Students can Download Maths Chapter 14 Introduction of Graphs Additional Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka State Syllabus Class 8 Maths Chapter 14 Introduction of Graphs Additional Questions

Question 1.

Choose the correct option.

(a) The point (4, 0) lie on the line _______

A. y – x = 0

B. y = 0

C. x = 0

D. y + x = 0

Solution:

B. y = 0

(b) The point (-5, 4) lie in ___________

A. the first qudrant

B. the second quadrant

C. the third quadrant

D. the fourth quadrant

Solution:

B. the second quadrant.

(c) If a straight pass through (0, 0) and (1, 5) then its equation is ________

A. y = x

B. y = 5x

C. 5y = x

D. y = x + 5

Solution:

C. 5y = x

(d) If a point P has coordinates (3, 4) in coordinate system X’OX ↔ Y’OY and if O has coordinates (4, 3) in another system X’_{1}OX_{1} ↔ Y_{1}‘0_{1}Y_{1} is __________

A. (3, 4)

B. (1, -1)

C. (7, 7)

D. (-1, 1)

Solution:

C. (7, 7)

(e) The coordinates of a point P in a system X’OX ↔ Y’OY are (5, 8). The coordinates of the same point in the system Y’OY ↔ XOX’ are

A. (-8, 5)

B. (8, 5)

C. (8, -5)

D. (-8, -5)

Solution:

C. (8, -5)

(f) The signs of the coordinates of a point in the third quadrant are

A. (+, -)

B. (-, +)

C. (+, +)

D. (-, -)

Solution:

D. (-, -)

(g) If a person moves either 1 unit in the direction of the positive x-axis or 1 unit in the direction of positive y-axis per step; then the number of steps he requires to reach (10, 12) starting from the origin (0, 0) in

A. 10

B. 12

C. 22

D. 120

Solution:

C. 22

(h) They coordinate of the point of 2. intersection of the line y = 3x + 4 with

x = 3 is

A. 4

B. 7

C. 10

D. 13

Solution:

B. 7

(i) The equation of the line which passes through (0, 0) and (1, 1) is

A. y = x

B. y = – x

C. y = 1

D. x = 1

Solution:

A . v = x

Question 2.

Find the quadrant in which the following points lie.

Solution:

(i) (5, 10) – I Quadrant

(ii) (-8, 9) – II Quadrant

(iii) (-800, -3000) – III Quadrant

(iv) (8, -100) – IV Quadrant

Question 3.

Match the following

(A) on the x axis – (i) x coordinate is negative

(B) In th second quadrant – (ii) cuts the y axis at (0, 4)

(C) The line y = 3x + 4 – (iii) coordinates of a point are of form (a, 0)

Solution:

(A) – (iii)

(B) – (i)

(C) – (ii)

Question 4.

Fill in the blanks

(a) They coordinate of a point on the x-axis in zero.

(b) The x coordinate is called abscissa.

(c) The x-axis and y-axis intersect at (0, 0)

(d) If a point (x, y) ≠ (0, 0) is in the third quadrant, then (x + y) has a negative sign.

(e) If a point (x, y) lies above the horizontal axis, then y is always positive.

(f) The point of intersection of x = y and x = – y is (0, 0)

(g) The line y = 4x + 5 intersects y-axis at the point (0, 5)

Question 5.

True or false

(a) The equation of the x-axis is x = 0.

Answer:

False

(b) The line x = 4 is parallel to the y-axis.

Answer:

True

(c) The line y 8 is perpendicular to the x-axis.

Answer:

False

(d) The line x = y and x = – y are perpendicular to each other.

Answer:

True

(e) The lines x = 9 and y = 9 are perpendicular to each other.

Answer:

True

(f) The graph of y = x2 is a straight line.

Answer:

False

(g) The line y = 3x + 4 does not in¬tersect x-axis.

Answer:

False

(h) In a rectangular coordinate system, the coordinate axes are chosen such that they form a pair of perpendicular lines.

Answer:

True

Question 6.

Determine the equation of the line which passes through the points (0, -8) and (7, 0).

Solution:

Let the equation be y = ax + b. The line passes through (0, -8). When

x = 0 y = – 8

y = ax + b

– 8 = a (0) + b

– 8 = 0 + b

∴ b = – 8

The lines passes through (7, 0). When x = 7, y = 0

y = ax + b

0 = a (7) + b, 0 = 7a + b

0 = 7a – 8 (∴ b = – 8)

8 = 7a

a = \(\frac{8}{7}\)

y = ax + b

y = \(\frac{8}{7}\) x + (-8)

7y = 8x – 56

56 = 8x – 7y

∴ The equation is 8x – 7y = 56

Question 7.

Determine the equation of the line in each of the following graph.

(i)

Solution:

From the graph we find that the line passes through the points (3, 8) and (-3, -3).

Let the equation be y = ax + b

When x = 3, y = 8

∴ 8 = a (3) + b

8 = 3a + b

∴ b = 8 – 3a

When x = – 3, y = – 3

∴ y = ax + b

– 3 = a(-3) + b

– 3 = – 3a + b

– 3 = – 3a + 8 – 3a [∴ b = 8 – 3a]

6a = 8 + 3

a = \(\frac{11}{6}\)

b = 8 – 3a

6y = 11x + 15

(ii)

From the graph we find that the line passes through the points (-4, 2) and (12, -3)

Let the equation is y = ax + b

When x = – 4, y = 2

∴2 = a(-4) + b 2 = – 4a + b

b =2+ 4a

When x = 12, y = – 3, y = ax + b

– 3 = a(12) + b – 3 = 12a + b

– 3 = 12a + 2 + 4a [∴ b = 2 + 4a]

– 3 = 16a + 2 – 3 – 2 =16a

y = \(\frac{-5 x+12}{16}\)

16y = – 5x + 12

16y + 5x + = 12

Question 8.

A point P has coordinates (7, 10) in a coordinate system X’OX ↔ Y’OY. Suppose it has coordinates (10, 7) in another coordinate system X’_{1}O_{1}X_{1} ↔ Y’_{1}O_{1}y_{1} with X’OX || X’_{1}O_{1}X_{1}.

Find the coordinates of O_{1} in the system X’OX ↔ Y’OY.

Solution:

Given P (x, y) = (7, 10) and (x’, y’) = (10, 7)

Let the coordinates of O1 in X’OX – Y”OY be (a, b)

x = a + x’ y = b + y’.

7 = a + 10

10 – b + 7

7 – 10 = a

10 – 7 = b

– 3 = a 3 = b

a = – 3 b = 3

∴The coordinates of O1 are (-3, 3)

Question 9.

Sketch the region {(x, y) : x > 0, y > 0, x + 2y < 4} in a coordinate system set up by you. Solution: Given x > 0, y > 0 x + 2y < 4

Let x = 0 and y = 1, .

then x + 2y = 0 + 2(1) = 0 + 2 = 2 < 4

x = 1 and y = 1,

then x + 2y + 1 + 2(1) = 1 + 2 = 3 < 4

x = 2 and y = 0, then x + 2y = 2 + 2(0) = 2 + 0 = 2 <4

x = 2 and y = 1

then x + 2y = 2 + 2(1) = 2 + 2 = 4

x = 0 and y = 0 then x + 2y = 0 + 2(0) = 0 + 0 = 0

Question 10.

Draw the graphs of lines 3x = 4x – 4 and 2x = 3y + 4 and determine the point at which these lines meet.

Solution:

3y = 4x – 4

2x = 3y + 4

y = \(\frac{4 x-4}{3}\)

2x – 4 = 3y

y = \(\frac{2 x-4}{3}\)

Question 11.

If a * b = ab + a + b, draw the graph of y = 3 * (x +1) * 2

Solution:

3 * (x +1) * 2

Find 3 * (x + 1) Given a * b = ab + a + b

∴ 3 * (x + 1) = 3( x + 1) + 3 + (x + 1)

= 3x +1 + 3 + x +1

3 * (x +1) = 4x + 5

Now find 3 * (x +1) * 2

= (4x + 5) * 2 = (4x + 5)2 + 4x + 5 + 2

= 8x + 10 + 4x + 7

3 * (x + 1) * 2 = 12 x + 27

∴ y = 12x + 27

When x = – 2, = 12 (-2) + 27

= – 12 + 27 – 15

When x = – 3 y = 12 (-3) + 27

= – 36 + 27 = – 9