In a sport of cricket captain sets the field according to the plan. He instructs
the players to take a position at a particular place. There are two reasons to
set a cricket field- to take wickets and to stop scoring runs.
The following graph shows the position of players during a cricket match.
(a) The distance between the points showing the players at the Gully A(1, 0) and
wicketkeeper B(4, p) is 5m then p =
(i) 4 m (ii) 8 m (iii) 6 m (iv) 9 m
(b) Suppose the length of a line segment joining the players of mid-off and mid-on by
10 units. If the coordinates of one end are (2, -3) and the abscissa of the other end
is 10 units then its ordinate is:
(i) 9, 6 (ii) 3, -9 (iii) -3, 9 (iv) 9, -6
(c) The coordinates of the point on x-axis which is equidistant from the points
representing the players at cover P(-3, 4) and midwicket Q(2, 5) is
(i) (20, 0) (ii) (-23, 0) (iii) (4/5 , 0) (iv) None
(d) The ratio in which (4, 5) divides
Answers
Answer:
a--2
10-- 1
d-- 1:3
is the answer
p = 4; the ordinate is 3 or -9; the coordinates of the midpoint on the x-axis are (4/5, 0).
Given:
Position of the cricketers as shown in the figure below.
To Find:
a. value of p
b. ordinate if abscissa is 10 and the point is at a distance of 10 units from (2, -3)
c. coordinates of x
Solution:
a. Distance between 2 points is given as:
√[(x₁ - x₂)² + (y₁ - y₂)²]
Thus, AB = √[(1 - 4)² + (0 - p)²]
⇒ 5² = 9 + p²
p = √16
p = 4
b. Similarly, here too, using the distance formula we can find the distance between (2, -3) and (10, y) as follows-
10 = √[(2 - 10)² + (-3 - y)²]
100 = 64 + (-3 - y)²
36 = (-3 - y)²
6 = -3-y or -6 = -3-y
y = -9 or 3
c. The point on the x axis will have y coordinate as 0. Let the x coordinate be x. Then, we can form the following equation using the distance formula-
√[(-3 - x)² + (2 - 0)²] = √[(2 - x)² + (5 - 0)²]
(-3 - x)² + 4 = (2 - x)² + 25
9 + x² + 6x + 4 = 4 + x² - 4x + 25
9 + 6x = -4x + 25
10x = 16
x = 4/5
Thus, p = 4; the ordinate is 3 or -9; the coordinates of the midpoint on the x-axis are (4/5, 0).
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