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Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.


Let the given points be A(0, 0) and B(36, 15) then,


AB space space equals space square root of open parentheses 36 minus 0 close parentheses squared plus left parenthesis 15 minus 0 right parenthesis squared end root
space space space space space space equals space space square root of open parentheses 36 close parentheses squared plus left parenthesis 15 right parenthesis squared end root
space space space space space space space equals space square root of 1296 plus 225 end root equals square root of 1521
space space space space space space space space equals space 39

Yes, we can find the distance between the two towns A and B discussed in section 7.2 and this distance = 39 km.
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Find the distance between the following pairs of points :
(– 5, 7), (– 1, 3)

Let the given points be A(-5, 7) and B(-1, 3).
Let the given points be A(-5, 7) and B(-1, 3).

AB space equals space square root of open parentheses straight x subscript 2 minus straight x subscript 1 close parentheses squared plus open parentheses straight y subscript 2 minus straight y subscript 1 close parentheses squared end root
rightwards double arrow space AB space equals square root of left curly bracket negative 1 minus left parenthesis 55 right parenthesis right curly bracket squared left parenthesis negative 3 minus left parenthesis negative 1 right parenthesis right curly bracket squared end root
space space space space space space space space space space space space equals space square root of 16 plus 16 end root equals square root of 32
space space space space space space space space space space space space space equals 4 square root of 2
1641 Views

Find the distance between the following pairs of points :
(a, b), (– a, – b)

Let the given points be A (a, b) and B(-a, -b)
We know that the distance between two points A (a, b) and B(-a, -b) is given by

AB space equals space square root of open parentheses straight x subscript 2 plus straight x subscript 1 close parentheses squared plus left parenthesis straight y subscript 2 minus straight y subscript 1 right parenthesis squared end root
therefore space space AB space equals space square root of open parentheses negative straight a minus straight a close parentheses squared plus left parenthesis negative straight b minus straight b right parenthesis squared end root
space space space space space space space space space space space equals space square root of left parenthesis negative 2 straight a right parenthesis squared plus left parenthesis negative 2 straight b right parenthesis squared end root
space space space space space space space space space space space equals space square root of 4 straight a squared plus 4 straight b squared end root
space space space space space space space space space space space space equals space square root of 4 left parenthesis straight a squared plus straight b squared right parenthesis end root
space space space space space space space space space space space space equals space square root of 4 left parenthesis straight a squared plus straight b squared right parenthesis end root
space space space space space space space space space space space space space space equals space 2 square root of straight a squared plus straight b squared end root

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Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear


Let the given points be A( 1, 5), B(2, 3) and C(-2, -11). Then


Let the given points be A( 1, 5), B(2, 3) and C(-2, -11). Then
Here,

Here, we see that AB + BC ≠ AC, BC + AC ≠ AB and AB + AC ≠ BC.
Hence, the points A, B and C are not collinear.

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Find the distance between the following pairs of points :
(2, 3), (4, 1)


Let the given points be A(2, 3) and B(4, 1).
We know that the distance between two points A(x1, y1) and B(x2, y2) is given by

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