KSEEB Solutions for Class 8 Maths Chapter 14 Introduction of Graphs Ex 14.1

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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 X¹ O X → Y¹ OY. Let X¹₁ O₁ X₁ ↔ Y¹OY be another system with X¹OX  ||  X¹₁O₁X₁ and O₁ has coordinates (9,5) with respect to X¹OX ↔ Y¹O Ywhat are the coordinate of p in the system X^1₁ O₁X₁ Y¹₁O₁Y₁?
Answer:
Let (x¹ ,y¹) be the coordinates point p in X¹₁ O₁ X₁ ↔ Y¹OY.
Then x = a + x¹ and y = b + y¹
but (a, b) = (9, 5)
∴ – 8 = 9 +x¹      y = b + y¹
– 8 – 9 = x¹         3 = 5 + y¹
-17 = x¹             3 – 5 = y¹
∴ (x¹ , y¹ ) = (-17,-2)
∴ Coordinatesof pin X¹₁ O₁ X₁ ↔ Y¹OY 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’₁O₁ X↔Y’₁O₁X₁. Determine the coordinates of O with respect to the system X’₁ O₁ X₁ || Y’₁ O₁Y₁’.
Answer:
Given p(x , y) = (10 , 2), (x’, y’) = (-3, -6)
Let the coordinates of O, in
X’₁O₁ X₁ ↔ Y’₁O₁ Y₁ be (a,b) then
x = a + x¹           y = b + y¹
10 = a + (-3)     2 = b + (-6)
10 = a – 3          2 = b – 6
10 + 3 = a       2 + 6 = b
a = 13                    b = 8
Coordinates of O₁ in
X’₁ O ₁Y₁, ↔ Y₁ O ₁Y’₁  is (13 , 8) we should find the coordinates of O
(0, 0) in X’₁O ₁X₁↔ Y’₁O₁Y₁
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 ₁Y₁, ↔ Y₁ O ₁Y’₁ are (-13 , -8).