Students can Download Maths Chapter 14 Introduction of Graphs Ex 14.1 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 Board Class 8 Maths Chapter 14 Introduction of Graphs Ex 14.1

Question 1.

Fix up your own coordinate system on a graph paper and locate the following points on the sheet,

i. P(-3,5)

ii. Q(0,-8)

iii. R(4,0)

iv. S (-4, -9)

Answer:

Question 2.

Suppose you are given a coordinate system to determine the quadrant in which the following points lie.

Answer:

i. A (4, 5) – I Quadrant

ii. B (-4, -5) – III Quadrant

iii. C (4, -5) – IV Quadrant

Question 3.

Suppose p is a point with coordinate (-8, 3) with respect to a coordinate system X^{1 }O X → Y^{1 }OY. Let X^{1}_{1} O_{1} X_{1} ↔ Y^{1}OY be another system with X^{1}OX || X^{1}_{1}O_{1}X_{1} and O_{1} has coordinates (9,5) with respect to X^{1}OX ↔ Y^{1}O Ywhat are the coordinate of p in the system X^{11} O_{1}X_{1} Y^{1}_{1}O_{1}Y_{1}?

Answer:

Let (x^{1 },y^{1}) be the coordinates point p in X^{1}_{1} O_{1} X_{1} ↔ Y^{1}OY.

Then x = a + x^{1 }and y = b + y^{1}

but (a, b) = (9, 5)

∴ – 8 = 9 +x^{1} y = b + y^{1}

– 8 – 9 = x^{1} 3 = 5 + y^{1}

-17 = x^{1} 3 – 5 = y^{1}

∴ (x^{1} , y^{1} ) = (-17,-2)

∴ Coordinatesof pin X^{1}_{1} O_{1} X_{1} ↔ Y^{1}OY are (-17,-2)

Question 4.

Suppose P has coordinates (10,2) in a coordinate system X’OX ↔ Y’OY and (-3,-6) in an other coordinates system X’_{1}O_{1 }X↔Y’_{1}O_{1}X_{1}. Determine the coordinates of O with respect to the system X’_{1} O_{1} X_{1 }|| Y’_{1} O_{1}Y_{1}’.

Answer:

Given p(x , y) = (10 , 2), (x’, y’) = (-3, -6)

Let the coordinates of O, in

X’_{1}O_{1} X_{1} ↔ Y’_{1}O_{1} Y_{1} be (a,b) then

x = a + x^{1} y = b + y^{1}

10 = a + (-3) 2 = b + (-6)

10 = a – 3 2 = b – 6

10 + 3 = a 2 + 6 = b

a = 13 b = 8

Coordinates of O_{1} in

X’_{1} O _{1}Y_{1}, ↔ Y_{1} O _{1}Y’_{1 } is (13 , 8) we should find the coordinates of O

(0, 0) in X’_{1}O _{1}X_{1}↔ Y’_{1}O_{1}Y_{1}

Let (x , y) = (0 , 0) , (x’, y’) = (13 , 8) .

(a ,b) = ? y = b + y’

0 = a + 13 0 = b + 8

-13 = a -8 = b

(a = – 13) b= -8

The coordinates of O w.r.t to

X’_{1} O _{1}Y_{1}, ↔ Y_{1} O _{1}Y’_{1} are (-13 , -8).