedanta Optional Mathematics Teacher's Guide ~ 10 161

Hence the function f(x) is continuous at x = 0.

(b) f(x) = 2x2 + 1, for x < 2 at x = 2.
9 , for x=2
4x +1, for x > 2.

Solution 2x2 + 1, for x < 2 at x = 2.
Here, f(x) = 9, for x=2
4x +1, for x > 2.

For functional value at x = 2, we take
f(2) = 9

For right hand limit at x = 2, we have

lim f(x) = lim 4x + 1

x 2+ x 2+

= 4×2 +1

=9

For left hand limit at x =2, we have,

lim f(x) = lim 2x2+1

x 2- x 2-

= 2.22 +1 = lim f(x)
x 2-
=9
lim f(x)
x 2+

The limit of the function f(x) exists at x = 2.

lim f(2) = 9

x2

Now, f(2) = lim f(x)
x 2

The function f(x) is continuous at x = 2.

4. Show that the following functions are continuous at the points mentioned.

x3 -8 , for x ≠ 2
x-2
at x = 2
f(x) =
12 , for x = 2

162 Vedanta Optional Mathematics Teacher's Guide ~ 10

Solution x3 -8 , for x ≠ 2
Here,f(x) = x-2
at x = 2

12, for x = 2

For x ≠ 2, we take limit of the function f(x) at x = 2.

For limit of f(x) at x = 2 we factorize the denominator and numerator of the function.

lim f(x) = lim x3-8
x2 x-2
x2

= lim x3 - 23
x2 x-2

= lim (x-2)(x2+2x+4)
x2 x-2

= lim (x2+2x+4)
x 2

=22 +2.2+4
=4+4+4
=12
Functional value at x = 2 is given as 12
i.e. f(2) =12

f(2) = lim f(x)
x2

The function is continuous at x = 2.

5. A function is defined as follow

3 + 2x, for - 3 ≤ x< 0
2 3

f(x) = 3 - 2x, for 0≤ x < 2
3
-3-2x , for x ≥ 2
3
Show that f(x) is continuous at x =0 and discontinuous at x = 2 .

Solution 3 + 2x, for - 3 ≤ x < 2
Here,f(x) = 2 3

3 - 2x , for 0≤ x < 2
3
-3-2x , for x ≥ 2

First let us discuss the continuity of f(x) at x = 0

Vedanta Optional Mathematics Teacher's Guide ~ 10 163

Functional value for x =0 is given by

f(0)=3 – 2×0 =3 (as x ≥0)

Left hand limit of f(x) at x =0 is given by

lim f(x) = lim (3+2x) ( as x < 2)

x 0- x 0-

=3 + 2×0

=3

Right hand limit of f(x) at x =0 is given by

lim f(x) = lim (3–2x) (as x ≥ 0)

x 0+ x 0+

= 3 – 2×0

=3

lim f(x) = lim 4x–3

x 0+ x 0-

i.e. the limit of the function f(x) exists at x = 0. ie. lim f(x) = 3
x0

Finally, we have,

f(0)=lim f(x)
x0

f(x) is continuous at x=0.proved

Again let us discuss the continuity of the function f(x) at x = 3 .
2
3
Functional value of f(x) at x= 2 is given by

f(x) =–3–2x (as x≥ 3 )
2
33
f( 2 ) = –3–2× 2

=–6

3
Left hand limit of f(x) at x= 2 is given by
lim f(x) lim (3-2x)
=x ( as x≤ 3 )
x 3 - 3 - 2
2 2

=3 – 2. 3
2
=0

Right hand limit of f(x) at x = 3 is given by
2

lim f(x) lim f(-3-2x) 3
= ( as x≥ 2 )
x 3 x 3 +
2 + 2

= –3 – 2× 3
2
=–6

164 Vedanta Optional Mathematics Teacher's Guide ~ 10

lim f(x) lim f(4x-3)

x 3– ≠x 3 +
2 2
3
i.e. the limit of the function does not exists at x = 2 .

The function is discontinuous at x = 3 . proved.
2

6. Find the value of m of f(x) is continuous at x =5

x2 -2x , for x≠5
x-2
at x = 5
f(x) =
m , for x = 5

Solution x2 -2x , for x≠5
Here, f(x) = x-2
at x = 5

m, for x = 5

Functional value of f(x) at x =5 is m.

i.e. f(5) =m

For limit of the function at x = 5, we take

lim f(x) = lim x2-2x
x5 x-2
x5

= lim x(x-2)
x5 (x-2)

= lim x
x 2

=5

Since the given function is continous at x =5, we take
lim f(x) =f(5)
x2

ie. 5=m
m=5

7. Discuss the continuity of given function at x =2

x2 -4 , for x≠2
x-2

f(x) = 5, for x = 2

If f(x) is not continuous, how can your make it continuous at the point x =2.
Vedanta Optional Mathematics Teacher's Guide ~ 10 165

Solution: x2 -4 , for x≠2
Here, f(x) = x-2

5, for x = 2

Functional value of f(x) at x =2 is 5.

i.e. f(2) =5

For limit of the function, we have

lim f(x) = lim x2-4 ( 0 forms)
x2 x-2 0
x2

= lim (x+3)(x-2)
x2 (x-2)

= lim (x+2)
x 2

=2+2

=4

f(2) ≠ lim f(x)
x2

i.e. the function is discontinous at x =2. To make the above function continuous at x =2,
we can redefine at as follows

x2 -4 , for x≠2
x-2

Here, f(x) = 4 , for x = 2

Questions for practice

1. Discuss the continuity of the following
(a) Growth of plants for certain period of time.
(b) A snake crawling on a ground.
(c) Motion of wheels of a motorcycle on the road when it is in motion.
(d) A frog jumping on a ground.
(e) traces of feet of a man when he walking on the road.
(AM:(a) Continuous (b) Continuous (c) Continuous (d) discontinuous (e) discontinuous
(f) Write the continuity of number line of set of real numbers. (continuity)
(g) Write the continuity of number in a number line.

166 Vedanta Optional Mathematics Teacher's Guide ~ 10

2. State the continuity or discontinuity of the function from the given graph.
(a) y (b) y

x' –4 x x' x
4 2

y' y'
(c) y (d) y

x' x x' x
–1 1 –1 O 2
O

y' y'
(e) y (f) y

x' x x' x

–4 –2 O 35 O4

y' y'

Vedanta Optional Mathematics Teacher's Guide ~ 10 167

3. Let f be a real valued function defined by f(x)=x +8

(a) What are the values of f(x), for x=1.9, 1.99,1.999,1.9999.
(b) What are the values of f(x), for x=2.1,2.01,2.001,2.0001
(c) What are the left hand and right hand limit of the above function ? Write the limit of the

function.
(d) Does the limit of the function exists at x =2 ?
(e) Can you say the function f(x) is continuous at x =2 ?
Also state the reason.

4. For a real valued function f(x)=2x+3
(a) Find the values f(2.95), f(2.99), f(2.999), f(3.01), f(3.001), f(3.0001), f(3).
(b) Is the function continuous for x=3 ?
(c) Write the conditions of continuity of above functions ?

5. For a real valued function f(x)=2x +3,
(a) Find f(4,9), f(4.99), f(4.999), f(4.9999).
(b) Write the left hand limit of the f(x) at x=5 with symbol.
(c) Find the values of f(5.1), f(5.01), f(5.001), f(5.0001)
(d) Write the right hand limit of the f(x) at x=5 with symbol.
(e) Find f(5).
(f) what conclusion can you draw from above ?

6. Given that f(x) = 3x + 1, for x < 1
4, for x = 1
5x - 1, for x ≥ 1.

(a) Find the left hand and right hand limit of the function at x=1.
(b) Find the value of f(x) at x=1.
(c) What is the meaning of existence of limit of a function at x=1?
(d) Is the above function continuous at x=1 ?
Give your reasons ?

168 Vedanta Optional Mathematics Teacher's Guide ~ 10

UNIT

seven Matrices

1. Objectives Objectives
S,N. Level
To define determinant
(i) Knowledge(k)
To define determinant of order 2 × 2 .

To define singular and non – singular matrices.

To define inverse matrix of a given matrix

To define Cramer's rule.

To find determinant of order 2 × 2.

To check given matrices singular or non – singular.
a b
(ii) Understanding(U) To write formula of inverse of matrix c d .

Write Cramer's rule to solve simultaneous equations with two
variables.

To solve problems of determinant of order 2 × 2.

(ii) Application(A) To solve simultaneous equation with two variables by using
inverse matrix method .

To solve simultaneous equation of two variables by Cramer's
rule..

Higher Ability To solve verbal problems in two variables by
(HA)
(iv) – inverse matrix method

– Cramer's rule.

2. Required Teaching Materials:
Chart papers with definitions

– determinants
– inverse matrix
– Cramer's rule.

2. Teaching learning strategies:
– Review definitions of a matrix.
– Discuss on definition of determinants.
– To calculate determinants of order 1 × 1 and 2 × 2 with suitable examples.
– To differentiate det. | – 7| and absolute value |– 7| .
– Discuss singular and non – singular matrices with examples.

Vedanta Optional Mathematics Teacher's Guide ~ 10 169

ab
– If c d ≠ 0, discuss and derive the formula

A1 = | 1 | a b , where, A = a b
A c d c d

– Solve at least two examples of solution of simultaneous equations with two variables
by matrix inverse method.

– Discuss about Cramer's rule to solve simultaneous equation with two variables x and y.
– Demonstrate solution of simultaneous equation in two variables by Cramer's rule.

Notes :

1. A square matrix A is called a singular matrix if its determinant is zero, ie. |A|=0,
otherwise it is non – singular.
a b
2. If A = c d is non – singular matrix, then inverse of A is given by

A–1= 1 d –b , where, A = a b = ad – bc ≠0
|A| –c a c d

3. |A|=0, then A–1 does not exist.

4. If AB = BA =I, then A and B are said to be inverse of each other.

5. Let two simultaneous equations be

a1x + b1y = c1 a1 b1 ≠ 0.
a2x + b2y = c2 and D = a2 b2

then solution will be, x = D1 where, D1 = c1 b1
D c2 b2

x = D2 where, D2 = a1 c1
D a2 c2

Some solved problems

Determinants

1. If P = 1 0 ,Q= 3 1 , find the determinant of P + Q + I.
3 1 5 3

Solution 1 0
0 1
: Here, I =

Now, P + Q + I

170 Vedanta Optional Mathematics Teacher's Guide ~ 10

= 1 0 + 3 1 + 1 0
3 1 5 3 0 1

= 1+3+1 0 +1 +0
3 +5 +0 1+3+1

= 5 1
8 5

Now, |P + Q +I|
5 1
= 8 5 = 25 – 8

= 17 a2 + ab + b2 b2 + bc + c2
a–b
2. Evaluate = b–c

Solution

We have, a3 – b3 = ( a – b) (a2 – ab + b2)

Here, = a2 + ab + b2 b2 + bc + c2
a–b
b–c

= (a3 – b3 ) – (b3 – c3 )
= a3 – b3 – b3 + c3
= a3 – 2b3 + c3

3. Show 42that–3|1ABa|n=d B|=A||B05 | –1
if A= 2

Solution 2 3 0 –1
4 –1 5 2
Here, AB = ×

= 0 + 15 –2 + 6
0–5 –4–2

= – 15 4
= 9–05+ 20– 6

= – 70 2 3 = – 2 – 12 = – 14
Again, |A| = 4 1

|B| = 0 –1 =0+5=5
5 2

|A||B| = – 14 × 5 = – 70

Vedanta Optional Mathematics Teacher's Guide ~ 10 171

|AB| = |A||B| proved.

4. If M = 1 22 3
4 5 and N = 3 5 , find the determinants of

(a) MT + NT (b) (MN)T

Solution 1 4 and NT = 2 3
Here, Mt = 2 5 3 5

Now, Mt+Nt = 1+2 4+3
2+3 5+5

= 3 7
5
10

|Mt + Nt | = 3 7 = 30 – 35 = – 5
5 10

(b) Here, MN = 1 2 × 2 3
4 5 3 5

= 2+6 3 + 10
8 + 15 12 + 25

= 8 13
23 37

Now, |MN| = 8 13 = 296 – 299 = – 3
23 37

5. Find the value of x if x x =9
3x 4x
Solution
x x
Here, =39x 4x = (4x2 – 3x2) = 9
or, x2

x =±3

6. If P = 1 2 and Q = 2 3
4 5 4 5

is |(P + Q)2 |=| P2 + 2PQ + Q2 |. 3
25
Solution 1 2 2 5
4 51 14
Here,P = and Q = 4
2
Now, P2 = P.P = 4
5

172 Vedanta Optional Mathematics Teacher's Guide ~ 10

= 1+8 2 + 10
4 + 20 8 + 25

= 9 12
24 33

Q2 = Q. Q = 2 3 2 3
4 5 4 5

= 4 + 12 6 + 15
8 + 20 12 + 25

= 16 21
28 37

PQ = 1 2 2 3
4 5 4 5

= 2+8 3 + 10
8 + 20 12 + 25

= 10 12
28 37

2PQ = 2 10 12
28 37

= 20 24
56 74

P+Q= 1 2 + 2 3 = 3 5
4 5 4 5 8 10

(P + Q)2 = (P + Q)(P + Q)

Now, (P + Q) (P + Q) 3 5 3 5
8 10 8 10

= 9 + 40 15 + 50
24 + 80 40 + 100

= 49 65
104 140

Now, LHS = | (P + Q )2|

= 45 57 49 65
108 144 104 140

Vedanta Optional Mathematics Teacher's Guide ~ 10 173

= 6860 – 6760

= 100

Also. P2 + 2PQ + Q2
9 12 20 24 16 21
= 24 33 + 56 74 + 28 37

= 45 57
108 144

RHS =| P2 + 2PQ + Q2 |
45 57

= 108 144

= 6480 – 6156
= 324

(P + Q)2 ≠ | P2 + 2PQ + Q2 |

7. If P = 0 –2 , find the determinant of 2P2 – 5P + 4I, where, I = 1 0
3 4 0 1.
Solution

Here, 0 –2
3 4

Now, P2 = P.P = 0 –2 × 0 –2
3 4 3 4

= 0 –6 0–8
0+ 12 – 6 + 16

= –6 –8
12 10

2P2 = 2 –6 –8
12 10

= – 12 – 16
24 20

5P = 5 0 –2 = 0 – 10
3 4 15 20

4I = 4 1 0 = 4 0
0 1 0 4

Now, 2P2 – 5P + 4I

174 Vedanta Optional Mathematics Teacher's Guide ~ 10

= – 12 – 16 – 0 – 10 + 4 0
24 20 15 20 0 4

= –12 – 0 + 4 – 16 + 10 + 0
24 – 15 + 0 20 – 20 + 4

= –8 –6
9 4

| 2P2 – 5P + 4I | = –8 –6 = – 32 + 54
9 4

= 22

Inverse Matrix 10 5
2 3
1. Find the adjoint matrices of A = .

Solution 10 5
2 3
Here, A =

3 –5
Adjoint of A = – 2 10

2. Find the inverse of given matrices.

(a) A = 4 6 (b) C = 10 5
2 3 12 3

Solution 4 6 , |A|= 4 6 = 12 –12 = 0
(a) Here, A = 2 3 2 3

|A| = 0, the inverse of matrix A ie. A – 1 does not exists.

(b) Here, C = 10 5
12 3

|C| = = 10 5 = 30 – 60 = – 30.
12 3

Since |C | ≠ 0, 3 –5
C – 1 exists – 12 10

adjoint of C =

1 3 –5
= –30 – 12 10

Vedanta Optional Mathematics Teacher's Guide ~ 10 175

–1 1
10 6
=2
5 – 1
3

3. Show that 31 2 –1
5 2 and – 5 3 are inverse to each other.
Solution
3 1 2 –1
Let A = 5 2 and B = –5 3

Now, AB = 3 1 . 2 –1
5 2 –5 3

= 6–5 –3+3
10 – 10 –5+6

= 1 0
0 1

2 –1 3 1
Also, BA = – 5 3 5 2

= 6–5 2–2
–15 + 15 –5+6

= 1 0
0 1

AB = BA = I
By definition A and B are inverse of each other

Alternative Method
3 1 2 –1
Let A = 5 2 and B = –5 3

|A| = 3 1 =6–5=1
5 2

|A| ≠ 0.

Hence inverse of martix A exists.
2 –1
Adjoint of A = –5 3

1
A – 1 = |A| adjoint of A

176 Vedanta Optional Mathematics Teacher's Guide ~ 10

= 1 2 –1
1 –5 3

= 2 –1 =B
–5 3

B is the inverse of A. Similarly , we can show that B is the inverse of A. proved

4. If A = 5 3 , then show that
3 2

i) (A –1) – 1 = A

ii) A –1A = AA –1 = I

Solution
i) (A –1) – 1 = A

Let us find A –1 3
5 2 5 3
Here, A = 3 , |A| = 3 2 = 10 – 9
=1
11 2 –3
A – 1 = |A| adjoint of A = 1 –3 5
Again, let A – 1 = B

Let us find B – 1 = ie. (A – 1 ) – 1
2 –3
|B| = –3 5 = 10 – 9 = 1

1
B – 1 = |B| adj. B

1 53
=1
32
5 3
=3 2 = A.

B – 1 = (A – 1 ) – 1 =A proved

ii) A –1A = AA –1 = I 5 3
2 –3 3 2
LHS = A –1A = –3 5

= 10 – 9 6–6
–15 + 15 – 9 + 10

= 1 0
0 1

Vedanta Optional Mathematics Teacher's Guide ~ 10 177

Again, AA –1 = 5 3 2 –3
3 2 –3 5

= 10 – 9 –15 – 15
6– 6 – 9 + 10

= 1 0
0 1

A –1 A = AA –1 = I proved.

5. For what value of x, the product of matrix 3 2 2 –1 does not have
its inverse matrix. x 4 3 2

Solution 3 2 2 –1
: Here, x 4 32

= 6+6 –3+4
2x + 12 –x+8

= 12 1
2x + 12 8–x

Now, 12 1
2x+12 8–x

= 96 – 12x – 2x – 12 = 84 – 14x

If = 12 1 = 0, the inverse matrix does not exists.
2x+12 8–x

ie. 84 – 14x = 0

84
x = 14 = 6

For x = 6, the inverse of given matrix does not exist.

Solution of system of Linear Equations by Inverse matrix method
2x + 4y
1. Factorize : 5x + y

Solution 2x + 4y
5x + y
Here,

178 Vedanta Optional Mathematics Teacher's Guide ~ 10

= 2 4 x
5 1 y

2. If A = a b , then answer the following question :
c d

(a) Find the determinant of matrix A

(b) Under which condition A does not have its inverse ?

Solution ab
cd
(a) Here, A =
= ad – bc
|A| = a b
c d

(b) If |A| ≠ 0, then inverse of matrix A exists. It means that if |A| = 0, then the given matrix A does not
have its inverse.

3. If 1 0 x = 4 , find the values of x and y.
0 1 y 5

Solution 1 0x 4
0 1y 5
Here, =

x +0 4
or, = 0 +y = 5

or, x = 4
y 5

Equating the corresponding elements of equal matrices
x = 4 and y = 5.

4. Check whether the system of linear equations have unique solution or not

(a) 4x + 2y = 8
x–y=1

Solution
4x + 2y = 8 ie. 2x + y = 4 .........(i)

and x – y = 1 .........(ii)

writing teh above equations in the matrix form, we get

2 1 x = 4
1 –1 y 1

or, AX = C,
where,

Vedanta Optional Mathematics Teacher's Guide ~ 10 179

A= 2 1 , X= x and C = 4
1 –1 y 1

Now, |A| = 2 1 =–2–1=–3
1 –1

Since |A| = – 3 ≠ 0, A – 1 exists and the given system of linear equations have unique solution

5. Solve the following system of linear equations by matrix method.

(a) x = 2y – 1 and y = 2x

Solution
Here, x = 2y – 1

or, x – 2y = – 1 ............(i)

and y = 2x

or, 2x – y = 0..........(ii)

Writing the above equation in matrix form, we get,

1 –2 x = –1
2 –1 y 0

or, AX = C ............(iii)
1 –2 x –1
A= 2 –1 ,X= y ,C= 0

where, 1 –2 =–1+4=3
|A| = 2 –1

Since |A| = 3 ≠ 0, A 1 exists. There is a unique solution of given system of linear
equations.
From (ii), we get,

X= A 1 C

To find A 1 , we have |A| = 3,
–1 2
adj. A = –2 1

A1 1 1 –1 2
= A – 1 = |A| adj . A = 3 –2 1

Now, X = A – 1 C 2 –1
–1 1 0
1 –2
=3

= 1 1 +0
3 2 +0

180 Vedanta Optional Mathematics Teacher's Guide ~ 10

x 1
y 3
= 2
3

Equating the corresponding elements of equal matrices, we get,
1 2
x== 3 and y = = 3 ,

3x y
(b) 2 – 3 =1
xy
3 – 3 =1

Solution :
3x

Here, 2 + 2y =1 or 3x + 4y = 2 ...........(i)
xy
3 – 3 =1 or x – y = 3......(ii)

Writing above equation in matrix form, we get,
3 4 x 2
1 –1 y = 3

or, AX = c or X = A – 1 C ..........(iii),

where, 3 4 x 2
1 –1 y 3
A= ,X= and C =

where, 3 4
1 –1
|A| = = – 3 – 4 = –7

|A| = –7 ≠ 0.

Hence A 1 exists and there is a unique solution of given system of linear equaiton.

From (ii), we get,
1 1 –1 –4
A – 1 = |A| adj . A = –7 – 1
3

Now, X = A – 1 C 2
–1 –4 3
1 –1 3
= –7

=– 1 – 2 – 12
7 –2+9

Vedanta Optional Mathematics Teacher's Guide ~ 10 181

=– 1 – 14
7 7

x 2
y = –1

x = 2 and y = –1

43 32 1
(c) x – y =1 and x – y = 24

Solution
Writing the given equation in matrix form, we get

43 11
3– x
1= 1
y 24

or, AX = C

or, X = A – 1 C

where,

4 3 11
3 – x
A= ,X= 1 and C = 1
y 24

where, 4 –3 = – 8 – 9 = – 17
|A| = 3 –2

Now, from X = A – 1 C –2 –3
11 –3 4

A – 1 = |A| adj . A = –1 7
Now, X = A – 1 C

1 –2 –3 1
= –1 7 1
–3 4 24
1 1
= –1 7 –2 – 6

–3 + 1
8

= 1 – 17
–1 7 8

– 17
6

182 Vedanta Optional Mathematics Teacher's Guide ~ 10

= – 1
8

– 1
6

= 1 = – 1
x 8

1 – 1
y 6
1
ie. x = 1 or x = 8
8

and 1 = 1 or x = 6
y 6

x = 8, y = 6

x2 4
(d) 3 + y = 2 and x + y = 5

Solution
Given equations can be written in the matrix form,

12 x 2
3 1= 5
14 y

or, AX = C

or, X = A – 1 C

where,

12 x 2
5
A= 3 , X = y and C =

14

where,

1 2 4 –2
|A| = 3 3 3
= –2=

14

Since |A| ≠ 0, A 1 exists and there is a unique solution of given system of equations.
1

A – 1 = |A| adj . A

Vedanta Optional Mathematics Teacher's Guide ~ 10 183

= –2 42
3 –1 1

3

3 42
2
=– –1 1

3

Now, X = A 1 C

3 42 2
2 5
=– –1 1

3

8 – 10

=– –2 + 5
3

=– 3 –2
2 –1
3

x3

==

11
y2
1
ie. x = 3 and y= 1 or y = 2.
2

(e) 3y + 4x = 2xy 7y + 5x = 29 xy

Solution
Here, 4x + 3y = 2xy

dividing both sides by xy we get,
43
y + x =2

34
or, x + y = 2 .......(i)
and 5x + 7y = 29xy

57
or, y + x = 29

or, 7 + 5 = 29 ..........(ii)
x y

184 Vedanta Optional Mathematics Teacher's Guide ~ 10

Writing above equations in matrix form,

34 12
75 x
1 = 29

y

or, AX = C or, X = A – 1 C

where X = 1 ,A= 3 4 and C = 2
3 x 7 5 29
1
|A| = 7 y

4 = 15 – 28 = – 13
5

A 1 exists and there is a unique solution of given system of equations.

1
A – 1 = |A| adj . A

= 1 5 –4
–13 –7 3

Now, X = 1 5 –4 2
–13 –7 3 29

= 1 10 –116
–13 – 14 + 87

= 1 – 106
–13 73

1 106
x = 13
1 y–=17–33
ie. x =y 13 , 13
106 73

(f) 3x + 5y = 7x + 3y =4
4 5

Solution
3x + 5y 7x + 3y
Here, 4 = 5 =4

3x + 5y
Taking 4 = 4

Vedanta Optional Mathematics Teacher's Guide ~ 10 185

or, 3x + 5y = 16............(i)

and taking 7x + 3y =4
5

and 7x + 3y = 20............(ii)

Writing above equations in matrix form,we get,
35x 16
7 3 y = 20

or, AX = C or, X = A – 1 C ............(iii)
3 5 x 16
where X = 7 3 ,A= y and C = 20

3 4 = 9 – 35 = – 26
|A| = 7 5

|A|≠ 0, A 1 exists and there is a unique solution of given system of equations.
1

Now, = |A| adj . A

= 1 3 –5
– 26 –7 3

From (iii), we get

Now, X = 1 3 –5 16
– 26 –7 3 20

= 1 48 –100
– 26 – 112 + 60

= 1 – 52 = 2
– 26 – 52 2

x=2,y=2.

(g) 2x + 4 =y= 40 – 3x =4
5 4

Solution

Here, 2x + 4 = y= 40 – 3x =4
5 4

2x + 4
Taking 5 = y

or, 2x + 4 = 5y

186 Vedanta Optional Mathematics Teacher's Guide ~ 10

2x – 5y = –4.........(i)

and taking y = 40 – 3x
4

or, 4y = 40 – 3x

3x + 4y = 40..........(ii)

and 7x + 3y = 20............(ii)

Writing above equations in matrix form,we get,
2 –5 x –4
3 4 y = 40

or, AX = C
X = A – 1 C ............(iii)

where X = 2 –5 ,A= x and C = –4
3 4 y 40

2 –5 = 8 + 15 = 23
|A| = 3 4

Since |A|≠ 0, A 1 exists

Now, A1 1
= |A| adj . A

= 1 4 5
23 –3 2

From (iii), we get, 4 5 –4
1 –3 2 40
Now, X = 23

= 1 –16 + 200
23 12 + 80

184 8
= 23 = 84

x92
y23 = 4

ie. x = 8 and y = 4 .

6. Equations of pair of lines are 2x – y = 5 and x – 2y =1
(a) Write the equation in matrix form.
(b) Is there unique solution of above given equations.
(c) Solve the equations.
(d) Check the solutions to show that the values of x and y so obtained are true.

Vedanta Optional Mathematics Teacher's Guide ~ 10 187

Solution

(a) Given equations are
2x – y = 5

x – 2y =1

writing above equations in matrix form,
2 –1 x 5
1 –2 y = 1

or, AX = C

X = A – 1 C ............(iii) x 5
2 –1 y 1
where X = 1 –2 ,A= and C =

(b) |A| = 2 –1
1 –2 = –4 + 1 = – 3

Since |A|≠ 0, A 1 exists and there is unique solution of given equations.

Now, A1 1
= |A| adj . A

= 1 –2 1
–3 –1 2

Now, X = A – 1 C = 1 2 –1 5
–3 1 –2 1

= 1 –10 + 1
–3 – 5+ 2

= 1 –9
–3 –3

= 3
1

x3
y =1

ie. x = 3 and y = 1 .

(d) put x = 3 and y = 1 in above equation from equation 2x – y =5
or, 2.3 – 1 = 5

5 = 5 (True).
Again from equation, x – 2y = 1
3 – 2.1 =1

188 Vedanta Optional Mathematics Teacher's Guide ~ 10

1 = 1 (True)
x = 3 and y = 1 are true.

Cramer's Rule

1. Equations are a1x + b1y = c1 and a2x + b2y = c2
What are the determinants represented by D, Dx and Dy.

Solution constants

here, a1x + b1y = c1
and a2x + b2y = c2
coefficient of x coefficient of y

a1 b1 c1
a2 b2 c2

D= a1 b1 = a1b2 –a2b1
Dx = a2 b2
Dy =
c1 b1 =b2c1 – b1c2
c2 b2
a1
a2 c1 =a1c2 – a2c1
c2

2. If D =4, Dx = 1 , Dy = 1 , find the values of x and y.
2 4

ting D, Dx and Dy.

Solution 1 , Dy = 1
Here, D =4, Dx = 2 4

Dx 1 1
D 2 8
x = = 4 =

Dy 1 1
D 4 = 16
y = =4

11
x = 8 , y = 16

3. Solve the following system of equations by using cramer's rule.
(a) 3x – 2y = 1 and – x + 4y = 3

Solution constants
coefficient of x coefficient of y

Vedanta Optional Mathematics Teacher's Guide ~ 10 189

3 – 2 1
3
– 1 4
3 –2
D= –1 –4 = 12 – 2 = 10

D1 = 1 –2
D2= 3 4 = 4 + 6 = 10

3 1 = 9 + 1 =10
–1 3

x = D1 = 10 =1
D 10

y = D2 = 10 =1
D 10

(b) 3 + 5 = 1, 4 + 3 = 29
x y x y 30

Solution 1 1
x y
coefficient of coefficient of constants
5 1
3 3

4 3
3 5
D= 4 3 = 9 – 20 = –11

15

D1 = 29 3 =3– 29 ×5=– 11
30 30 6

31

D2= 4 29 =3× 29 –4 =– 11
30 30 10

D1 –11 1
D =–116 =6
1 =
x

or, x = 6 and

D2 –11 1
D 10 = 10
1 = = –11 y = 10
y

(c) 10 – 2y = –1 and 4 + 3y = 11
x x

190 Vedanta Optional Mathematics Teacher's Guide ~ 10

Solution coefficient of y constants
coefficient of x –1

10 – 2 11

4 3
10 –2
D= 4 3 = 30 + 8 = 38

–1 – 2
D1 = 11 3 = – 3 + 22 = 19

10 –1
D2= 4 11 = 110 + 4 =114

1 = D1 = 19 = 1 x=2
x D 38 2

y = Dy 114 =3
D = 38

x = 2 and y = 3.

(d) 7 (x –y) = x + y and 5(x + y) = 35 ( x – y)

Solution
Here, 7 (x –y) = x + y

or, 7x – 7y = x + y

or, 6x – 8y = 0 ..........(i)

and 5(x + y) = 35 ( x – y)

or, x + y = 7x – 7y

or, 6x – 8y = o ...........(ii)

coefficient of x coefficient of y constants
0
6 – 8 0

6 – 8

D = 6 –8 = –48 + 48 = 0
6 –8

Since D = 0, there is no solution of given equations.

(e) 2(3x – y) =5(x – 2) and 3( x +4y) = 2(y – 3)

Solution
Here, 2(3x – y) =5(x – 2)
or, 6x – 2y – 5x = –10
or, x – 2y = – 10 ........(i)
and 3( x +4y) = 2(y – 3)

Vedanta Optional Mathematics Teacher's Guide ~ 10 191

or, 3x + 12y= 2y – 6

or, 3x + 10y = – 6 ......(ii)

coefficient of x coefficient of y constants
– 10
1 – 2 –6

3 –2 10
1 10
Now, D = 3 = 10 + 6 = 16

–10 –2
D1 = –6 10 = – 100 – 12 = – 112

1 –10
D2= D2 = 3 –6 = – 6 + 30 =24

x = D1 = –112 =–7
D 16

y = D2 24 3
D = 16 =2

3
x = – 7 and y = 2

4(a) A helicopter has 4 seats for passengers. Those willing to pay first class fares can take
60 kg of baggage each but tourist class passengers are restricted to 20 kg each. The
helicopter can carry only 120 kg baggage all together. To find the number of passengers
of each kind, use Cramer's rule.

Solution
Let x and y be the number of passengers of first class and tourist class respectively. Then
by question, we get,

x + y = 4 ........i)

60x + 20y = 120

or, 3x + y = 6 ..........ii

To solve equations (i) and (ii) by Cramer's rule

coefficient of x coefficient of y constants

1 1 4

3 1 6
1 1
Now, D = 3 1 = 1 – 3 = –2

4 1 =4–6=–2
D1 = 6 1

1 4
D2= 3 6 = 6 – 12 =– 6

192 Vedanta Optional Mathematics Teacher's Guide ~ 10

x = D1 –2 =1
D =–2

y = D2 –6 =3
D =–2

x = 1 and y = 3

Questions for practice

Determinants

1. Find the determinants of given matrices
1 2 –4 –5
(a) A = 4 5 , (b) P = 6 7

2. If P = 1 2 and Q = 3 2 ,find the determinants of
4 5 4 5

a) P + Q b) 2P + 3Q (c) P + Q + I.

3. If M = 4 2 and N = 3 2 , then find the determinants of MN and NM.
1 2 6 3

4. If A = 2 3 and B = 4 2 , then verify that
–2 4 3 6

5|A. BIf|P==|A||B42| . 1 –2 3
3 1 4
and Q = , then find the determinants of (a) (P + Q)T (b) PT + QT .

6. If P = 2 1 , then find the determinant of P2 + 4P – 5I.
3 4

7. If A = 1 2 and B = 2 1 , is |(A + B)|2 = |A2 + 2AB + B2| .
3 2 4 2

8. If A = 2 1 , show that |A|2 = |A|2 .
–2 –1

Inverse Matrix 6 4 .
2 3
1. Find the adjoint matrix of

2. Find the inverse of the following matrices.

(a) A = 2 4 (b) P = 2 4 (c) R = 5 7
3 5 6 7 5 7

Vedanta Optional Mathematics Teacher's Guide ~ 10 193

3. Show that 3 2 and 5 –2 are inverse to each other.
7 5 –7 3

4. If A = 5 4 and AB = I, then find the matrix B.
6 7

5. If P = 1 3 , then show the following
2 4

a) PP –1 = P –1 P = I

b) (P –1) – 1 =2P 3
6. If A= 1 –4
and B = 2 –2 , then show that (AB)– 1 = B – 1 A – 1 .
–2 3

7. Find the matrix P when 1
1 0 0 0
(a) (3P – 1 ) = 0 1 (b) (3P)– 1 = 2 1

(c) (I + 3P) = 2 0
0 2

8.Solve for matrix A under the following conditions.

(a) –3 2 A= 6
6 5 7

(b) 4 2 A= 1 2
–6 2 3 4
Solution of system of Linear Equations by Inverse matrix

method. 4x + 3y
6x + 5y
1. Factorize

2. Solve the following matrices by inverse matrix method.

(a) 3x + 2y = 20 and 2x – y =4.

xy
(b) 4x – 5y = 2 and 4 + 3 = 4

3x + 5y 5x – 2y
(c) 8 = 5 =4

3x + 5y 7x + 3y
(d) 4 = 5 =4

5 10
(e) x + 3y = 7 and 7y – x = 12

194 Vedanta Optional Mathematics Teacher's Guide ~ 10

xy
(f) 4 + 3 = 2 , x + y = 7

(g) 1 + 2 = 2, 3 + 4 =5
x y x y

Cramer's rule :

Solve the following equations using Cramer's rule.

1. x + y = 5 and x – y = 3

2. 2x + 3y = 5 and 3x – 2y = 1

3. 5x – 4y = 1 and 4x + 5y = 9

xy x 3y
4. 4 + 10 = 1 and 5 + 25 = 1

11 23
5. x + y = 2 and x – y = 5

6. 3 – 7 = 1 and 5 + 4 = 17
x y x y

46 34 17
7. x + y = 0 and x – y = – 6

45 73
6. x + y = 58 and x + y = 67

Vedanta Optional Mathematics Teacher's Guide ~ 10 195

UNIT

eight Coordinate Geometry

Angle between two lines

Estimated Teaching periods : 7

1. Objectives :

S.N. Level Objectives

(i) Knowledge (K) To tell formula of angle between two lines.
To tell conditions for parallelism and perpendicularity of two lines.

To find angle between two lines when their slopes are given.
(ii) Understanding(U) To identify given lines parallel or perpendicular when their

slopes are given.

(iii) Application(A) To use formula to find angle between two lines when their
equations are given

To derive formula to find angle between the lines y=m1x +c1
and y=m2x + c2.
(iv) Higher Ability
(HA) Find conditions for parallelism and perpendicularity of two

lines.

2. Teaching Materials
Chart papers with angle between two lines, condition of parallelism and perpendicularity.

3. Teaching Learning Activities
– First review the formula of equation straight lines that the students have studied in
class 9.

– Draw figure to derive the formula to find angle between two lines y=m1x +c1 and y
=m2x +c2 and derive formula to find the angle between them.

– Discuss the conditions for parallelism and perpendicularity of two lines.

– Discuss the meaning of m1 = m2 and m1.m2 = –1
– Let the students do some problems after the teacher solved some problems as

examples.

Notes :

1. Three standard forms of equations of straight lines are
(i) Slope–intercept form y=mx + c .
((iiiii))DNoourbmlealin/ tpeerrcpeepnt dfoicrmularxafo+rmby =1

196 Vedanta Optional Mathematics Teacher's Guide ~ 10

xcos + ysin =p

2. Slope of general equation of straight line ax + by +c =0 is given by

34m..=SSpl–oepcCCeiaoooleeffffaoffiircclmiiieensnnettojfoooiffenqxyiunagt=itow–noobapfositnrtasig(mht)=linyxe22.--yx11
iii))sTlowpoe––ppooiinnttsfoformrmyy–y–y1 1==myx(22x--yx–11x1) (x–x1)

5(a) Angle between two lines y=m1x +c1 and y =m2x + c2 is given by

=tan–1 m1-m2
1 +m1m2

(i) Condition for parallelism, m1 = m2
(ii) Condition for parallelism, m1 – m2 = –1

(b) Equation of straight line parallel to the line ax +by +c =0 is ax +by +k =0, where k
is a constant

(b) Equation of a straight line perpendicular to the line ax +by +c =0 is bx – ay +k =0

Some solved problems

1. Show that the lines mx +my +p=0 and 2mx + 2ny +r =0 are parallel to each other.

Solution Coefficient of x =– m
Given equations of lines are Coefficient of y n
mx +my +p=0 ............(i)
2mx + 2ny +r =0..........(ii)

From equation (i), its slope(m1)=–

From equation (ii), its slope(m2)=– 2m =– m
2n n

since m1 =m2, the given two lines are parallel to each other.

2.Find the slope of line parallel to 4x +3y +12=0 .

Solution

Given equations of lines 4x +3y +12=0..............(i)

its slope(m1)=– Coefficient of x =– -4
Coefficient of y 3

Vedanta Optional Mathematics Teacher's Guide ~ 10 197

Equation of any line parallel to (i) has the same slope. Hence the required slope is

(m1)=– -4 2m m
3 2n n

From equation (ii), its slope(m2)=– =–

since m1 =m2, the given two lines are parallel to each other.

3.Find the slope of line perpendicular to 3x +2y +20=0 .

Solution

Given equations of lines 3x +2y +20=0..............(i)

slope of given line is (m1)=– Coefficient of x =– 3
Coefficient of y 2

Let m2 be the slope of the line perpendicular to line (i) then

m1 .–m32 = –1
or, 2 .m2 =–1

m2 = 2
3

4. Find the acute angle between the two given lines.

(a) y – 2- 3 x =5 and y= 2+ 3 x+8

Solution
Equation of the given lines are
2- 3 x – y+5=0..........(i)

and 2+ 3 x–y+8=0.......(ii)

From equation (i), its slope is (m1)=– Coefficient of x =– 2- 3 =2- 3
Coefficient of y -1

From equation (ii), its slope is (m2)=– 2- 3 =2+ 3
-1

Let be the angle between the lines (i) and (ii), we get

tan =± m1–m2
1 +m1m2

(2– 3 –2– 3)

1 +(2– 3 )(2+ 3)

=± 1 –2 3 =± 3
+4–3

198 Vedanta Optional Mathematics Teacher's Guide ~ 10

we require acute angle , taking positive sign,
tan = 3
=tan60º

=60º

(b) xcos +ysin = p and xsin – ycos =q

Solution Coefficient of x = -cos
Given equation of line are Coefficient of y sin
xcos +ysin – p=0...........(i)
xsin – ycos – q=0.........(ii)

From equation(i),slope (m1)=–

From equation (ii), slope (m2) =––scions = sin
cos

Let be the angle between the given lines.

Then, tan =± m1-m2
1 +m1m2
cos sin
- sin - cos
=± –cos sin
1+ sin . cos
=±∞

Taking positive sign, we get tan =∞

tan =tan90˚

=90˚

Hence the angle between lines (i) and (ii) as 90˚.

5. Find the obtuse angle between the lines .

(a) y+ 3 x+8=0 and y+20=0

Solution
The given equations of lines are

y+ 3 x+8=0 ...........(i)

y+20=0..........(ii)

Slope of line (i), m1=– Coefficient of x =– 3 =– 3
Coefficient of y 1

Slope of line (ii), m2=– Coefficient of x =– 0 = 0
Coefficient of y 1

Let be the angle between the lines. Then,
tan =± m1-m2

1 +m1m2

Vedanta Optional Mathematics Teacher's Guide ~ 10 199

=± - 3 -0
1+0

=± 3
Taking negative sign, we get

tan =– 3
tan =tan120˚

=120˚

Hence the angle between lines (i) and (ii) as 120˚.

(b) 2x+y = 3 and x +2y =1

Solution
Given equations are

2x+y +3=0 ...........(i)

x +2y –1=0..........(ii)

From equation (i),slope of line (i), m1=– Coefficient of x =– 2 =–2
Coefficient of y 1

From equation (ii), slope of line (ii), m2=– Coefficient of x =– 3
Coefficient of y 2

Let be the angle between the lines, (i) and (ii)

Then, tan =± m1-m2
1 +m1m2

-2 + 3
2
=± 1+ 2 . 3
2

=± 1
2×4

=± 1
8

Taking positive sign, we get

tan =– 1 =172.87˚
=tan–1 8–1

8

6. Show that the lines x – y +2 =0 and the line joining the points (4,6) and (10,12) are
parallel to each other.

Solution
Given equations of lines is x – y+2=0

200 Vedanta Optional Mathematics Teacher's Guide ~ 10


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