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प्रश्न
The distance between the points (0, 5) and (–5, 0) is ______.
पर्याय
5
`5sqrt2`
`2sqrt(5)`
10
उत्तर
The distance between the points (0, 5) and (–5, 0) is `underlinebb(5sqrt2)`.
Explanation:
Distance formula: d2 = (x2 – x1)2 + (y2 – y1)2
According to the question,
We have,
x1 = 0, x2 = – 5
y1 = 5, y2 = 0
d2 = ((– 5) – 0)2 + (0 – 5)2
d = `sqrt((-5 - 0)^2 + (0 - 5)^2`
d = `sqrt((-5)^2 + (-5)^2`
d = `sqrt(25 + 25)`
d = `sqrt(50) = 5sqrt(2)`
So the distance between (0, 5) and (–5, 0) = `5sqrt(2)`
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संबंधित प्रश्न
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.
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(i) A(9,3) and B(15,11)
Prove that the following set of point is collinear :
(5 , 5),(3 , 4),(-7 , -1)
A point A is at a distance of `sqrt(10)` unit from the point (4, 3). Find the co-ordinates of point A, if its ordinate is twice its abscissa.
Calculate the distance between A (7, 3) and B on the x-axis whose abscissa is 11.
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
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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`
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