Functions and their graphs

1. Scanned by CamScanner

2. BY ASSOCIATE PROFESSOR NADEEM UDDIN
Distance
A path (Shortest distance) between two points is called distance. The distance between two
points A and B having coordinates(x1,y1) and (x2,y2) is ( ) ( )2 2
2 1 2 1
D= x x y y− + −
Example-3
Find the distance between the points (3,9) and (6,5).
Solution:
𝑥1 = 3 𝑎𝑛𝑑 𝑦1 = 9 ; 𝑥2 = 6 𝑎𝑛𝑑 𝑦2 = 5
( ) ( )
( ) ( )
( ) ( )
where x 3 and x 6
1 2
y 9 and y 5
1 2
2 2
D 6 3 5 9
2 2D 3 4
D 9 16
D 25
D 5
2 2
D= x x y y
2 1 2 1
= =
= =
= − + −
= + −
= +
=
=
− + −

3. Example-4
In the following graph, find the length of the line segment AB.
Solution:
𝑥1 = 4 𝑎𝑛𝑑 𝑦1 = 6
𝑥2 = −2 𝑎𝑛𝑑 𝑦2 = −2
|𝐴𝐵| = √(𝑥2 − 𝑥1)2 + (𝑦2 − 𝑦1)2
|𝐴𝐵| = √(−2 − 4)2 + (−2 − 6)2
|𝐴𝐵| = √(−6)2 + (−8)2
|𝐴𝐵| = √36 + 64 = √100 = 10 units

4. Example-5
In the following graph, the movement of a person is shown. If the person moves from A to B and
then from B to C, find the total distance covered by the person.
Solution:
D = |𝐴𝐵| + |𝐵𝐶|
D = √(5 − 5)2 + (8 − 3)2 + √(8 − 5)2 + (12 − 8)2
D = √(0)2 + (5)2 + √(3)2 + (4)2
D = √0 + 25 + √9 + 16
D = √25 + √25 = 5+ 5 =10 units

5. Example-6
In the given diagram if |𝑊𝑈| = |𝑊𝑉|, then show that (a – b)(a + b) = 0.
Solution:
|𝑊𝑈| = |𝑊𝑉|
√(a − a)2 + (0 − b)2 = √(0 − a)2 + (b − b)2
√(0)2 + (−b)2 = √(−a)2 + (0)2
√0 + 𝑏2 = √𝑎2 + 0
√𝑏2 = √𝑎2
Squairing both sides
b2
= a2
a2
- b2
= 0
(a – b) (a + b) = 0

6. Example-7
(a) Find, in term of D, the distance between points A(1+D, - 1) and
B(D, - 1- D)
(b) If the distance between points A and B is √10 units, then find the possible values of
D.Also write the coordinates of points A and B.
Solution:
(a)
|𝐴𝐵| = √(𝑥2 − 𝑥1)2 + (𝑦2 − 𝑦1)2
|𝐴𝐵| = √(D − 1 − D)2 + (−1 − D + 1)2
|𝐴𝐵| = √(−1)2 + (−D)2
|𝐴𝐵| = √1 + 𝐷2
(b)
|𝐴𝐵| = √1 + 𝐷2
√10 = √1 + 𝐷2
Squairing both sides
10 = 1 + 𝐷2
𝐷2
= 9
D = ± 3
Where coordinates are
If D = 3 then A(1+3, -1) and B(3, -1-3) = A(4, -1) and B(3, -4)
If D = - 3 then A(1-3, -1) and B(- 3, -1+3) = A(- 2, -1) and B(- 3, 2)