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प्रश्न
Find distance between point A(7, 5) and B(2, 5)
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उत्तर
Let A(x1, y1) = A(7, 5) and B(x2, y2) = B(2, 5)
∴ By distance formula,
d(A, B) = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`
= `sqrt((2- 7)^2 + (5 - 5)^2`
= `sqrt((-5)^2 + 0^2)`
= `sqrt(25)`
= 5 cm
∴ The distance between points A and B is 5 cm.
संबंधित प्रश्न
Prove that the points (–3, 0), (1, –3) and (4, 1) are the vertices of an isosceles right angled triangle. Find the area of this triangle
Find the values of x, y if the distances of the point (x, y) from (-3, 0) as well as from (3, 0) are 4.
Find the centre of the circle passing through (6, -6), (3, -7) and (3, 3)
Find the co-ordinates of points of trisection of the line segment joining the point (6, –9) and the origin.
Find value of x for which the distance between the points P(x,4) and Q(9,10) is 10 units.
If P (x , y ) is equidistant from the points A (7,1) and B (3,5) find the relation between x and y
Find the distance between the following pair of points.
L(5, –8), M(–7, –3)
Show that the points A(1, 2), B(1, 6), C(1 + 2`sqrt3`, 4) are vertices of an equilateral triangle.
If A and B are the points (−6, 7) and (−1, −5) respectively, then the distance
2AB is equal to
Find the distance between the following point :
(sec θ , tan θ) and (- tan θ , sec θ)
Find the relation between a and b if the point P(a ,b) is equidistant from A (6,-1) and B (5 , 8).
Find the point on the x-axis equidistant from the points (5,4) and (-2,3).
The points A (3, 0), B (a, -2) and C (4, -1) are the vertices of triangle ABC right angled at vertex A. Find the value of a.
Show that each of the triangles whose vertices are given below are isosceles :
(i) (8, 2), (5,-3) and (0,0)
(ii) (0,6), (-5, 3) and (3,1).
Find distance between points O(0, 0) and B(– 5, 12)
Find distance between point A(–1, 1) and point B(5, –7):
Solution: Suppose A(x1, y1) and B(x2, y2)
x1 = –1, y1 = 1 and x2 = 5, y2 = – 7
Using distance formula,
d(A, B) = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`
∴ d(A, B) = `sqrt(square +[(-7) + square]^2`
∴ d(A, B) = `sqrt(square)`
∴ d(A, B) = `square`
The point which lies on the perpendicular bisector of the line segment joining the points A(–2, –5) and B(2, 5) is ______.
Case Study -2
A hockey field is the playing surface for the game of hockey. Historically, the game was played on natural turf (grass) but nowadays it is predominantly played on an artificial turf.
It is rectangular in shape - 100 yards by 60 yards. Goals consist of two upright posts placed equidistant from the centre of the backline, joined at the top by a horizontal crossbar. The inner edges of the posts must be 3.66 metres (4 yards) apart, and the lower edge of the crossbar must be 2.14 metres (7 feet) above the ground.
Each team plays with 11 players on the field during the game including the goalie. Positions you might play include -
- Forward: As shown by players A, B, C and D.
- Midfielders: As shown by players E, F and G.
- Fullbacks: As shown by players H, I and J.
- Goalie: As shown by player K.
Using the picture of a hockey field below, answer the questions that follow:
What are the coordinates of the position of a player Q such that his distance from K is twice his distance from E and K, Q and E are collinear?
Point P(0, 2) is the point of intersection of y-axis and perpendicular bisector of line segment joining the points A(–1, 1) and B(3, 3).
A circle has its centre at the origin and a point P(5, 0) lies on it. The point Q(6, 8) lies outside the circle.
