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प्रश्न
Find the distance between the following pairs of point.
W `((- 7)/2 , 4)`, X (11, 4)
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उत्तर
Let the co-ordinates of point W are (x1, y1) and of point X are (x2, y2)
`((-7)/2,4)` = (x1, y1)
(11, 4) = (x2, y2)
d (W, X) = `sqrt((x_2-x_1)^2+(y_2-y_1)^2)`
= `sqrt((11-(-7/2))^2+(4-4)^2`
= `sqrt((11+7/2)^2+0)`
= `(11 + 7/2)`
= `11/1+7/2`
= `(22+7)/2`
= `29/2`
= 14.5
∴ Distance between points W and X is 14.5.
संबंधित प्रश्न
Show that the points (a, a), (–a, –a) and (– √3 a, √3 a) are the vertices of an equilateral triangle. Also find its area.
Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.
If Q (0, 1) is equidistant from P (5, − 3) and R (x, 6), find the values of x. Also find the distance QR and PR.
An equilateral triangle has two vertices at the points (3, 4) and (−2, 3), find the coordinates of the third vertex.
If A (-1, 3), B (1, -1) and C (5, 1) are the vertices of a triangle ABC, find the length of the median through A.
Find the distance between the following pair of point.
T(–3, 6), R(9, –10)
Find the distance between the following pairs of point in the coordinate plane :
(4 , 1) and (-4 , 5)
Find the distance between the following point :
(p+q,p-q) and (p-q, p-q)
Find the distance of a point (13 , -9) from another point on the line y = 0 whose abscissa is 1.
Find the value of a if the distance between the points (5 , a) and (1 , 5) is 5 units .
Find the point on the x-axis equidistant from the points (5,4) and (-2,3).
A point P (2, -1) is equidistant from the points (a, 7) and (-3, a). Find a.
The length of line PQ is 10 units and the co-ordinates of P are (2, -3); calculate the co-ordinates of point Q, if its abscissa is 10.
Calculate the distance between A (5, -3) and B on the y-axis whose ordinate is 9.
Use distance formula to show that the points A(-1, 2), B(2, 5) and C(-5, -2) are collinear.
The distance between points P(–1, 1) and Q(5, –7) is ______
Find distance between point Q(3, – 7) and point R(3, 3)
Solution: Suppose Q(x1, y1) and point R(x2, y2)
x1 = 3, y1 = – 7 and x2 = 3, y2 = 3
Using distance formula,
d(Q, R) = `sqrt(square)`
∴ d(Q, R) = `sqrt(square - 100)`
∴ d(Q, R) = `sqrt(square)`
∴ d(Q, R) = `square`
If the distance between the points (4, P) and (1, 0) is 5, then the value of p 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:
If a player P needs to be at equal distances from A and G, such that A, P and G are in straight line, then position of P will be given by ______.
The distance of the point (5, 0) from the origin is ______.
