The document discusses distance and how to calculate it between two points with given coordinates. It provides the distance formula and works through several examples of finding distances between points on a graph. In one example, it calculates the total distance traveled between three points by adding the distances between each point. Another example shows that if the distance between two points W and V is equal, then the expression (a – b)(a + b) must equal 0.

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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)