Vedanta Opt. Excel in Mathematics Book 8 Final (2078)

Coordinates Geometry

Exercise 8.4

Short Questions :

1. (a) Do the points (2, 3) and (5, – 1) lie in the locus with equation 2x + 3y = 13?

(b) Which of the points (4, 5), (–5, –4), and (4, –5) lie on the locus with equation
x2 + y2 = 41?

2. (a) Find the value of k is that the point (3, 2) lies on the locus

x2 + y2 – ky = 22.

(b) Find the value of m so that the point (2, 2) lies on the locus with equation
x2 + y2 + mx + 4y = 64.

(c) If the point (5, 2) lies on the locus with equation ax – 4y = 7, find the value
of a.

3. Find the locus of a point which moves such that

(a) its distance from (0, 0) is 5 units.

(b) its distance from (3, 4) is 5 units.

(c) its distance from (4, 6) is 8 units.

(d) its distance form (a, b) is 13 units.

4. Find the equation of a point which moves such that its distance is equal from
the points.

(a) (3, 4) and (–3, –4) (b) (4, 5) and (6, 0)

(c) (–3, 4) and (5, 6) (d) (p, q) and (q, p)

5. Find the equation of locus such that its distance.

(a) from X-axis is always 7 (b) from X-axis is always 6

(c) from Y-axis is always 5 (d) from Y-axis is always –8

Long Questions :

6. (a) If (2, 3) is a point on the locus y2 = ax, find the value of a. Also show that
the point (18, 9) also lies on the locus.

(b) (4, 5) is a point on the locus ax + 3y = 19 ,find the value of a and also show
that the point (10, 3) also lies on the locus.

7. (a) Find the equation of locus of a point which moves such that its distance
from (2, 4) is double of its distance from (–2, 3).

(b) Find the equation of locus of a point which moves such that its distance
fixe point (3, 4) is three times its distance from the point (–3, –4).

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Coordinates Geometry

(c) Find the equation of the locus of a point which moves so that its distance
from the point (4, 3) is half a its distance from X-axis.

(d) Find the equation of locus of a point which moves so that its distance from
the Y-axis is half of its distance from the origin.

8. (a) The coordinates of two fixed points A and B are (a, 0) and (–a, 0) respectively.
Find the equation of the locus of a moving point P in the following case.

(i) PA = 2PB (ii) PA2 + PB2 = AB2

(b) Find the equation of the locus of a point such that Ap2 + Bp2 = 25, where
A(4, 5) and B(3, 1).

(c) If M(–4, 3) and N(1, 0) are two fixed points and P(x, y) is a moving point
which moves such that PM2 – PN2 = 20, find the locus of the point P.

(d) Let M(2, 1) and N(–1, –1) be two fixed point. Find the equation of the locus
PM 2
of a point such at that PN = 3

1. (a) (2, 3) lies but (5, – 1) does not lie] (b) All points lie on the locus

2. (a) – 9 (b) 24 (c) 3
2 (b) x2 + y2 – 6x – 8y = 0
3. (a) x2 + y2 = 25

(c) x2 + y2 – 8x – 12y = 12 (d) x2 + y2 – 2ax – 2by + a2 + b2 = 169

4. (a) 3x + 4y = 0 (b) 4x – 10y + 5 = 0 (c) 4x + y = 9 (d) x = y

5. (a) y = 7 (b) y = 6 (c) x = 5 (d) x = –8

6. (a) 9 (b) a = 1
2
7. (a) 3x2 + 3y2 + 20x – 16y + 32 = 0 (b) 2x2 + 2y2 + 15x + 20y + 50 = 0

(c) 4x2 + 3y2 – 32x – 24y + 100 (d) 3x2 = y2

8. (a) (i) 3(x2 + y2) + 10ax + 3a2 = 0 (ii) x2 + y2 = a2

(b) 2x2 + 2y2 – 14x – 12y + 17 = 0 (c) 5x – 3y + 20 = 0

(d) 5x2 + 5y2 – 44x – 62y + 109 = 0]

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Equations of Straight Lines

Equations of Straight 9
Lines

9.1 Equations of Straight Lines Parallel to Coordinate Axes

(i) Equations of Straight Lines Parallel to Y-axis. Y B
a x=a
In the given figure AB is a line parallel to Y-axis, the
distance between the Y-axis and the line AB is 'a' Y' X
units. Hence every point on the line has x-coordinate A
a. Hence the equation of the line AB is x = a. If the line
AB is right of Y-axis, the value of a is positive and if it X'
is in the left of Y-axis, the value of a is negative.

Y
Q
U5 (3, 4)
4 (3, 3)
(-3, 3) 3 (3, 2)
(-3, 2) 2 (3, 1)
(-3, 1) 1 (3, 0)
(-3, 0)
X' O 1 2 3 4(35, -16) 7 8 X
-8 -7 -6 -5 -4 -3 -2 -1
-1
-2
-3 (3, -3)
-4
-5
-6
P (3, -6)
V -7

Y'

For example, in the above graph line PQ is parallel Y-axis and the points on the
line PQ are (3, 0), (3, 1), (3, 2), (3, 4), (3, 1), (3, –3), (3, –6), and so on. We note that
x-coordinate is 3 at every point of the line. Hence, the equation of MN is: x = 3.

In the same graph, the line UV is parallel to Y-axis at 3 unit left of it. Hence, its
equation is: x = –3, or x + 3 = 0.

Example 1. Find the equation of a straight line parallel to Y-axis and at 5 units
Solution: Y
right from the Y-axis. Show the line in the figure. 5 N

Here, equation of any line parallel to Y-axis is x = a. 4 x=5
The distance of the line from Y-axis = a = 5. 3
2
1
X' -2 -1-O1 1 2 3 4 5 X
Hence, the required equation of straight line is -2 M

Y'

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Equations of Straight Lines

x=a i.e. x = 5.

In the figure, MN is a straight line parallel to Y-axis, whose equation is
x = 5.

Example 2. Find the equation of the straight line parallel to Y-axis and passing
through the point (–5, 8).

Solution : Here, the equation of a line parallel to Y-axis is x= a

This line passes through the point (–5, 8)

Hence a = –5 i.e. x = –5 or x + 5 = 0 is the required equation of a
straight line.

(ii) Equation of Straight Line Parallel to X-axis. A Y B
X' y=b X
In the figure AB is parallel to X-axis. The distance
between AB and X-axis is b units. Hence, every point b
on the line has y-coordinate b. Hence, the equation of O
the line AB is y = b.

Y

A y=3 B Y'

X' O X
C y = -3 D

Y'

In the above graph, AB and CD are parallel to X-axis. Every point on line AB has
y-coordinate 3. It means that AB is at 3 units above X-axis. Hence, its equation is
y = 3. Again, every point on line CD has y-coordinate –3. It means that CD is at 3
units below X-axis. Hence, its equation is y = –3 or y + 3 = 0.

Example 3. Find the equation of a straight line parallel to X-axis at 3 units below
Solution :
X-axis. Show it in figure. Y

Here, equation of a line parallel to X-axis is X' O X
y = b. y = –3

The line is 3 units below X-axis,

hence, b = –3. Y'

Hence, the required equation of straight
line is:

y = –3 or y + 3 = 0

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Example 4. Find the equation of straight line parallel to X-axis and passing
through the following points. (i) (2, 3) and (ii) (–3, –5)

Solution : (i) Let the equation of a line parallel to X-axis be y = b, as the line
passes through the point (2, 3), b= 5. Hence, the required equation
of line is b = 5.

(ii) Let the equation of straight line parallel to X– axis be y = b, as
the line passes through the point (–3, –5), b= –5. Hence, the
required equation of straight line is y = –5 or y + 5 = 0.

Collinear Points YB
R
A set of points lying on the same straight line are called
collinear points. In the adjoint figure, P, Q, and R points Q
are on the same straight line AB. Hence, they are collinear P

points. If P, Q, and R are collinear points, slope of X' T X
O
PQ = slope of QR = slope of PR. A
Y'
Hence the slopes of the line segments of the same line

are equal to each other.

For examples : Y

In the adjoint graph, P(2, 3), Q(4, 5), and R(6, 7) are 9
8
7
collinear points. 6 Q(4, 5) R(6, 7)

5–3 2 5
4–2 2 4
Slope of PQ = = =1 3
2 P(2, 3)
7 – 5 2
Slope of QR = 6 – 5 = 2 = 1 1
X' -3 -2 -1-O1 1 2 3 4 5 6 7 8 9 X

Slope of PR = 7 – 3 = 4 = 1 -2
6 – 2 4
Hence, slope of PQ = slope of QR = slope of PR. Y'

Intercepts Y
B(0, b)
Let MN be a straight line. It cuts X-axis at A and Y-axis N
at B. O
Y'
Then,

(i) OA is called x-intercept of line MN and it is X' A(a, 0)
denoted by a. X

M

? x-intercept of MN = OA = a

(ii) OB is called y-intercept of MN and it is denoted by b.

? y-intercept of MN = OB = b

(iii) OA and OB are called intercepts of the line MN.

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Example 5. Find the intercepts of the following straight lines as shown in the
figures:

(a) Y (b) Y

B (0, 4) X' A(–4, 0) O X

(6, 0) B(0, -5)
X' O A X

Y' Y'

Solution : (a) In the figure, (b) In the figure,
x-intercept (OA) = a = 6 x-intercept (OA) = a = –4
y-intercept (OB) = b = 4 y-intercept (OB) = b = –5

9.2 Equation of straight line in slope intercept from
y=mx + c

To find the equation of straight line in form of y = mx + c.

Let AB be a straight line. It makes an intercept OC on Y-axis.

Then, let OC = c and let T be the inclination of AB. Y P(x, y) B
Then, slope of the line AB(m) = tan T.
Let P(x, y) be any point on the line AB.

PM is drawn perpendicular to X-axis from P. CT N
Then, OC = MN = c, MP = y

OM = CN = x, X' Q T MX
NP = MP – MN = y – c. A
Since CN is parallel to X-axis, O
‘PCN = ‘CQM = T. Y'

From right angled triangle PCN,

or, tan T = PN
m=y– cCN

x
or, mx = y – c

? y = mx + c

? y = mx + c is the required equation of straight line in slope intercept form.

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Note :
If the line AB passes through the origin, y-intercept c = 0, the equation of the
straight line passing the origin is y = mx.

Example 6. Find the equation of straight line whose y-intercept is –4 and slope is
Solution: 5
6 .

Here, y-intercept (c) = –4

slope (m) = 5
6

Now, equation of the straight line is

y = mx + c

or, y= 5 x + (–4) or, 6y = 5x – 24
6
? 5x – 6y = 24 is the required equation of the straight line.

Example 7. Find the slope and y-intercept of line y = 6x + 5.
Solution: Here, comparing equation y = 6x + 5 with y = mx + c,

Example 8. y-intercept (c) = 5
Solution: slope (m) = 6
Find the equation of straight line passing through the origin whose
inclination is 45°.
Here, angle of inclination = 45°
slope (m) = tan 45° = 1
Equation of the straight line passing through the origin is
y = mx
or, y = 1 . x

? y = x is the required equation of the straight line.

Exercise 9.1

Very Short Questions

1. Find the equation of straight line from the following figures:

(a) Y (b) Y

M N X' O X
5 N

X' O X M A(0, –5)
Y' Y'

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Equations of Straight Lines (d) Y E

(c) Y A X X' O (5, 0) X

6

X' O

Y' B Y' F

(e) Y (f) Y B

B (0,6)

X' O 30° X 60°
A Y' O
X' X
A Y'

2. Find the equation of straight lines parallel to Y-axis passing through the
following points:

(a) (3, 5) (b) (6, 8)

3. Find the equation of straight lines parallel to X-axis and passing through the
following points:

(a) (–3, 4) (b) (6, 5)

4. (a) Find the equation of straight lines parallel to X-axis and at a distance of

(i) 6 units above X-axis (ii) 8 units below X-axis

(b) Find the equation of straight lines parallel to Y-axis and at a distance of

(i) 5 units right of Y-axis (ii) 7 units left of Y-axis

5. Find the slope of lines passing through the points:

(a) (P(4, 6) and Q(7, 10) (b) M(2, 3) and N(6, 10)

(c) R(–4, –5) and S(6, 6) (d) T(6, 4) and S(7, 8)

6. Fill slope and y-intercept of given lines in the table:

Equations Slope (m) y-intercept (c)
(a) y = 4x + 2
(b) y = 3x + 2
(c) y = 3x + 4
(d) y = – 3x + 2

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7. Find the equation of straight line under the given conditions:

Slope (m) y-intercept (c)
(a) 2 4
(b) – 2 2
(c) 4 –6
(d) 6
(e) tan 45° 0 (passing through origin)
(f) tan 60° 2
–3

Long Questions :

8. (a) Find the equation of a straight line cutting the y-intercept 2 from the axis
of y and inclined at 45° with the positive direction of X-axis.

(b) Find the equation of a straight line making an angle of 60° with the positive
direction of X-axis and cutting of an intercept 6 from the negative Y-axis.

9. (a) Find the equation of a straight line passing through the point (0, 8) which
makes an angle of 60° with positive direction of X-axis.

(b) Find the equation of a straight line which makes an angle of 30° with X-axis
and passes through the point (0, –5).

10. Derive the equation of straight lines in the slope intercept form.

1. (a) y = 5 (b) y + 5 = 0 (c) x = 6 (d) x = 5

(e) y = 1 x (f) y = 3x + 6 2.(a) x = 3 (b) x = 6
3
3. (a) y = 4 (b) y = 5 4.(a) (i) y = 6 (ii) y + 8 = 0

(b) (i) x = 5 (ii) x + 7 = 0

5. (a) 4 (b) 7 (c) 11 (d) 4
3 4 10
6. (a) m = 4, c = 2 (b) m = 3, c = 2 (c) m = 3, c = 4 (d) m = – 3, c = 2

7. (a) y = 2x + 4 (b) 2x + y = 2 (c) y = 4x – 6 (d) y = 6x

(e) y = x + 2 (f) y = 3x – 3 8.(a) y = x + 2 (b) y = 3x + 6

9. (a) y = 3x + 8 (b) x – 3y = 5 3 10. y = mx + c

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9.3 Equation of a Straight Line x y
a b
in double intercepts form : + = 1

To find the equation of a straight line in the double intercept form of x + y = 1.
Let MN be a straight line. It cuts X-axis at A and Y-axis at B such that, a b

x-intercept = OA = a Y
y-intercept = OB = b N
Then, the coordinates of A and B are respectively
(a, 0) and (0, b) respectively. B(0, b)

Let P(x, y) be any point on the line AB. Then, P(x, y)

slope of AP = y – 0 = y X' O A(a, 0) X
slope of PB = bx –– ay = xb –– ay Y' M
0 – x
–x
Since A, P, and B are collinear, we have

slope of AP = slope of PB

or, x y a = b– y
– –x
or, – xy = bx – xy – ab + ay

or, ay + bx = ab

Dividing both sides by ab, we get,

ay + bx = ab
xaaaby+a+bbyba=bx ab

or, =1
?
1 is the required equation of the straight in double intercepts form.

Worked Out Examples

Example 1. Find the equation of a straight line whose x-intercept and y-intercepts
Solution: are respectively 2 and 3.

Here, x-intercept (a) = 2

y-intercept (b) = 3

Equation of required straight line is:

or, x + y = 1 or, 3x + 2y = 1
xa + by = 1 6
2 3
? 3x + 2y = 6, is the required equation.

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Example 2. Find the x-intercept and y-intercept of a straight line 6x + 5y = 30.
Solution:
Here, 6x + 5y = 30

or, 6x + 5y = 30
30 30 30
x y
Comparing it to a + b = 1, we get

x-intercept (a) = 5

y-intercept (b) = 6

Example 3. Find the equation of a straight line which passes through the point
Solution: (4, –5) and cuts equal intercepts on the axes but opposite in sign.

Let x-intercept (a) = k then, y-intercept (b) = –k

The equation of the straight line is:

x + x = 1
a b
x x
or, k + –k = 1

or, x – y = k ............................. (i)

It passes through the point (4, 5), we get,

4–5=k or, k = – 1

Substituting the value of k in equation (i), we get,

x–y=–1 or, x – y + 1 = 0

? x – y + 1 = 0 is the required equation of the line.

Example 4. Find the length of a line intercepted between the coordinate axes if
Solution: the equation of the line is 4x + 3y = 12.

Here, equation of the given line is 4x + 3y = 12 Y

Now, 4x + 3y = 12 B(0, 4)
12 12
4x 3y
or, 12 + 12 = 1

or, x + y = 1 X' O X
3 4 A(3, 0)
x y
Comparing it to a + b = 1 Y'

x-intercept (a) = OA = 3

y-intercept (b) = OB = 4
The coordinates of A and B are (3, 0) and (0, 4).

Now, AB = (x2 – x1)2 + (y2 – y1)2 = (0 – 3)2 + (4 – 0)2
= 9 + 16 = 25 = 5 units.

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Exercise 9.2

Short Questions

1. (a) In an equation 4x + 3y = 24, put x = 0 and y = 0 successively. What do
the values obtained of x and y represent?

(b) Find the intercepts of given line 6x + 5y = 30.

(c) Write down the equation 6x – 5y – 30 = 0 in the form of x + y = 1.
a b

2. Find the equation of straight lines from the following figures:

(a) Y (b) Y

X' O X
A(7, 0)

B(0, 3)

X' O A(7, 0) B(0, –6)
Y' X Y'

(c) Y (d) Y
B(0, 7)
X' A(–6, 0)
X
O

B(0, –7) A(–3, 0)
Y' X' O X

Y'

3. Find the equation of straight lines having intercepts as given below in table:

x-intercept (a) y-intercept (b)

(a) 2 4

(b) – 6 5

(c) 4 5

(d) – 6 –3

4. Find the intercepts of the following equations:

(a) 3x – 4y = 12 (b) 7x + 5y – 35 = 0

(c) 8x + 7y – 56 = 0 (d) 4x + 5y – 20 = 0

(e) 5x + 6y = 60

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Long Questions

5. (a) Find the equation of straight line which cuts intercepts 5 from X-axis and
6 from the Y-axis.

(b) Find the equation of straight line which cuts intercepts 6 from X-axis and
–8 from Y-axis.

6. (a) Find the equation of a straight line which cuts equal intercepts on the axes
and passes through the point (7, 7).

(b) Find the equation of straight line which cuts equal intercepts on the axes
and passes through the point (5, 8).

7. (a) Find the equation of a straight line which passes through the points (2, 3)
and makes y-intercept twice as long as that on x-intercept.

(b) Find the equation of a straight line which passes through the point (2, 4)
and makes intercepts on the X-axis thrice as long that on Y-axis.

8. (a) Find the equation of straight line which cuts equal intercepts on the axes
but opposite in sign and passes through the point (5, 6).

(b) Find the equation of straight line which cuts equal intercepts on the axes
and opposite in sigh and passes through the point (–6, 8).

9. (a) Find the equation of the straight line whose portion intercepted is bisected at (2, 3).

(b) Find the equation of a straight line which passes through the point (4, 4) and
the portion of the line intercepted between the axis is bisected at the point:

10. Find the lengths of line intercepted between the coordinates axis of the following lines:

(a) 6x + 3y = 24 (b) 4x + 3y = 60

11. Find the equation of straight line in the form of x + y = 1.
a b

1. (a) x-intercept = 6, y-intercept = 8 (b) x-intercept = 5, y-intercept = 6

(c) x – y =1 2. (a) 3x + 7y = 21 (b) 6x – 7y = 42
5 6
(c) 7x + 6y + 42 = 0 (d) 7x – 3y + 21 = 0

3. (a) 2x + y = 4 (b) 5x – 6y + 30 = 0 (c) 5x + 4y = 20 (d) x + 2y + 6 = 0

4. (a) x-intercept = 4, y-intercept = –3 (b) x-intercept = 5, y-intercept = 7

(c) x-intercept = 7, y-intercept = 8 (d) x-intercept = 5, y-intercept = 4

(e) x-intercept = 12, y-intercept = 10 5.(a) 6x + 5y = 30 (b) 4x – 3y = 24

6. (a) x + y = 14 (b) x + y = 13 7.(a) 2x + y = 7 (b) x + 3y = 14

8. (a) x – y + 1 = 0 (b) x – y + 14 = 0 9.(a) 3x + 2y = 12 (b) x + y = 8

10. (a) 4 5 units (b) 25 units

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Standard forms of equations of straight line.

The following three forms of equations of a straight line are called standard forms.

(a) Slope intercept form : y = mx + c

(b) Intercepts form : x + y = 1
a b
(c) Normal form : x cos D + y sin D = p

we have already derived equations of a straight line in slope -intercept form and
double intercepts form. Here, we derive equation of a straight line in normal or
perpendicular form.

9.4 Equation of a Straight Line in Normal or
Perpendicular Form (x cos D + y sin a = p)

To find the equation of a straight line in the form of
x cos D + y sin D = p

Let a straight line MN cut the X-axis at A and Y-axis at B such that

x-intercept = OA

y-intercept = OB

Let OQ = p be perpendicular distance of line Y
MN from the origin and ‘AOQ = D. NB

Then, ‘OBQ = 90° – ‘BOQ = 90° – (90° – D) = D

From right-angled 'OAQ, Q
p
OA OA
sec D = OQ = p X' D AX
O M
? OA = OQ sec D = p sec D

Again, from right-angled triangle OBQ, Y'

cosec D = OB OpB
OQ
? OB = OQ cosec D = p cosec D

Now, equation of the straight line MN is given by,

or, pOxAsexc+DOy+B =1 D = 1

x cos y
p p cosec
y sin
or, D + p D = 1

? x cosD + y sinD = p is the required equation of a straight line in normal /

perpendicular form.

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Note :
(i) When D = 0°, then x = p, which is equation of a line parallel to Y-axis.
(ii) When D = 90°, then y = p, which is equation of a line parallel to X-axis.

Worked out Examples

Example 1. The perpendicular distance of a straight line from the origin is 4
Solution: units and the angle made by the perpendicular with X-axis is 60°.
Find the equation of the straight line.

Here, length of the perpendicular (p) = 4 Y

The angle made by perpendicular with B

X-axis (D) = 60° p=4 Q

Now, the equation of straight line is X' O 60° A
x cos D + y sin D = p X

or, x cos 60° + y sin 60° = 4 Y'

or, x × 1 + y × 3 = 4
2 2

or, x + 3y = 8

? x + 3y = 8 is the required equation of straight line.

Example 2. A straight line passes through the point (5, 5) and the inclination of
Solution: perpendicular is 45°. Find the equation of the straight line.

Here, length of perpendicular (p) is to be found.

The inclination of the perpendicular (D) = 45°

Now, the equation of required straight line is

x cos D + y sin D = p Y
B
or, x cos 45° + y sin 45° = p
Q
or, x× 1 + y × 1 = p .............. (i)
2 2 p
45° A
It passes through the point (5, 5) X' O X

5× 1 + 5 × 1 = p Y'
2 2

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or, 5 + 5 = p
2 2
5 + 5
or, p = 2

or, p = 10
2

or, p = 5 2

? p=5 2

Substituting the value of p in equation (i), we get,

x + y = 5 2
2 2
or, x + y = 5 2 × 2

= 5 × 2 = 10

? x + y = 10 is the required equation.

Exercise 9.3

Short Questions
1. Write down the equations of the straight lines from the following figures.

(a) Y (b) Y
Q
N

P 60°
O
p=7 X' X
p=4
X' 45° X
O M

Y' (d) P
Y'
(c) Y X' Y

T V
P
45°
X' 60° X Pp=6 O X
O R U
Y'
Y'

166 vedanta Excel in Additional Mathematics - Book 8

Equations of Straight Lines

2. If p is the perpendicular distance of a line from the origin and D the angle
of inclination with X-axis. Find the equation of straight under the following
conditions:

p D
(a) 1 45°
(b) 2 60°
(c) 2 30°
(d) 5 90°

Long Questions

3. (a) If the perpendicular distance of a straight line from the origin is 6 units and
the inclination of the perpendicular with X-axis is 45°. Find the equation of
the line.

(b) The length of perpendicular distance of a straight line from the origin is
3 3 units and the slope of inclination of perpendicular is 3. Find the
equation of the line.

4. (a) The length of perpendicular drawn from the origin on a straight line is 3

units and the perpendicular is inclined at an angle of 60° to the X-axis. Find

the equation of the straight line. Also, prove that the line passes through
23
the point 4, 3 .

(b) Find the equation of a straight line whose length of perpendicular drawn

from the origin on the straight line is 4 units and the perpendicular is

inclined at an angle of 60° with X-axis. Also, prove that it passes through

the point (5, 3).

5. Derive the equation of straight line in the form of x cos D + y sin D = p.

1. (a) x + y = 4 2 (b) y = 3x (c) x + 3y = 14 (d) x – y + 6 2 =0
2. (a) x + y = 2 (b) x + 3y = 4 (c) 3x + y = 4 (d) y = 5
3. (a) x + y = 6 2 (b) x + 3y = 6 3
4. (a) x + 3y = 6 (b) x + 3y = 8

vedanta Excel in Additional Mathematics - Book 8 167

Matrix

10Matrix

10.1 Introduction

Let us take three students Ajit, Bijaya, and Chandan of class 8. They got the following
marks in C. Mathematics, Science, and Optional Mathematics.

Name of Students Science Marks Obtained in Opt. Maths
60 C. Maths 93
Ajit 68 90 90
Bijaya 45 85 70
Chandan 65

The above information can be written in the form of

Science C. Maths Opt. Maths

Ajit 60 90 93
Bajaya 68 85 90
Chandan 45 65 70

This rectangular arrangements of numbers in rows and columns is called matrix.
Matrix is one of powerful tool in mathematics. It has wide application in multiple
fields such as Economics, Statistics, Engineering, Physics, etc. Matrices are generally
denoted by capital matters and their corresponding elements are denoted by small
letters.

10.2 Definition of Matrix

A rectangular arrangement of numbers in rows and columns enclosed within square
brackets or round brackets is known as matrix. Usually, a matrix is represented by a
capital letter. Its plural form is matrices.

eg A= 1 3
2 4

B= 1 2 3
4 5 6

C= –1 2
3 5

168 vedanta Excel in Additional Mathematics - Book 8

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10.3 Rows and Columns of a Matrix

As a rectangular arrangement of numbers, a matrix contains elements in the form of
rows and columns. A row is usually a horizontal line and column is a vertical line.

1 4 2 4 6
For Example : A = 2 5 ,B= 7 8 9

3 6

The matrix A has 3 rows and 2 columns and matrix B contains 2 rows and 3 columns.

10.4 Order of a Matrix

The number of rows and columns present in a matrix gives the order of a matrix.

For Example : where, R1 = first rows
R2 = second row
A= 2 4 6 R1 C1 = first column
7 9 10 R2 C2 = second column
C3 = third column
C1 C2 C3

Matrix A has 2 rows and 3 columns. So, its order is 2 × 3. It is read as 2 by 3.
Again, let us take a matrix B.

12 3 R1
R2
B= 4 5 6 R3
7 8 9

C1 C2 C3
It has 3 rows and 3 columns. Its order is 3 × 3.

Worked out Examples

Example 1. Determine the order of the following matrices.
Solution:
(a) A = –2 5 –3 4 2 3 4
6 –8 (b) B = 5 6 (c) C = 7 8 5 (d) D = [8]
2 0 2
8 1
169
(a) Here, A = –2 5
6 –8
Number of rows in matrix A = 2

Number of columns in matrix A = 2
Hence, order of A is 2 × 2.

–3 4
(b) Here. B = 5 6

82
Number of rows in matrix B = 3
Number of columns in matrix B = 2
Hence, order of B is 3 × 2.

vedanta Excel in Additional Mathematics - Book 8

Matrix

234
(c) Here, C = 7 8 5

102
Number of rows in matrix C = 3

Number of columns in matrix C = 3

Hence, order of C is 3 × 3.

(d) Here, D = [8]

Number of row in matrix D = 1
Number of column in matrix D = 1
Hence, order of D is 1 × 1.

Example 2. Construct a matrix with the order specified. (Elements can be chosen
Solution:
randomly.)

(a) A, order 2 × 2 (b) B, order 2 × 3

(c) C, order 3 × 2 (d) O, order 3 × 3

(e) E, order 1 × 2

(a) Matrix A has order 2 × 2

It means A should have 2 rows and 2 columns.

Let, A = 1 2
4 8
(b) Matrix B has order 2 × 3.

It means matrix B should have 2 rows and 3 columns.

Let, B = 6 7 8
9 10 12
(c) Matrix C has order 3 × 2

It means matrix C must have 3 rows and 2 columns.

xy
Let, C = z w

xy
(d) Matrix D has order 3 × 3.

It means matrix D must have 3 rows and 3 columns.

456
Let, D = 7 8 9

10 11 12
(e) Matrix E has order 1 × 2

It means matrix E must have 1 row and 2 columns

Let, E = [2 7]

170 vedanta Excel in Additional Mathematics - Book 8

Matrix

10.5 Types of Matrices

(i) Row matrix
A matrix having only one row is called a row matrix. It can have any number of
columns.

For Example : A = [2 4 6]

It has only one row. A is a matrix of order 1 × 3.

(ii) Column matrix

A matrix having only one column but any number of rows is called a column
matrix.

For Example : P = 4 , Q= 6
5 7
8

Here, P is a matrix of order 2 × 1, Q is a matrix of order 3 × 1.

(iii) Rectangular matrix

A matrix having unequal number of rows and columns is called a rectangular
matrix. In short, number of rows and number of columns should not be equal.

For Example : P = 1 3 –3
2 1 3
Here, matrix P has 2 rows and 3 columns which is unequal so it is rectangular

matrix. P is rectangular matrix of order 2 × 3.

2 –4
For Example : B = 6 7

53
Here, Q is a rectangular matrix of order 3 × 2.

(iv) Zero or null matrix

A matrix having each of the element zero (0) is called a null or zero matrix.

For Example : O = 0 0 0
0 0 0
This is a null matrix of order 2 × 3.

(v) Square matrix

A matrix whose number of rows and columns are equal is known as a square
matrix.

For Example : A = 2 3 , B= 6 7 8
4 5 2 3 4
9 10 12

Here, A is a square matrix of order 2 × 2 and B is a square matrix of order 3 × 3.

vedanta Excel in Additional Mathematics - Book 8 171

Matrix

Example 3. State the type of matrices given below. Also state its order.

(a) A = [2 4] (b) B= 6
–7

–4 –6 00
5 7
(c) P= (d) O = 0 0

28 00
(e) R = –5 6
Solution:
34
(a) Here, A = [2 4]

It is a row matrix. Its order is 1 × 2.

(b) Here, B = 6
–7
It is a column matrix. Its order is 2 × 1.

(c) Here, P = –4 –6
5 7
It is a square matrix. It is order is 2 × 2.

00
(d) Here, O = 0 0

00
It is a null matrix. Its order is 3 × 2.

28
(e) Here, R = –5 6

34
It is a rectangular matrix. Its order is 3 × 2.

10.6 Equal Matrices

The matrices having same order and equal corresponding elements are said to be
equal matrices.

For Example : A = 2 4 B= 2 4
6 8 6 8
Here, matrices A and B are said to be equal matrices as their order is same as well

as their corresponding elements are equal.

Example 4. Find the values of x, y, z and w if the matrices are equal.
Solution:
(a) x 7 = 8 z (b) 16 3x = 4z 6
10 y w 2 6y 1 12 2w

8
(a) Here, as the matrices are equal, their corresponding elements

are also equal.

So, x = 8, z = 7, w = 10, and y = 2

172 vedanta Excel in Additional Mathematics - Book 8

Matrix

(b) Since the matrices are equal their corresponding elements are
also equal.

So, 16 = 4z or, z = 4

3x = 6 or, x = 2

6y = 12 or, y = 2

and, 1 = 2w or, w = 1
8 16
1
? x = 2, y = 2, z = 4 and w = 16

Example 5. Find the values of x and y if x–2 8 = 2 8
Solution: 10 y+2 10 6

Here, x–2 8 = 2 8
10 y+2 10 6
Since the matrices are equal, their corresponding elements are

also equal.

So, x – 2 = 2

or, x = 2 + 2 = 4

Again, y + 2 = 6

or, y = 6 – 2

or, y = 4

Hence, x = 4 and y = 4

Example 6. Find the values of a and b if a+b 5 = 4 5 .
Solution: 2 a–b 2 1

Here, a+b 5 = 4 5
2 a–b 2 1
Equation the corresponding elements of equal matrix.

a + b = 4 ..................... (i)

a – b = 1 ...................... (ii)

Adding equation (i) and (ii), we get,
a+b=4

a – b=1

2a = 5

? a = 5
2

Put the value of a in equation (i), we get,

b = 4 – a = 4 – 5
2

vedanta Excel in Additional Mathematics - Book 8 173

Matrix or, b = 8 – 5 = 3
2 2
5 3
? a = 2 and b = 2

Exercise 10.1

Short Questions :

1. Define the following matrices

(a) Row matrix (b) Equal matrices

(c) Column matrix (d) Null matrix

(e) Square matrices (f) Rectangular matrices

2. Determine the order of the following matrices:

(a) 2 4 (b) [8 2] (c) 2
6 7 –4

(d) [2 6 7] (e) [12] (f) 2 4 –8
62 8
3. Construct the following matrices with the orders specified. (Elements may be

chosen randomly.)

(a) A1 × 1 (b) B2 × 1 (c) C1 × 2

(d) D2 × 2 (e) E2 × 3 (f) F3 × 2

(g) G1 × 3 (h) H3 × 1 (i) I3 × 3

4. State the type and order of the following matrices:

(a) [6] (b) 2 (c) [3 7 8]
4

(d) 2 2 00 (f) 234
4 5 (e) 0 0 678

00

5. Find the values of a, b, c, and d from the following pair of equal matrices:

(a) a b = 2 4 (b) 12 2a = 6c 6
c d 6 3 4b 12 8 4d
6. Find the values of x and y if -

(a) x–4 = 3 (b) 5x + 1 2 = 21 2
y+3 6 4 3y – 2 4 13

(c) x–4 2 = 2x – 4 2 (d) x+4 = 10
4 8 4 y+4 x–y 6

(e) x+y 3 = 7 3 (f) x–y 4 = 6 4
4 x–y 4 3 20 x+y 20 8

174 vedanta Excel in Additional Mathematics - Book 8

Matrix

a11 a12 a13 123

7. If a21 a22 a23 = 7 8 9 , find values of a11, a12, a13, ....., a33.
a31 a32 a33 10 11 12

2. (a) 2 × 2 (b) 1 × 2 (c) 2 × 1 (d) 1 × 3

(e) 1 × 1 (f) 2 × 3 3. Show to your teacher.

4. (a) square matrix, 1 × 1 (b) column matrix, 2 × 1

(c) row matrix, 1 × 3 (d) square matrix, 2 × 2

(e) null matrix, 3 × 2 (f) rectangular matrix, 2 × 3

5. (a) a = 2, b = 4, c = 6, d = 3 (b) a = 3, b = 2, c = 2, d = 3

6. (a) x = 7, y = 3 (b) x = 4, y = 5 (c) x = 0, y = 4 (d) x = 6, y = 0

(e) x = 5, y = 2 (f) x = 7, y = 1

7. a11 = 1, a12 = 2, a13 = 3 a21 = 7, a22 = 8, a23 = 9
a31 = 10, a32 = 11, a33 = 12

10.7 Operation of Matrices - Addition and Subtraction

Addition, subtraction, and multiplication are operations of matrices. Here, we study
only Addition and Subtraction of two matrices.

Let A= 1 2 and B = 2 4
4 7 3 2
Here, as the order of both the matrices is 2 × 2. We can add then.

So, A + B = 1 2 + 2 4
4 7 3 2

= 1+2 2+4
4+3 7+2

= 3 6
7 9

Two or more matrices of same orders can be added.

Again, B – A = ?

Now, B – A = 2 4 – 1 2
3 2 4 7

= 2–1 4–2
3–4 2–7

= 1 2
–1 –5

vedanta Excel in Additional Mathematics - Book 8 175

Matrix

Worked Out Examples

Example 1. Can the following matrices be added and subtracted? Give reasons for
Solution:
your answers.

(a) A = [4 5], B = [7 8] (b) P= 4 2 , Q = 4 3
3 7 3 4
2 3 2
(c) X= 2 4 , Y = 1

(a) Yes, they can be added and subtracted as their order is the same.

(b) Yes, they can be added and subtracted as their order is the same.

(c) No, they cannot be added or subtracted as orders are different.

Example 2. Perform as indicated. A = 2 ,B= 6
3 7
Find A + B, A – B, B – A, B + A.

Solution: A+B= 2 + 6 = 2+6 = 8
3 7 3+7 10

A–B= 2 – 6 = 2–6 = –4
3 7 3–7 –4

B–A= 6 – 2 = 6–2 = 4
7 3 7–3 4

B+A= 6 + 2 = 6+2 = 8
7 3 7+3 10

10.8 Multiplication of Matrix by a Scalar

Let A = 2 4
3 6
Multiplying all the elements of matrix A by a scalar quantity (number) is called

scalar multiplication.

For example : If A = 2 4 , find 2A.
3 6

Solution: Here, A = 2 4
3 6

? 2A = 2 2 4 = 2×2 2×4 = 4 8
3 6 2×3 2×6 6 10

Example 3. P= 1 5 , Q= 4 7 , Find 2P – 3Q.
Solution: 2 6 5 8

2P – 3Q =2 1 5 –3 4 7
2 6 5 8

= 2 10 – 12 21
4 12 15 24

176 vedanta Excel in Additional Mathematics - Book 8

Matrix

= 2 – 12 10 – 21 = – 10 – 11
4 – 15 12 – 24 – 11 – 12

10.9 Transpose of a Matrix

Let A = 2 5 be a matrix then a matrix obtained by simply interchanging the
6 8
elements of rows and columns is called transpose of a matrix. It is denoted by AT or

A' or At. 2 6
5 8
Here, AT =

In the example it can be clearly seen that the elements of row have been changed

into columns and vice versa.

Example 4. If A = 1 2 and B = 6 7
3 4 8 9
Find (a) A' (b) B' (c) A' + B' (d) (A + B)'

Solution: Here,

(a) A= 1 2 A' = 1 3
3 4 2 4

(b) B= 6 7 B' = 6 8
8 9 7 9

(c) A' + B' = 1 3 + 6 8
2 4 7 9

= 1+6 3+8 = 7 11
2+7 4+9 9 13

(d) (A + B)'

For (A + B)', let's find out (A + B) first,

A+B = 1 2 + 6 7
3 4 8 9

= 1+6 2+7 = 7 9
3+8 4+9 11 13

So, (A + B)' = 7 11
9 13

Example 5. If A = 1 2 show that (A')' = A.
Solution: 3 4

Here, A = 1 2
3 4

A' = 1 3
2 4

(A')' = 1 2 =A Proved.
3 4

vedanta Excel in Additional Mathematics - Book 8 177

Matrix

Exercise 10.2

Short Questions :

1. Which of the following matrices can be added?

P= 1 2 , Q= 1 2 3 , R= 1 4 , S= 2 6
3 4 4 5 6 6 7 5 4

1 2 U= 1 4 6 , 4 5
T= 3 4, 2 4 8 V= 6 7
9
6 7 8

2. Perform as indicated:

(a) A = [1 2], B = [3 2], find A + B and A – B.

(b) P= 3 , Q = 7 , find Q – P and Q + P
6 2

(c) A= 1 4 , B= 2 4 , find A + B, B – A and A – B.
6 4 3 3

3. Find the transpose of the following matrices:

(a) P= 1 (b) Q= 2 4 (c) R = [4 2]
2 6 8

(d) S= 5 6 7 235
8 9 10 (e) T = 0 2 1

321

4. (a) If A = 2 4 and B = 2 5 , find 4A – 3B.
6 5 6 4

(b) If A = 3 4 and B = 8 2 ,
6 5 1 6

Find (i) 2A + 3B (ii) 1 A + 1 B.
2 2

5. (a) A= 3 5 ,B= 1 0 ,
6 7 2 4

Find (i) A' (ii) B' (iii) A' + B' (iv) (A + B)'

(b) If A = 2 3 , B = 2 1 ,
6 7 5 6

Find (i) A' (ii) B' (iii) A' – B' (iv) (A – B)'

6. (a) If A = 4 5 , verify that (A')' = A.
6 7

2 6
(b) If A = 3 7 , prove that (A')' = A

4 9

178 vedanta Excel in Additional Mathematics - Book 8

Matrix

2. (a) [4 4], [–2 0] (b) 4 , 10 (c) 3 8 , 1 0 , –1 0
–4 8 9 7 –3 –1 3 1

3. (a) [1 1] (b) 2 6 (c) 4 5 8 2 0 3
4 8 2 (d) 6 9 (e) 3 2 2
10 1 1
7 5

2 1 30 14 11 3
6 8 5 29
4. (a) (b) (i) (ii) 2 11
7 2

2

5. (a) (i) 3 6 (ii) 1 2 (iii) 4 8 (iv) 4 8
5 7 0 4 5 11 5 11

(b) (i) 2 6 (ii) 2 5 (iii) 0 1 (iv) 0 1
3 7 1 6 2 1 2 1

10.10 Properties of Matrix Addition

Addition of matrices satisfies the following properties.

Let A, B, and C be of same order matrices.

(a) Closure Property :

If A and B are two matrices of same orders, their sum A + B is also of same
order as that of A and B.

Let A = 1 2 and B = 2 6 be two matrices of order 2 × 2.
3 4 3 5

Then A + B = 1 2 + 2 6
3 4 3 5

= 1+2 2+6
3+3 4+5

= 3 8 , which is of order 2 × 2.
6 9

Hence, matrix addition satisfies closure property.

(b) Commutative property :

If A and B are two matrices of same orders, A+B = B+A.

Let A= 1 2 and B = 1 4
6 7 3 6
1 2 1 4
Then, A + B = 6 7 + 3 6

= 2 6
9 13

vedanta Excel in Additional Mathematics - Book 8 179

Matrix

and B+A= 1 4 + 12
3 6 67
2 6 26
= 9 13 9 13

? A+B=B+A

Hence, the matrices addition satisfies commutative properly.

(c) Associative properly :

If A, B, and C are three matrices of same orders,

then (a + B) + C= A + (B + C).

Let A= 1 2 ,B= 2 1 and C = 2 4
3 4 5 6 6 2

Now, A + B = 1 2 + 21
3 4 56

= 3 3
8 10

(A + B) + C = 3 3 + 2 4
8 10 6 2

= 5 7
14 12

Again, B + C = 2 1 + 2 4
5 6 6 2

= 4 5
11 8

A + (B + C) = 1 2 + 4 5
3 4 11 8

= 5 7
14 12

? (A + B) + C = A + (B + C)

Hence, matrix addition satisfies associative properly.

(d) Existence of Identity Element :

If A is any matrix and O be a null matrix of same order as that of A, then

A+O=O+A=A

Let, A= 1 6 and O = 0 0
3 7 0 0

Now, A + O = 1 6 + 0 0
3 7 0 0

= 1 6
3 7

and O+A = 0 0 + 1 6 = 1 6
0 0 3 7 3 7

180 vedanta Excel in Additional Mathematics - Book 8

Matrix

? A+O=O+A=A

Here, O is called additive identity of matrix A. Hence, null matrix is additive
identity .

(e) Existence of Additive Inverse : If A is a matrix of any order, then there exists
another matrix (–A) of same order as that of A.

Then A + (–A) = O
where O is a null matrix of same order as that of A .

Let A= 2 5 and (–A) = –2 –5
7 8 –7 –8
2–2 5–5
= 7–7 8–8

= 0 0 =O
0 0
Here, (–A) is called additive inverse of matrix A.

Worked Out Examples

Example 1. Let P = 1 2 3 and Q = 3 2 4 , verify that P + Q = Q + P.
Solution: 4 5 6 1 2 5

Example 2. Here, P = 1 2 3 and Q = 3 2 4
Solution: 4 5 6 1 2 5

P+Q = 1 2 3 + 3 2 4
4 5 6 1 2 5

= 1+4 2+2 3+4
+1 5+2 6+5

= 4 4 7
5 7 11

Also, Q + P = 3 2 4 + 1 2 3
1 2 5 4 5 6

= 3+1 2+2 4+3
1+4 2+5 5+6

= 4 4 7
5 7 11

? P + Q = Q + P Proved.

Find the additive inverse of A = 4 5 .
9 10

Here, A = 4 5 , Let B = a b be additive inverse of A.
9 10 c d

Then, by definition, we get,

vedanta Excel in Additional Mathematics - Book 8 181

Matrix

A + B = O [where O is a null matrix of order 2 × 2]

or, 4 5 + a b = 0 0
9 10 c d 0 0

or, 4+a 5+b = 0 0
9+c 10 + d 0 0

Equating the corresponding elements of equal matrices, we get,

4+a=0 Ÿa=–4

5+b=0 Ÿb=–5

9+c=0 Ÿc=–9

10 + d = 0 Ÿ d = – 10

? B= –4 –5
–9 – 10

Example 3. If P = 2 3 , find the matrix Q such that P + Q = I, where I is a unit
Solution: 4 5
matrix of order 2 × 2.
Example 4.
Solution: Here, P + Q = I

or, Q = I – P

= 1 0 – 2 3
0 1 4 5

= 1–2 0–3
0–4 1–5

= –1 –3
–4 –4

? Q= –1 –3
–4 –4

If 6 5 + 4 y = z 6 , find the values of x, y, z, and p.
2 x 2 1 p 5

Here, 6 5 + 4 y = z 6
2 x 2 1 p 5

or, 6+4 5+y = z 6
2+2 x+1 p 5

Equating the corresponding elements, we get,

6+4=z Ÿ z = 10

5+y=6 Ÿy=6–5=1

2+2=p Ÿp=4

x+1=5 Ÿx=5–1=4

? x = 4, y = 1, z = 10 and p = 4

182 vedanta Excel in Additional Mathematics - Book 8

Matrix

Exercise 10.3

Short Questions :

1. Let P = 1 2 , Q = 4 5 , R = 2 1 , then verify the following :
3 4 2 1 –2 –3

(a) P + Q = Q + P

(b) P + R = R + P

(c) Q + R = R + Q

2. Find the additive inverse of the following matrices:

(a) P= 1 2 (b) Q= 2 3
3 4 4 5

(c) R= 2 5 (d) S= –2 –3
–4 –6 –4 –5

3. Find the matrix Q under the following conditions if P = 4 5 :
7 3

(a) P + Q = I

(b) P + Q = 2I

(c) P + Q = 1 4 6
2 2 2

4. If P = 2 3 and Q = 2 4 , then verify the following :
4 5 4 1

(a) (P + Q)' = Q' + P' (b) 2P' + 2Q' = 2(P' + Q')

5. If P = 2 3 , then find the additions identity of matrix P.
6 7

Long Questions :

6. (a) Let P = 1 2 , Q = 2 4 and R = 2 –2 then verify that
3 4 6 7 –3 1

P + (Q + R) = (P + Q) + R

(b) Let X = –2 –3 , Y = 2 4 and Z = –2 –3 then verify that
2 4 6 7 6 7

(X + Y) + Z = X + (Y + Z)

7. (a) If 4p – 1 8 + 7 r = 1 0 , find the values of p, q, r and s.
q+3 4 6 s 0 1

(b) If a 2 + 4 x = 6 3 , find the value of a, b, x and y.
b 5 8 y 10 5

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Matrix

8. (a) If A = 5 6 and B = 2 4 , find 2A + 3B.
8 4 0 3

(b) If P = 5 8 and Q = 2 5 and k = 3, then verify that
–1 0 6 4

(i) P + Q = Q + P

(ii) k(P + Q) = kP + kQ

9. If P = 1 2 , Q = 3 2 and R = 2 1 , then verify that
3 4 1 2 4 3

(a) (P + Q) + R = P + (Q + R)

(b) (P + Q + R)' = P' + Q' + R'

10. (a) If P + Q = 5 2 and P – Q = 3 6 find the matrices P and Q.
0 9 0 –1

(b) If A + B = 5 7 and A – B = -3 -3 find the matrices A and B.
10 12 -2 -2

11. If M = 2 4 6 and N = 1 2 4 , is (M – N)' = M' – N'? Justify your
3 2 1 2 0 3

answer.

2. (a) –1 –2 (b) –2 –3 (c) –2 –5 (d) 2 3
–3 –4 –4 –5 4 6 4 5

3. (a) –3 –5 (b) –2 –5 (c) –2 –2 5. 0 0
–7 –2 –7 –1 –6 –2 0 0

7. (a) p = – 54, q = – 9, r = – 8, s = – 3 (b) a = 2, b = 2, x = – 1, y = 0

8. (a) 19 26
24 18

10. (a) P = 4 4 ,Q= 1 -2 (b) A = 1 2 ,B= 4 5
0 4 0 5 4 5 6 7

11. Yes.

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Vectors

11Vectors

11.1 Introduction

In our daily life we use many types of physical quantities which can be measured
by using numbers with units. Some physical quantities require magnitudes only
whereas some physical quantities require both magnitudes and directions.

For Example :

(a) A car travelled 10 km north from Kathmandu.

Here, 10 km is magnitude and north is direction. Hence, 10 km north is a vector
quantity.

(b) Simon suffered from fever whose temperature was 102°F.

Here, temperature does not need direction. It has only magnitude. Hence,
temperature is a scalar quantity.

The physical quantities which need magnitude only are called scalars. Examples
of scalars are distance, area, volume, speed, temperature, etc.

The physical quantities which need both the magnitude and direction are called
vectors. Examples of vectors are displacement velocity, force, acceleration, etc.

11.2 Directed Line Segment

A 10 cm B

In figure, we can see that AB is a line segment of 10 cm length. Along with it, its
starting point A (initial point) and the ending point B (terminal point) are known,
giving us the idea about the line segment and its direction. This is called vector AB.

Here, initial point is A

terminal point is B

It is written as AoB and read as vector AB,

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Vectors

If BoA is written, it notifies that the initial point is B and the terminal point is A. In
figure,

B 10 cm A

Usually, vectors are written with both capital letters with an arrow head on top or
small alphabets with an arrow head.

For Example : AoB, CoD, MoN, oa , ob , oc , etc.

Note : x2 – x1
If A(x1, y1) and B(x2, y2) are two points, then AoB = y2 – y1

11.3 Components of a Vector

Let AoB be a vector. A vector always has components or can be expressed in terms of

components. Y

For Example : AoB = 4 B
5 5

This explains us that AoB is a vector which has 4 units as

x-component and 5 units as y-component. This can be

represented in the graph as shown. X' A 4 X
O 6
Here, to reach B from A, we have to move 4 units to the right Y
and 5 units upward. P Y'

So, AoB = 4 5
5

Here, we are moving 5 units on the right and 6 units

down.

As going down is referred to negative. PoQ = 5 X' QX
–6
O
So, if we have components, we can represent the Y'

vector in a graph paper. It is also not important

where the initial point is to be kept. Whenever we

keep the initial point, the terminal point can be obtained.

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11.4 Magnitude of a Vector

The magnitude of a vector means the length of the vector. It is always a positive
value as if refers to length. The magnitude of any vector is given by

(x-component)2+(y-component)2

If AoB is a vector, its magnitude is denoted by |AoB|. If AoB = 4
3
Then, |AoB| = 42 + 32 = 16 + 9 = 25 = 5 units.

Note :

If oa = x , then its magnitude is denoted by |oa | (read as modulus of oa )
y
|oa | = (x-component)2+(y-component)2 = x2 + y2

If magnitude of a vector is unity, then it is called unit vector.

Worked Out Examples

Example 1. Find the magnitude of AoB = 6
Solution: 8
Here, AoB = 6
8
x-component = 6

y-component = 8

Magnitude of AoB, |AoB| = (x-component)2+(y-component)2

= 62 + 82 = 36 + 64 = 100 = 10 units.

Example 2. From the vector given below, find the initial point and terminal
Solution:
point. AoB and magnitude of AoB. B

Here, initial point is A 12

terminal point is B A5

AoB = 5
12
As we have to move 5 units right and 12 units up to reach B from A.

x-component = 5 , y- component = 12

|AoB| = (x-component)2+(y-component)2

= 52 + 122 = 25 + 144 = 169 = 13 units.

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Vectors Types of Vector

9.5

Column vector :

A vector whose components are arranged in form of a column is known as column

vector. For example: AoB = 1 , AoB = 4 etc.
2 5

Simply, AoB = x-component = x
y-component y

Row Vector

A vector whose components are written in row is called a row vector.

For example : AoB = (4, 5), AoB = (6 7).

Simply, AoB = (x- component, y-component). Z A
B
Position Vector z a

A vector whose initial point is at origin and terminal P 0
point is other than origin is called a position vector. It is P b
denoted by OoA or oa , OoB or ob , OoP or op .

Equal Vector

Two vectors are said to be equal if they have same magnitude and same direction.

Example, oa = 1 and ob = 1 are equal vectors.
2 2

Unit Vector

A vector whose magnitude is 1 unit is known as unit vector.

For example : OoA = 1 , OoB = 0 , OoC = 1
0 1 2
1
2

Example 3. Show that OoP = 1 is a unit vector.
0

Solution: Here, x-component = 1, y-component = 0

|OoP| = (x-component)2+(y-component)2

= 12 + 02 = 1 = 1 unit

? OoP is a unit vector.

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Vectors

Null vector

A vector whose magnitude is zero is called a null vector.

For example : AoA = 0
0

11.6 Operations of Vectors

Addition of two vectors

The vectors can be added by just adding their corresponding components.

Example 4. If oa = 4 , ob = 6 , find oa + ob .
Solution: 5 7
Here, oa + ob 4 6
= 5 + 7

= 4+6 = 10
5+7 12

Subtraction of two vectors

The vectors can be subtracted by just subtracting the corresponding components.

Example 5. If oa = 8 , ob = 5 , find oa – ob .
Solution: 3 2
Here, oa – ob 8 5
= 3 – 2

= 8–5 = 3
3–2 1

Scalar Multiplication

Let oa = 2 . Then multiplying all the elements of oa by a scalar is known as scalar
3

multiplication.

Let oa = a ,
b

koa = k a = ka
b kb

Example 6. If oa = 4 , find 2 oa .
Solution: 5

Here, oa = 4
5

2 oa = 2 4 = 8
5 10

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Vectors

Example 7. Find the vector AoB when the initial point and the terminal points are
Solution: given A(2, 4) and B(7, 8) respectively. Find its magnitude.

Here, A = (2, 4) = (x1, y1)

B = (7, 8) = (x2, y2)

Here, AoB = x2 – x1 = 7–2 = 5
y2 – y1 8–4 4
Now, |AoB| = (x-component)2+(y-component)2

= 52 + 42 = 25 + 16 = 41 units.

Example 8. Given P(3, 3), Q(7, 5), R(5, 5) and S(9, 7) are four points. Show that
Solution: PoQ = RoS.

Here, P(3, 3), Q(7, 5), R(5, 5), and S(9, 7) are four points.

PoQ = x2 – x1 = 7–3 = 4
y2 – y1 5–3 2
RoS = x2 – x1 9–5 4
y2 – y1 = 7–5 = 2

Hence, PoQ = RoS Proved.

Example 9. If A is (6, 4) and B is (x, y), and AoB = 8 . Find the values of x and y.
9
Solution: Here, A = (6, 4), B = (x, y)

But AoB = 6
4

So, x–6 = 8
y–4 9
Equating the corresponding components

x–6=8 or, x = 8 + 6 = 14

y–4=9 or, y = 9 + 4 = 13

? x = 14 and y = 13

Exercise 11

Short Questions :

1. Define the following vectors with an example of each

(a) Position vector (b) Row vector

(c) Column vector (d) Unit vector

(e) Null vector

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2. Determine PoQ from the graphs given below.

(a) Y (b) Y
PQ
Q

X' P X'

(c) Y' (d) Y'
X' X'
Y Y
Q Q

P P

Y' Y'
3. Determine AoB from the following where the initial and the end points are given?

(a) Y (b) Y
A(7, 8)
B(7, 9)

A(3, 4) O X
X' X'
X
O
B(-2,-2)
Y'
Y'
(c) Y
(d) Y
B(–3, 8)
A(-2, 4)

A(4, 2) X' O X
X B(6, -5)
X' O
191
Y' Y'

4. Present the following vectors in a squared grid paper.

(a) AoB = 2 (b) CoD = 6
4 0

vedanta Excel in Additional Mathematics - Book 8

Vectors

(c) EoF = –4 (d) GoH = 2
5 5

(e) oIJ = 3 (f) KoL = 7
–4 0

(g) MoN = 5 (h) PoQ = 2
6 4

5. From the vectors determined in question no. 3, find the magnitudes of all the

vectors.

6. (a) Prove that OoA = (0, 1) is a unit vector.

(b) Show that AoB = 3 , 4 is a unit vector.
5 5

7. Find the sum of following vectors.

(a) PoQ = 2 and RoS = 7
6 8

(b) MoN = 3 and PoT = 6
2 4

8. From the following vector oa and ob , find oa – ob .

(a) oa = 4 and ob = 2
6 3

(b) oa = 2 and ob = 6
4 2

9. If oa = 4 , ob = 7 , oc = 2 , find the following:
5 8 1
(a) oa + ob + oc (b) oa + 2ob + 5oc (c) oa – ob + oc

(d) oa + ob – oc (e) oa + 3ob + oc

10. (a) Given that A(2, 7), B(5, 2), C(2, 5), and D(5, 0) are four points.

Show that AoB = CoD

(b) If A(2, 3), B(4, 7), C(5, 1) and D(7, 5), prove that AoB = CoD.

11. (a) If A(x, 2), B(6, y), and AoB = 6 , find the values of x and y.
4
(b) If A(–5, –5), B(x, y), and AoB = (9, 5), find the values of x and y.

12. If oa = 3 , ob = 6 , then find |oa + ob | and |oa – ob |.
4 4
13. (a) If A(–2, –5), B(4, 5) and C(4, – 5) are four points, find AoB, BoC and AoC.

Verify that 'ABC is a right-angled triangle. Also draw 'ABC in graph.

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(b) If P(3, 3), Q(9, 0) and R(12, 21) are from points, find PoQ, QoR and PoR. Verify
that 'PQR is a right-angled triangle. Also draw 'PQR in graph.

14. If P(4, 6), Q(9, 0), R(14, 7) and S(9, 13) are four points, plot quadrilateral PQRS
in graph is /PoQ/ = /SoR/ ? Is PQRS is a parallelogram? Justify your answer.

2. (a) 4 (b) 4 (c) –4 (d) 0
4 0 4 4

3. (a) 4 (b) –9 (c) –7 (d) 8
5 – 10 6 –9

4. Show to your teacher

5. 41 units (b) 181 units (c) 85 units (d) 145 units

7. (a) 9 (b) 9
14 6

8. (a) 2 (b) –4
3 2

9. (a) 13 (b) 28 (c) –1
14 26 –2

(d) 9 (e) 27
12 30

10. (a) x = 0, y = 6 (b) x = 4, y = 0

11. (a) 145 units (b) 3 units

14. Yes.

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Transformations

12Transformations

12.1 Introduction

A process in which an object changes its M

size, shape, orientation, or position is called A P A'

a transformation. A transformation forms

images which are congruent or similar to given

objects. Transformation is also called special R
type of mapping which maps each point of a C C'

plane with a definite point in the same plane. B Q B'
In transformation, each point of a plane has

exactly one image point and each image point N

has exactly one pre-image point in the same plane. Hence, transformation is a one

to one mapping from a plane onto itself.

If T is a transformation and A' is the image of point A under T,

then it is denoted by T(A) = A' or T : A o A' or simply A o A'.

If another transformation brings the image A' back to the object (pre-image) point A,
such a transformation is called the inverse of the transformation T and it is denoted
by T–1.

Transformation sometimes leaves certain points unchanged. These points are said
to be invariant. For example, in a rotations centre of rotation is invariant point. They
are left unchanged by the transformation such as distance, angle, parallelism, etc.
they are also described as invariant.

Thus, the rule under which the position or size (or both) of an object (or a geometrical
figure) may be changed is known as transformation.

There are four fundamental transformations:

(i) Reflection (ii) Rotation

(iii) Translation (or Displacement) (iv) Enlargement

Here, we shall discuss about reflection, rotation, and translation.

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12.2 Reflection

Reflection is a rule which shifts an object to the image of R Q

the same form, each point being at equal distance from P

a fixed line. The fixed line is called line of reflection. P' B

The line of reflection is also called the axis of reflection. A Q'

In the given figure, 'PQR is reflected by the axis of

reflection AB to form the image 'P'Q'R'.

Properties of reflection R'

(i) The geometrical figure and its image are at equal distance from the axis of
reflection.

(ii) The areas of the geometrical figure and its image are equal.

(iii) The appearance of the image of a figure is opposite to the figure.

Now, let’s learn to draw the image of a geometrical figure reflected about an axis of

reflection. PR

In the figure, PQR is a triangle. MN is the axis of reflection. Q

(i) Draw perpendiculars PA, QB and RC from each vertex M A B C N
of 'PQR on the axis of reflection. Q'

(ii) Produce PA, QB, and RC so that PA = P'A, QB = Q'B P' R'
and RC = R'C.

(iii) Join P', Q' and R' by using ruler.

Thus, 'P'Q'R' is the image of 'PQR.

12.3 Use of Coordinates in Reflection

Let’s learn to find the coordinates of the images of geometrical figures formed due
to the reflection about x-axis and y-axis. Here, x-axis and y-axis are the axes of
reflection.

(i) Reflection on X-axis

Study the following illustrations and learn to find the coordinates of the image
of a point in different quadrants when x-axis is the axis of reflection.

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Transformations

The point A is in The point A is in The point A is in The point A is in
the 1st quadrant. the 2nd quadrant. the 3rd quadrant. the 4th quadrant.

Y Y Y Y

A(2,4) A'(-2,3) A'(4,4)
OX
A(-2,1)
A(4,–4)
X' O X X' O X X' O X X'

A'(-2,-1)

A'(2,-4) A(-2,-3)

Y' Y' Y' Y'
A(2, 4) o A'(2, –4) A(–2, 1)o A'(–2, –1) A(–2, –3)o A'(–2, 3) A(4, – 4)o A'(4, 4)
? P(x, y) o P'(x, –y) ? P(–x, y)o P'(–x, –y) ? P(–x, –y)o P'(–x, y) ?P (x, – y)o P'(x, y)

Only the sign of Only the sign of Only the sign of Only the sign of
y-coordinate is y-coordinate is y-coordinate is y-coordinate is

changed changed changed changed

From the above illustrations, it is clear that when a figure is reflected about x-axis,
and x-coordinate of the image of vertex remains the same and the sign of y-coordinate
of the image is changed.

Worked Out Examples

Example 1. A(3, 6), B(– 2, 4) and C(5, 1) are the vertices of ∆ABC. Find the
Solution:
coordinates of its image when it is reflected about X-axis. Also draw

the graphs of ∆ABC and its image. Y

Here, A(3, 6), B(– 2, 4), and C(5, 1) are the B(-2,4) A(3,6)
vertices of 'ABC. When 'ABC is
C(5,1)
reflected about X-axis, X' O Y
we have, P(x, y) X-axis P'(x, –y)
C'(5,-1)

A(3, 6) X-axis A'(3, – 6) B'(-2,-4) A'(3,-6)
B(– 2, 4) X-axis B'(– 2, – 4)
Y'

C(5, 1) X-axis C'(5, –1)

? A'(3, – 6), B'(– 2, – 4) and C'(5, – 1) are the vertices of the image
of 'ABC.

'ABC and its image 'A'B'C' are plotted in graph.

(ii) Reflection on Y-axis

Study the following illustrations and learn to find the coordinates of the image
of point in different quadrants when y-axis is the axis of reflection.

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Transformations

The point A is in The point A is in The point A is in The point A is in
the 1st quadrant. the 2nd quadrant. the 3rd quadrant. the 4th quadrant.

Y Y Y Y

A'(-4,1) A(4,1) A(-3,2) A'(3,2) X' O X
X' O X
X' O X A(-4,-2) A'(4,-2) X' O X

A'(-5,-1) A(5,-1)

Y' Y' Y' Y'

A(4, 1) o A'(– 4, 1) A(–3, 2) o A'(3, 2) A(–4, –2) o A'(4, –2) A(5, –1)o A'(–5, –1)

?P(x, y) o P'(– x, y) ? P(– x, y) o P'(x, y) ? P(–x,–y)o P'(x, –y) ? P(x,–y)o P'(–x, –y)

Only the sign of Only the sign of Only the sign of Only the sign of
x-coordinate is x-coordinate is x-coordinate is x-coordinate is

changed changed changed changed

Fromtheaboveillustrations,itisclearthatwhenafigureisreflectedabouty-axis,and the
y-coordinate of the image of vertex remains the same. The sign of x-coordinate of
the image is changed.

Example 2. M(0, – 2), N(6, – 4) and R(2, 5) are the vertices of ∆MNR. Find the

Solution: coordinates of its image when it is reflected about Y-axis. Also draw the

graphs of ∆MNR and its image. Y

Here, M(0, – 2), N(6, – 4), and R(2, 5) are R'(-2,5) R(2,5)
the vertices of 'MNR. When is 'MNR

reflected on Y-axis, we get, X' O M(0,-2) X
We have, P(x, y) Y-axis P'(–x, y)
M'(0,-2)
M(0, – 2) Y-axis M'(0, – 2)
N(6, – 4) Y-axis N'(– 6, – 4) N'(-6,-4) N(6,-4)
R(2, 5) Y-axis R'(– 2, 5)
Y'

? M'(0, – 2), N'(– 6, – 4) and R'(– 2, 5) are the vertices of the image of 'MNR.

'MNR and 'M'N'R' are plotted on graph.

Exercise 12.1

Short Questions :

1. Draw the image of the following figures about MN as axis of reflection.

(a) Q (b) A B

M

PR C
MN D
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N 197

Transformations

(c) N (d) P N

U

QR

M WM
V

2. Write the coordinates of images of given points.

(a) Axis of reflection is X-axis.

(i) P(x, y) (ii) P(–x, y) (iii) P(–x, –y) (iv) P(x, – y)

A(4, 5) A(–2, 5) A(–4, –5) A(4, –5)

B(6, 7) B(–4, 5) B(–6, 7) B(7, –4)

(b) Axis of reflection is Y-axis.

(i) P(x, y) (ii) P(–x, y) (iii) P(–x, –y) (iv) P(x, – y)

A(2, 4) A(–2, 4) A(–4, –5) A(4, –6)

B(5, 6) B(–7, 5) B(–6, –7) B(8, –7)

3. Copy the following figures in your own graph papers and draw their images
under the reflection about X-axis. Also, write the coordinates of the vertices of
images.

(a) Y (b) Y (c) Y
B
P R F
X' D
EA CO F X X' U QO PX
Q X X' R
O

DE V W N
M

Y' Y' X'

4. Copy the following figures in your own graph papers and draw their images

under the reflection about Y-axis. Also, write the coordinates of the vertices of

images.

(a) Y (b) Y (c) Y

PM U R
X X'
X' Q RN PS Q X
W X X' O P C
O
OD

U V R
Y' Q
A B
Y' Y'

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Transformations

5. Plot the following points in graph papers and draw triangles joining the points
in order. Draw the image of each triangle under the reflection about X-axis.
Write the coordinates of the vertices of images.

(a) A(2, 5), B(4, 8), C(3, 1)

(b) A (–1, 5), B(4, 5), C(–6, 8)

(c) U(–2, –3), V(–5, –4), R(2, 3)

(d) M(4, –4), N(3, 5), P(6, –2)

6. Plot the following points in graph papers and draw triangles joining the points
in order. Draw the image of each triangle under the reflection about Y-axis.
Write the coordinates of the vertices of images.

(a) A(2, 4), B(5, 8), C(9, 2)

(b) M(–2, 4), N(–4, –5), P(–6, 6)

(c) P(2, –8), Q(4, –3), R(0, –5)

(d) U(2, 4), V(5, –6), W(5, 7)

7. (a) A(2, –4), B(– 4, 8), and C(4, 3) are the vertices of 'ABC. Find the coordinates
of its image under the reflection about X-axis. Draw object and image on
same graph.

(b) M(6, 7), N(8, 4), and P(0, 5) are the vertices of 'MNP. Find the coordinates
of its image under the reflection about X-axis. Draw object and image on
same graph.

8. (a) A(5, 3), B(4, 7), and C(10, 4) are the vertices of ∆ABC. Find the coordinates
of the vertices of ∆A'B'C' under the reflection about Y-axis. Draw object and
image on same graph.

(b) P(8, –3), Q(–3, –4), and R(5, 7) are then vertices of ∆PQR. Find the
coordinates of the vertices of ∆P'Q'R' under the reflection about Y-axis.
Draw object and image on same graph.

1. - 4. Show to your teacher. (b) A'(–1, –5), B'(4, –5), C'(–6, –8)
5. (a) A'(2, –5), B'(4, –8), C'(3, –1) (d) M'(4, 4), N'(3, –5), P'(6, 2)
(b) M'(2, 4), N'(4, –5), P'(6, 6)
(c) U'(–2, 3), V'(–5, 4), R'(2, –3) (d) U(–2, 4), V'(–5, –6), W'(–5, 7)
6. (a) A'(–2, 4), B'(–5, 8), C'(–9, 2) (b) M'(6, –7), N'(8, –4), P'(0, –5)
(b) P'(–8, –3), Q'(3, –4), R'(–5, 7)
(c) P'(–2, –8), Q'(–4, –3), R'(0, –5)
7. (a) A'(2, 4), B'(–4, –8), C'(4, –3)
8. (a) A'(–5, 3), B'(–4, 7), C'(–10, 4)

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Transformations

12.4 Rotation - Introduction

Observe the following illustrations and investigate the idea of rotation of a point.

Rotation of a point P through 90° in anti-clockwise direction about P'

the centre of rotation O. It is also called positive quarter turn about

the origin. 90°
90°
Here, P' is the image of P. O P
P
Rotation of a point P through 90° in clockwise direction about the O

centre of rotation O. It is also called negative quarter turn about the

origin. P'
Here, P' is the image of P.

Rotation of a point P through 180° in anti-clockwise

direction about the centre of rotation O. It is also called 180°
O
positive half turn about the origin. P' P
O P
Here, P' is the image of P. 180°

Rotation of a point P through 180° in clockwise direction P'
about the centre of rotation O. It is also called negative half
turn about origin.

Here, P' is the image of P.

To rotate a geometrical figure, following three conditions must be given.

(i) Centre of rotation (ii) Angle of rotation (iii) Direction of rotation

We can rotate a figure in two directions.

(i) Anti-clockwise direction (Positive direction)

(ii) Clock-wise direction (Negative direction)

In anti-clockwise direction, we rotate a figure in opposite direction of the rotation
of hands a clock. In clockwise direction, we rotate a figure in the same direction of
the rotation of hands of a clock.

Geometrical Rotation C

Steps of rotation of an object in a given direction B' B
about given centre.
Let us consider a triangle ABC which is to be rotated A
through +90° about the centre O. Then, we do the O
following steps:
A'
(i) Join OA and measure the length OA. Take OA C'
as a radius, rotate the point A in anti-clockwise

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